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Index finger there, middle finger there, your left thumb, or your right thumb, I should say, will go straight up. So it goes in the right direction. So this would be an upward pointing vector. So we got the right orientation for our surface. Now let's think about what this is. And it's important to realize we could calculate all of the components of this, but then we're just going to take the dot product of that with k. So it's really we care about the k component only. But I'll work it out.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So we got the right orientation for our surface. Now let's think about what this is. And it's important to realize we could calculate all of the components of this, but then we're just going to take the dot product of that with k. So it's really we care about the k component only. But I'll work it out. So this is equal to k times a matrix, i, j, k, of the partial of t with respect to x. Well, the partial of t with respect to x, I'll do this in blue, is going to be 1, 0, 1, 0, and the partial of f2 with respect to x. And then the partial of t with respect to y is going to be 0, 1, 0, 1, and the partial of f2 with respect to y.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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But I'll work it out. So this is equal to k times a matrix, i, j, k, of the partial of t with respect to x. Well, the partial of t with respect to x, I'll do this in blue, is going to be 1, 0, 1, 0, and the partial of f2 with respect to x. And then the partial of t with respect to y is going to be 0, 1, 0, 1, and the partial of f2 with respect to y. And then, of course, we have to multiply times dA. And this is all going to be equal to k, unit vector k, dotted with, and I don't even have to even work it all out, it's going to be something times the i unit vector minus something, checkerboard pattern, minus something times something else, necessarily, times our j unit vector, plus, and now we can think about the k unit vector. So the k unit vector is going to be 1 times 1 minus 0.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then the partial of t with respect to y is going to be 0, 1, 0, 1, and the partial of f2 with respect to y. And then, of course, we have to multiply times dA. And this is all going to be equal to k, unit vector k, dotted with, and I don't even have to even work it all out, it's going to be something times the i unit vector minus something, checkerboard pattern, minus something times something else, necessarily, times our j unit vector, plus, and now we can think about the k unit vector. So the k unit vector is going to be 1 times 1 minus 0. So it's just going to be plus the k unit vector. We know it's 1 times the k unit vector. And so when you take this dot, and of course, we have our dA out right here.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So the k unit vector is going to be 1 times 1 minus 0. So it's just going to be plus the k unit vector. We know it's 1 times the k unit vector. And so when you take this dot, and of course, we have our dA out right here. But when you take this dot product, you only are left with the k components. And it's essentially just 1 times 1. You end up with a scalar quantity of 1.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And so when you take this dot, and of course, we have our dA out right here. But when you take this dot product, you only are left with the k components. And it's essentially just 1 times 1. You end up with a scalar quantity of 1. All of this business just simplified to dA. So now we can rewrite our surface integral. And we're going to rewrite it in the xy domain now, in our parameters domain.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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You end up with a scalar quantity of 1. All of this business just simplified to dA. So now we can rewrite our surface integral. And we're going to rewrite it in the xy domain now, in our parameters domain. So our surface integral right up here. So this will be good for this video. And then we'll do the same thing with this surface here, just making sure that we get the orientation right.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And we're going to rewrite it in the xy domain now, in our parameters domain. So our surface integral right up here. So this will be good for this video. And then we'll do the same thing with this surface here, just making sure that we get the orientation right. So this surface integral, S2, and I'll even rewrite a little bit. S2, which is a function, r is a function of x, y, and z, times k dot n ds. I just rewrote all of this right up here.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then we'll do the same thing with this surface here, just making sure that we get the orientation right. So this surface integral, S2, and I'll even rewrite a little bit. S2, which is a function, r is a function of x, y, and z, times k dot n ds. I just rewrote all of this right up here. Is equivalent to the double integral over our parameters domain, which is just d, of r of x, y, z times all of this business. All of this business just simplified to dA. And since I want to write it in terms of my parameters, I'll write it as r of x, y.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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I just rewrote all of this right up here. Is equivalent to the double integral over our parameters domain, which is just d, of r of x, y, z times all of this business. All of this business just simplified to dA. And since I want to write it in terms of my parameters, I'll write it as r of x, y. And while we're on that surface, z is equal to F2. So it's x, y, and F2 of x, y. And then all of this business we just saw simplified to dA.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And since I want to write it in terms of my parameters, I'll write it as r of x, y. And while we're on that surface, z is equal to F2. So it's x, y, and F2 of x, y. And then all of this business we just saw simplified to dA. So you might be saying, hey, Sal, it didn't look like you simplified it a lot. But at least now put it in terms of a double integral, instead of a surface integral. So at least in my mind, that is a simplification.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then all of this business we just saw simplified to dA. So you might be saying, hey, Sal, it didn't look like you simplified it a lot. But at least now put it in terms of a double integral, instead of a surface integral. So at least in my mind, that is a simplification. In the next video, we're going to do the exact same thing with this, just making sure that our vectors are oriented properly. And we could just introduce a negative sign to make sure that they are. And then we're going to think about the triple integrals and try to simplify those.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So imagine that we're sitting in the coordinate plane, and that I draw for you a whole bunch of little droplets, droplets of water, and then these are going to start flowing in some way. How would you describe this flow mathematically? So, at every given point, the particles are moving in some different way over here, they're kind of moving down and to the left. Here, they're moving kind of quickly up, over here, they're moving more slowly down. So what you might want to do is assign a vector to every single point in space, and a common attribute of the way that fluids flow, this isn't necessarily obvious, but if you look at a given point in space, let's say like right here, every time that a particle passes through it, it's with roughly the same velocity. So you might think over time, that velocity would change, and sometimes it does, a lot of times there's some fluid flow where it depends on time, but for many cases, you can just say, at this point in space, whatever particle is going through it, it'll have this velocity vector. So over here, they might be pretty high upwards, whereas here, it's kind of a smaller vector downwards, even though, and here I'll play the animation a little bit more here, and if you imagine doing this at all of the different points in space, and assigning a vector to describe the motion of each fluid particle at each different point, what you end up getting is a vector field.
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Fluid flow and vector fields Multivariable calculus Khan Academy.mp3
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Here, they're moving kind of quickly up, over here, they're moving more slowly down. So what you might want to do is assign a vector to every single point in space, and a common attribute of the way that fluids flow, this isn't necessarily obvious, but if you look at a given point in space, let's say like right here, every time that a particle passes through it, it's with roughly the same velocity. So you might think over time, that velocity would change, and sometimes it does, a lot of times there's some fluid flow where it depends on time, but for many cases, you can just say, at this point in space, whatever particle is going through it, it'll have this velocity vector. So over here, they might be pretty high upwards, whereas here, it's kind of a smaller vector downwards, even though, and here I'll play the animation a little bit more here, and if you imagine doing this at all of the different points in space, and assigning a vector to describe the motion of each fluid particle at each different point, what you end up getting is a vector field. So this here is a little bit of a cleaner drawing than what I have, and as I mentioned in the last video, it's common for these vectors not to be drawn to scale, but to all have the same length, just to get a sense of direction, and here you can see each particle is flowing roughly along that vector, so whatever one it's closest to, it's moving in that direction, and this is not just a really good way of understanding fluid flow, but it goes the other way around it's a really good way of understanding vector fields themselves. So sometimes, you might just be given some new vector field, and to get a feel for what it's all about, how to interpret it, what special properties it might have, it's actually helpful, even if it's not meant to represent a fluid, to imagine that it does, and think of all the particles, and think of how they would move along it. For example, this particular one, as you play the animation, as you let the particles move along the vectors, there's no change in the density, at no point do a bunch of particles go inward, or a bunch of particles go outward, it stays kind of constant, and that turns out to have a certain mathematical significance down the road, you'll see this later on, as we study a certain concept called divergence, and over here, you see this vector field, and you might want to understand what it's all about, and it's kind of helpful to think of a fluid that pushes outward from everywhere, and is kind of decreasing in density around the center, and that also has a certain mathematical significance.
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Fluid flow and vector fields Multivariable calculus Khan Academy.mp3
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So over here, they might be pretty high upwards, whereas here, it's kind of a smaller vector downwards, even though, and here I'll play the animation a little bit more here, and if you imagine doing this at all of the different points in space, and assigning a vector to describe the motion of each fluid particle at each different point, what you end up getting is a vector field. So this here is a little bit of a cleaner drawing than what I have, and as I mentioned in the last video, it's common for these vectors not to be drawn to scale, but to all have the same length, just to get a sense of direction, and here you can see each particle is flowing roughly along that vector, so whatever one it's closest to, it's moving in that direction, and this is not just a really good way of understanding fluid flow, but it goes the other way around it's a really good way of understanding vector fields themselves. So sometimes, you might just be given some new vector field, and to get a feel for what it's all about, how to interpret it, what special properties it might have, it's actually helpful, even if it's not meant to represent a fluid, to imagine that it does, and think of all the particles, and think of how they would move along it. For example, this particular one, as you play the animation, as you let the particles move along the vectors, there's no change in the density, at no point do a bunch of particles go inward, or a bunch of particles go outward, it stays kind of constant, and that turns out to have a certain mathematical significance down the road, you'll see this later on, as we study a certain concept called divergence, and over here, you see this vector field, and you might want to understand what it's all about, and it's kind of helpful to think of a fluid that pushes outward from everywhere, and is kind of decreasing in density around the center, and that also has a certain mathematical significance. And it might also lead you to ask certain other questions, like if you look at the fluid flow that we started with in this video, you might ask a couple questions about it, like it seems to be rotating around some points, in this case counterclockwise, but it's rotating clockwise around others still, does that have any kind of mathematical significance? Does the fact that there seem to be the same number of particles roughly in this area, but they're slowly spilling out there, what does that imply for the function that represents this whole vector field? And you'll see a lot of this later on, especially when I talk about divergence and curl, but here I just wanted to give a little warm up to that, as we're just visualizing multivariable functions.
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Fluid flow and vector fields Multivariable calculus Khan Academy.mp3
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So let's work on the curl of f. So the curl of f is going to be equal to, and I just remember it as the determinant. So we have our i, j, k components. It's really, you can imagine, it's the del operator crossed with the actual vector. So the del operator, and I'll write this in a different color just to ease the monotony. So it's the partial with respect to x, partial with respect to y, partial with respect to z, and then our vector field, I copy and pasted it right over here. It is just equal to negative y squared is our i component, x is our j component, and z squared is our k component. And so this is going to be equal to, this is going to be equal to i, is going to be equal to i times the partial of z squared with respect to y.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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So the del operator, and I'll write this in a different color just to ease the monotony. So it's the partial with respect to x, partial with respect to y, partial with respect to z, and then our vector field, I copy and pasted it right over here. It is just equal to negative y squared is our i component, x is our j component, and z squared is our k component. And so this is going to be equal to, this is going to be equal to i, is going to be equal to i times the partial of z squared with respect to y. Well, the z squared is just a constant with respect to y, so the partial of z squared with respect to y is just going to be zero. So it's going to be zero minus the partial of x with respect to z. Well, once again, this is just a constant when you think in terms of z, so that's just going to be zero.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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And so this is going to be equal to, this is going to be equal to i, is going to be equal to i times the partial of z squared with respect to y. Well, the z squared is just a constant with respect to y, so the partial of z squared with respect to y is just going to be zero. So it's going to be zero minus the partial of x with respect to z. Well, once again, this is just a constant when you think in terms of z, so that's just going to be zero. So that's nice simplification. And then we're gonna have minus j, we need our little checkerboard pattern, so we put a negative in front of the j, minus j, and so we'll have the partial of z squared with respect to x, well, that's zero again, and then minus the partial of negative y squared with respect to z. Well, that's zero again.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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Well, once again, this is just a constant when you think in terms of z, so that's just going to be zero. So that's nice simplification. And then we're gonna have minus j, we need our little checkerboard pattern, so we put a negative in front of the j, minus j, and so we'll have the partial of z squared with respect to x, well, that's zero again, and then minus the partial of negative y squared with respect to z. Well, that's zero again. And then finally, we have our k component, k, so plus k, and k, we're gonna have the partial of x with respect to x, well, that actually gives us a value, that's just going to be one, minus the partial of negative y squared with respect to y. So the partial of negative y squared with respect to y is negative two y, and we're subtracting that, so it's going to be plus two y. So curl of f simplifies to just, oh, this is just zero up here, it's just one plus two y times k, or k times one plus two y.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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Well, that's zero again. And then finally, we have our k component, k, so plus k, and k, we're gonna have the partial of x with respect to x, well, that actually gives us a value, that's just going to be one, minus the partial of negative y squared with respect to y. So the partial of negative y squared with respect to y is negative two y, and we're subtracting that, so it's going to be plus two y. So curl of f simplifies to just, oh, this is just zero up here, it's just one plus two y times k, or k times one plus two y. And so if we go back to this right up here, if we go back up to that, we are going to get, we rewrite the integral, so from zero to one, then that's our r, our r parameter's gonna go from zero to one, theta's gonna go from zero to two pi, and now curl of f has simplified to, and I won't skip any steps here, although it's tempting, it's one plus two y, and actually, instead of writing two y, let me write it in terms of the parameters. We saw it up here, y was r sine theta, if I remember correctly, right? Y was r sine theta, so let me write y that way.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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So curl of f simplifies to just, oh, this is just zero up here, it's just one plus two y times k, or k times one plus two y. And so if we go back to this right up here, if we go back up to that, we are going to get, we rewrite the integral, so from zero to one, then that's our r, our r parameter's gonna go from zero to one, theta's gonna go from zero to two pi, and now curl of f has simplified to, and I won't skip any steps here, although it's tempting, it's one plus two y, and actually, instead of writing two y, let me write it in terms of the parameters. We saw it up here, y was r sine theta, if I remember correctly, right? Y was r sine theta, so let me write y that way. Two times r sine theta k, and we're gonna dot this, we're gonna take the dot product of that with this right over here, with r times j plus r times k, d theta dr. And so when we take the dot product, this thing only has a k component. The j component is zero, so when you take the dot product with this j component, you're gonna get zero, and neither of them actually even have an i component, and so the inside is just going to simplify to, this piece right over here is going to simplify to, we just have to think about the k components, because everything else is zero, so it's gonna be r times this, and then we're done. So it's gonna be r plus two r squared sine theta, d theta dr. D theta dr. And once again, theta goes from zero to two pi, and r goes from zero to one, and now this is just a straight up double integral.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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Y was r sine theta, so let me write y that way. Two times r sine theta k, and we're gonna dot this, we're gonna take the dot product of that with this right over here, with r times j plus r times k, d theta dr. And so when we take the dot product, this thing only has a k component. The j component is zero, so when you take the dot product with this j component, you're gonna get zero, and neither of them actually even have an i component, and so the inside is just going to simplify to, this piece right over here is going to simplify to, we just have to think about the k components, because everything else is zero, so it's gonna be r times this, and then we're done. So it's gonna be r plus two r squared sine theta, d theta dr. D theta dr. And once again, theta goes from zero to two pi, and r goes from zero to one, and now this is just a straight up double integral. We just have to evaluate this thing. And so first we take the antiderivative with respect to theta. So the antiderivative with respect to theta is going to give us, so this is going to be giving, so we're gonna focus on theta first.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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So it's gonna be r plus two r squared sine theta, d theta dr. D theta dr. And once again, theta goes from zero to two pi, and r goes from zero to one, and now this is just a straight up double integral. We just have to evaluate this thing. And so first we take the antiderivative with respect to theta. So the antiderivative with respect to theta is going to give us, so this is going to be giving, so we're gonna focus on theta first. So the antiderivative of r with respect to theta is just r theta. You can just view r as a constant. And then the antiderivative of this, antiderivative of sine of theta is negative cosine of theta.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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So the antiderivative with respect to theta is going to give us, so this is going to be giving, so we're gonna focus on theta first. So the antiderivative of r with respect to theta is just r theta. You can just view r as a constant. And then the antiderivative of this, antiderivative of sine of theta is negative cosine of theta. So this is gonna be negative two r squared cosine of theta, and we're gonna evaluate it from zero to two pi, and then we have the outside integral, which I will recolor in yellow. Recolor in yellow. So we'll still have to integrate with respect to r, and r is gonna go from zero to one.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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And then the antiderivative of this, antiderivative of sine of theta is negative cosine of theta. So this is gonna be negative two r squared cosine of theta, and we're gonna evaluate it from zero to two pi, and then we have the outside integral, which I will recolor in yellow. Recolor in yellow. So we'll still have to integrate with respect to r, and r is gonna go from zero to one. But inside right over here, if we evaluate all of this business right over here at two pi, we get two pi r, two pi r, that's that right over there, minus, cosine of two pi is just one, so it's minus two r squared, and then from that we are going to subtract from that, we're going to subtract, this evaluated at zero. Well r times zero is just zero, and then cosine of zero is one, so it's just minus two r squared, or negative two r squared. Negative two r squared.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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So we'll still have to integrate with respect to r, and r is gonna go from zero to one. But inside right over here, if we evaluate all of this business right over here at two pi, we get two pi r, two pi r, that's that right over there, minus, cosine of two pi is just one, so it's minus two r squared, and then from that we are going to subtract from that, we're going to subtract, this evaluated at zero. Well r times zero is just zero, and then cosine of zero is one, so it's just minus two r squared, or negative two r squared. Negative two r squared. And this negative and this negative, you get a positive, and then you have a negative two r squared, and then a plus two r squared is just going to cancel out. That and that cancel out, and so this whole thing has simplified quite nicely to a simple definite integral, zero to one of two pi, two pi r dr. And the antiderivative of this is just going to be pi r squared. So we're just going to evaluate pi r squared from zero to one.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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Negative two r squared. And this negative and this negative, you get a positive, and then you have a negative two r squared, and then a plus two r squared is just going to cancel out. That and that cancel out, and so this whole thing has simplified quite nicely to a simple definite integral, zero to one of two pi, two pi r dr. And the antiderivative of this is just going to be pi r squared. So we're just going to evaluate pi r squared from zero to one. When you evaluate it at one, you get pi. When you evaluate it at zero, you just get zero. So you get pi minus zero, which is equal to, and now we deserve a drum roll, because we've been doing a lot of work over many videos, this is equal to pi.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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So we're just going to evaluate pi r squared from zero to one. When you evaluate it at one, you get pi. When you evaluate it at zero, you just get zero. So you get pi minus zero, which is equal to, and now we deserve a drum roll, because we've been doing a lot of work over many videos, this is equal to pi. So just to remind ourselves what we've done over the last few videos, we had this line integral that we were trying to figure out, and instead of directly evaluating the line integral, which we could do, and I encourage you to do so, and if I have time, I might do it in the next video. Instead of directly evaluating that line integral, we use Stokes' Theorem to say, oh, we could actually instead say that that's the same thing as a surface integral over a piecewise smooth boundary, or over a piecewise smooth surface that this path is the boundary of. And so we evaluated this surface integral, and eventually with a good bit of, little bit of calculation, we got to evaluating it to be equal to pi.
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Stokes example part 4 Curl and final answer Multivariable Calculus Khan Academy.mp3
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We've seen this before. dS is going to be equal to the magnitude of the cross product of the partial of R with respect to U crossed with the partial of R with respect to V, dU, dV. So first, let's take the cross product, and we'll do that with a 3 by 3 matrix. I'll do that right over here. We set up, and I'll just kind of fill in what R sub U and R sub V is in the actual determinant right over here. So first, we have our components, I, J, and K. And now first, let's think about what R sub U is, the partial of R with respect to U. Well, its I component is going to be 1, the partial of U with respect to U is just 1.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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I'll do that right over here. We set up, and I'll just kind of fill in what R sub U and R sub V is in the actual determinant right over here. So first, we have our components, I, J, and K. And now first, let's think about what R sub U is, the partial of R with respect to U. Well, its I component is going to be 1, the partial of U with respect to U is just 1. So its I component is 1. Its J component is going to be 0. Partial of V with respect to U is 0.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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Well, its I component is going to be 1, the partial of U with respect to U is just 1. So its I component is 1. Its J component is going to be 0. Partial of V with respect to U is 0. V does not change with respect to U. So this is going to be 0. This should be parentheses around here.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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Partial of V with respect to U is 0. V does not change with respect to U. So this is going to be 0. This should be parentheses around here. Partial of this with respect to U is just going to be 1 again. This is the partial of V squared with respect to U is just 0. So this is just 1 again.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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This should be parentheses around here. Partial of this with respect to U is just going to be 1 again. This is the partial of V squared with respect to U is just 0. So this is just 1 again. And then r sub v, the partial of r with respect to v, the i component is going to be 0. j component is going to be 1. And the partial of u plus v squared with respect to v is going to be 2v. So it's a pretty straightforward determinant.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So this is just 1 again. And then r sub v, the partial of r with respect to v, the i component is going to be 0. j component is going to be 1. And the partial of u plus v squared with respect to v is going to be 2v. So it's a pretty straightforward determinant. So let's try to evaluate. So the i component, it's going to be i. So we cross out this column, this row.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So it's a pretty straightforward determinant. So let's try to evaluate. So the i component, it's going to be i. So we cross out this column, this row. It's going to be 0 times 2v minus 1 times 1. So essentially, it's just going to be negative 1 times i. So we're going to have negative i.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So we cross out this column, this row. It's going to be 0 times 2v minus 1 times 1. So essentially, it's just going to be negative 1 times i. So we're going to have negative i. So this is equal to negative i. And then the j component, and we're going to have to put a negative out front, because remember, we do that checkerboard pattern. So cross out that row, that column.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So we're going to have negative i. So this is equal to negative i. And then the j component, and we're going to have to put a negative out front, because remember, we do that checkerboard pattern. So cross out that row, that column. 1 times 2v is 2v. Let me make sure I got that. 1 times 2v is 2v minus 0 times 1.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So cross out that row, that column. 1 times 2v is 2v. Let me make sure I got that. 1 times 2v is 2v minus 0 times 1. So this is going to be 2v. But since it was the j component, which is going to be negative, it's going to be negative 2vj. Let me make sure I did that right.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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1 times 2v is 2v minus 0 times 1. So this is going to be 2v. But since it was the j component, which is going to be negative, it's going to be negative 2vj. Let me make sure I did that right. That column, that row, 1 times 2v is 2v minus 0 is 2v. The checkerboard pattern, you'd have a negative j. So you have negative 2vj.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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Let me make sure I did that right. That column, that row, 1 times 2v is 2v minus 0 is 2v. The checkerboard pattern, you'd have a negative j. So you have negative 2vj. And then we have, find the k component. Cross out that row, that column, 1 times 1 minus 0. So it's going to be plus k. So if we want the magnitude of this, this whole thing right over here is just going to be the square root.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So you have negative 2vj. And then we have, find the k component. Cross out that row, that column, 1 times 1 minus 0. So it's going to be plus k. So if we want the magnitude of this, this whole thing right over here is just going to be the square root. I'm just taking the magnitude of this part right over here, the actual cross product. It's going to be of negative 1 squared, which is just 1, plus negative 2v squared, which is 4v squared, plus 1 squared, which is just 1. So this whole thing is going to evaluate 2.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So it's going to be plus k. So if we want the magnitude of this, this whole thing right over here is just going to be the square root. I'm just taking the magnitude of this part right over here, the actual cross product. It's going to be of negative 1 squared, which is just 1, plus negative 2v squared, which is 4v squared, plus 1 squared, which is just 1. So this whole thing is going to evaluate 2. 2 is going to evaluate 2. We have 2 plus 2v squared du dv. And if we want to, actually, I almost made a mistake.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So this whole thing is going to evaluate 2. 2 is going to evaluate 2. We have 2 plus 2v squared du dv. And if we want to, actually, I almost made a mistake. That would have been a disaster. 2 plus 4v squared du dv. And if we want, maybe it'll help us a little bit if we factor out a 2 right over here.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And if we want to, actually, I almost made a mistake. That would have been a disaster. 2 plus 4v squared du dv. And if we want, maybe it'll help us a little bit if we factor out a 2 right over here. This is the same thing as 2 times 1 plus 2v squared du dv. If you factor out the 2, you get this is equal to the square root of 2 times the square root of 1 plus 2v squared du dv. And I now think we are ready to evaluate the surface integral.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And if we want, maybe it'll help us a little bit if we factor out a 2 right over here. This is the same thing as 2 times 1 plus 2v squared du dv. If you factor out the 2, you get this is equal to the square root of 2 times the square root of 1 plus 2v squared du dv. And I now think we are ready to evaluate the surface integral. So let's do it. All right. So let me just write this thing down here.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And I now think we are ready to evaluate the surface integral. So let's do it. All right. So let me just write this thing down here. So I'm going to write everything that has to do with v, I'm going to write in purple. So I'm just going to write the ds part right over here. It is the square root of 2 times the square root of 1 plus 2v squared.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So let me just write this thing down here. So I'm going to write everything that has to do with v, I'm going to write in purple. So I'm just going to write the ds part right over here. It is the square root of 2 times the square root of 1 plus 2v squared. And then we're going to have du, which I'll write in green, and then dv. So this is just the ds part. Now we have the y right over here.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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It is the square root of 2 times the square root of 1 plus 2v squared. And then we're going to have du, which I'll write in green, and then dv. So this is just the ds part. Now we have the y right over here. y is just equal to v. So I'll write that in purple. So y is just equal to v. Let me make it very clear. And all of this is ds.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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Now we have the y right over here. y is just equal to v. So I'll write that in purple. So y is just equal to v. Let me make it very clear. And all of this is ds. ds. All of this is ds. And now I can write the bounds in terms of u and v. And so the u part, u is the same thing as x.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And all of this is ds. ds. All of this is ds. And now I can write the bounds in terms of u and v. And so the u part, u is the same thing as x. It goes between 0 and 1. And then v is the same thing as y. And y goes between, or v goes between 0 and 2.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And now I can write the bounds in terms of u and v. And so the u part, u is the same thing as x. It goes between 0 and 1. And then v is the same thing as y. And y goes between, or v goes between 0 and 2. And I now think we're ready to evaluate. And so the u and v variables are not so mixed up. So we can actually separate out these two integrals, make this a product of two single integrals.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And y goes between, or v goes between 0 and 2. And I now think we're ready to evaluate. And so the u and v variables are not so mixed up. So we can actually separate out these two integrals, make this a product of two single integrals. The first thing, if you look at it with respect to du, all this stuff in purple is just a constant with respect to u. So we can take it out of the du integral. We can take all of this purple stuff out of this du integral.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So we can actually separate out these two integrals, make this a product of two single integrals. The first thing, if you look at it with respect to du, all this stuff in purple is just a constant with respect to u. So we can take it out of the du integral. We can take all of this purple stuff out of this du integral. And so this double integral simplifies to the integral from 0 to 2 of, I'll write it as the square root of 2 v times the square root of 1 plus 2 v squared. And then so I factor it out all of this stuff. And then you have times the integral from 0 to 1 du.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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We can take all of this purple stuff out of this du integral. And so this double integral simplifies to the integral from 0 to 2 of, I'll write it as the square root of 2 v times the square root of 1 plus 2 v squared. And then so I factor it out all of this stuff. And then you have times the integral from 0 to 1 du. And then times d. And then you have dv. Now, if this was really complicated, I could say, OK, this is just going to be a function of u. It's a constant with respect to v. And you could factor this whole thing out and separate the integrals.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And then you have times the integral from 0 to 1 du. And then times d. And then you have dv. Now, if this was really complicated, I could say, OK, this is just going to be a function of u. It's a constant with respect to v. And you could factor this whole thing out and separate the integrals. But this is even easier. This integral is just going to evaluate to 1. So this whole thing just evaluates to 1.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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It's a constant with respect to v. And you could factor this whole thing out and separate the integrals. But this is even easier. This integral is just going to evaluate to 1. So this whole thing just evaluates to 1. And so we've simplified this into one single integral. So this simplifies to, and I could even take the square root of 2 out front, the square root of 2 times the integral from v is equal to 0 to v is equal to 2 of v times the square root of 1 plus 2 v squared dv. And so we really are in the home stretch of evaluating the surface integral.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So this whole thing just evaluates to 1. And so we've simplified this into one single integral. So this simplifies to, and I could even take the square root of 2 out front, the square root of 2 times the integral from v is equal to 0 to v is equal to 2 of v times the square root of 1 plus 2 v squared dv. And so we really are in the home stretch of evaluating the surface integral. And here, this is basic. This is actually a little bit of u substitution that we can do in our head. You have a function, or this kind of this embedded function right here, 1 plus 2v squared.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And so we really are in the home stretch of evaluating the surface integral. And here, this is basic. This is actually a little bit of u substitution that we can do in our head. You have a function, or this kind of this embedded function right here, 1 plus 2v squared. What's the derivative of 1 plus 2v squared? Well, it would be 4v. We have something almost 4v here.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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You have a function, or this kind of this embedded function right here, 1 plus 2v squared. What's the derivative of 1 plus 2v squared? Well, it would be 4v. We have something almost 4v here. We can make this 4v by multiplying it by 4 here and then dividing it by 4 out here. This doesn't change the value of the integral. And so now, this part right over here, it's pretty straightforward to take the antiderivative.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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We have something almost 4v here. We can make this 4v by multiplying it by 4 here and then dividing it by 4 out here. This doesn't change the value of the integral. And so now, this part right over here, it's pretty straightforward to take the antiderivative. The antiderivative of this is going to be, we have this embedded function's derivative right over here. So we can kind of just treat it like an x or a v. Take the antiderivative with respect to this thing right over here. And we get, this is essentially 1 plus 2v squared to the 1 half power.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And so now, this part right over here, it's pretty straightforward to take the antiderivative. The antiderivative of this is going to be, we have this embedded function's derivative right over here. So we can kind of just treat it like an x or a v. Take the antiderivative with respect to this thing right over here. And we get, this is essentially 1 plus 2v squared to the 1 half power. We increment it by 1, so it's 1 plus 2v squared to the 3 halves power. And then you divide by 3 halves, or you multiply by 2 thirds. So times 2 thirds.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And we get, this is essentially 1 plus 2v squared to the 1 half power. We increment it by 1, so it's 1 plus 2v squared to the 3 halves power. And then you divide by 3 halves, or you multiply by 2 thirds. So times 2 thirds. So this is the antiderivative of that. And then, of course, you still have all of this stuff out front, square root of 2 over 4. Square root of 2 over 4.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So times 2 thirds. So this is the antiderivative of that. And then, of course, you still have all of this stuff out front, square root of 2 over 4. Square root of 2 over 4. And we are going to evaluate this from 0 to 2. And actually, just to simplify it, let me factor out the 2 thirds so we don't have to worry about that at 2 and 0. So I'm going to factor out the 2 thirds.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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Square root of 2 over 4. And we are going to evaluate this from 0 to 2. And actually, just to simplify it, let me factor out the 2 thirds so we don't have to worry about that at 2 and 0. So I'm going to factor out the 2 thirds. So times 2 over 3. And actually, this will cancel out. That becomes a 1.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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So I'm going to factor out the 2 thirds. So times 2 over 3. And actually, this will cancel out. That becomes a 1. That becomes a 2. And so we are left, this stuff over here is 1 sixth times, and now if you evaluate this at 2, you have 2v squared. That's going to be 2 times 4 is 8, plus 1 is 9.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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That becomes a 1. That becomes a 2. And so we are left, this stuff over here is 1 sixth times, and now if you evaluate this at 2, you have 2v squared. That's going to be 2 times 4 is 8, plus 1 is 9. 9 to the 3 halves power. So 9 to the 1 half is 3. 3 to the 3rd is 27.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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That's going to be 2 times 4 is 8, plus 1 is 9. 9 to the 3 halves power. So 9 to the 1 half is 3. 3 to the 3rd is 27. So it's going to be equal to 27. And then minus this thing evaluated at 0. Well, this is evaluated at 0, so it's going to be 1.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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3 to the 3rd is 27. So it's going to be equal to 27. And then minus this thing evaluated at 0. Well, this is evaluated at 0, so it's going to be 1. 1 to the 3 halves is just 1. So minus 1. And so this gives us, oh, I almost made a mistake.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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Well, this is evaluated at 0, so it's going to be 1. 1 to the 3 halves is just 1. So minus 1. And so this gives us, oh, I almost made a mistake. There should be a square root of 2 up here. Square root of 2 over 6 times 27 minus 1. So drum roll, this gives us the value of our surface integral.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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And so this gives us, oh, I almost made a mistake. There should be a square root of 2 up here. Square root of 2 over 6 times 27 minus 1. So drum roll, this gives us the value of our surface integral. Let's see, 27 minus 1 is 26. So we get 26 times the square root of 2 over 6. And we can simplify a little bit more.
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Surface integral ex2 part 2 Evaluating integral Multivariable Calculus Khan Academy.mp3
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Hello everyone. So now that we have an intuition for what divergence is trying to represent, let's start actually drilling in on a formula. And the first thing I want to do is just limit our perspective to functions that only have an x component, or rather where the y component of the output is just zero. So this is some kind of vector field, and if there's only an x component, what this means it's gonna look like is all of the vectors only go left or right, and there's kind of no up or down involved in any of them. So in this case, let's start thinking about what positive divergence of your vector fields might look like near some point x, y. So if you have your point, you know, this is the point x, y, somewhere sitting off in space. Two cases where the divergence of this might look positive are one, where nothing happens at the point, right?
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Divergence formula, part 1.mp3
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So this is some kind of vector field, and if there's only an x component, what this means it's gonna look like is all of the vectors only go left or right, and there's kind of no up or down involved in any of them. So in this case, let's start thinking about what positive divergence of your vector fields might look like near some point x, y. So if you have your point, you know, this is the point x, y, somewhere sitting off in space. Two cases where the divergence of this might look positive are one, where nothing happens at the point, right? So in this case, p would be equal to zero at our point, but then to the left of it, things are moving to the left, meaning p, the x component of our vector-valued function, is negative, right? That's why the x component of this vector is negative, but then to the right, vectors would be moving off to the right, so over here, p would be positive. So this would be an example of kind of a positive divergence circumstance where only the x component is responsible, and what you'll notice here, this would be p starts negative, goes zero, then becomes positive.
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Divergence formula, part 1.mp3
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Two cases where the divergence of this might look positive are one, where nothing happens at the point, right? So in this case, p would be equal to zero at our point, but then to the left of it, things are moving to the left, meaning p, the x component of our vector-valued function, is negative, right? That's why the x component of this vector is negative, but then to the right, vectors would be moving off to the right, so over here, p would be positive. So this would be an example of kind of a positive divergence circumstance where only the x component is responsible, and what you'll notice here, this would be p starts negative, goes zero, then becomes positive. So as you're changing in the x direction, p should be increasing. So a positive divergence here seems to correspond to a positive partial derivative of p with respect to x. And if that seems a little unfamiliar, if you're not sure how to think about, you know, partial derivatives of a component of a vector field, I have a video on that, and you can kind of take a look and refresh yourself how you might think about this partial derivative of p with respect to x.
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Divergence formula, part 1.mp3
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So this would be an example of kind of a positive divergence circumstance where only the x component is responsible, and what you'll notice here, this would be p starts negative, goes zero, then becomes positive. So as you're changing in the x direction, p should be increasing. So a positive divergence here seems to correspond to a positive partial derivative of p with respect to x. And if that seems a little unfamiliar, if you're not sure how to think about, you know, partial derivatives of a component of a vector field, I have a video on that, and you can kind of take a look and refresh yourself how you might think about this partial derivative of p with respect to x. And once you do, hopefully it makes sense why this specific positive divergence example corresponds with a positive partial derivative of p. But remember, this isn't the only way that a positive divergence might look. You could have another circumstance where, let's say, your point, x, y, actually has a vector attached to it. So this here, again, represents our point, x, y, and in this specific example, this would be kind of p is positive, p of x, y is positive at your point there.
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Divergence formula, part 1.mp3
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And if that seems a little unfamiliar, if you're not sure how to think about, you know, partial derivatives of a component of a vector field, I have a video on that, and you can kind of take a look and refresh yourself how you might think about this partial derivative of p with respect to x. And once you do, hopefully it makes sense why this specific positive divergence example corresponds with a positive partial derivative of p. But remember, this isn't the only way that a positive divergence might look. You could have another circumstance where, let's say, your point, x, y, actually has a vector attached to it. So this here, again, represents our point, x, y, and in this specific example, this would be kind of p is positive, p of x, y is positive at your point there. But another way that positive divergence might look is that you have things coming in towards that point and things going away, but the things going away are bigger than the ones coming in. But again, this kind of exhibits the idea of p increasing in value. You know, p starts off small, it's a positive but small component, and then it gets bigger, and then it gets even bigger.
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Divergence formula, part 1.mp3
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So this here, again, represents our point, x, y, and in this specific example, this would be kind of p is positive, p of x, y is positive at your point there. But another way that positive divergence might look is that you have things coming in towards that point and things going away, but the things going away are bigger than the ones coming in. But again, this kind of exhibits the idea of p increasing in value. You know, p starts off small, it's a positive but small component, and then it gets bigger, and then it gets even bigger. So once again, we have this idea of positive partial derivative of p with respect to x, because changes in x, as you increase x, it causes an increase in p, seems to correspond to positive divergence. And you can even look at it if you go the other way, where you have a little bit of negative component to p here, so p is a little bit negative, but to the left of your point, it's really negative, and then to the right, it's not nearly as negative. And in this case, it's kind of like, as you're moving to the right, as x is increasing, you start off very negative, and then only kind of negative, and then barely negative.
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Divergence formula, part 1.mp3
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You know, p starts off small, it's a positive but small component, and then it gets bigger, and then it gets even bigger. So once again, we have this idea of positive partial derivative of p with respect to x, because changes in x, as you increase x, it causes an increase in p, seems to correspond to positive divergence. And you can even look at it if you go the other way, where you have a little bit of negative component to p here, so p is a little bit negative, but to the left of your point, it's really negative, and then to the right, it's not nearly as negative. And in this case, it's kind of like, as you're moving to the right, as x is increasing, you start off very negative, and then only kind of negative, and then barely negative. And once again, that corresponds to an increase in the value of p as x increases. So what you'd expect is that a partial derivative of p, that x component of the output with respect to x, is gonna be somewhere involved in the formula for the divergence of our vector field at a point x, y. And in the next video, I'm gonna go a similar line of reasoning to see what should go on with that y component.
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Divergence formula, part 1.mp3
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And to do that, let's imagine. So let me draw my axes. So that's my z-axis. That is my x-axis. And then that is my y-axis. And now let's imagine a region in the xy-plane. So let me draw it like this.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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That is my x-axis. And then that is my y-axis. And now let's imagine a region in the xy-plane. So let me draw it like this. So let's say this is my region in the xy-plane. I will call that region R. And I also have a boundary of that region. And let's say we care about the direction that we traverse the boundary.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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So let me draw it like this. So let's say this is my region in the xy-plane. I will call that region R. And I also have a boundary of that region. And let's say we care about the direction that we traverse the boundary. And let's say we're going to traverse it in a counterclockwise direction. So we have this path that goes around this region. We can call that C. So we'll call that C. And we're going to traverse it in the counterclockwise direction.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And let's say we care about the direction that we traverse the boundary. And let's say we're going to traverse it in a counterclockwise direction. So we have this path that goes around this region. We can call that C. So we'll call that C. And we're going to traverse it in the counterclockwise direction. And let's say that we also have a vector field F that essentially its i component is just going to be a function of x and y. And its j component is only going to be a function of x and y. And let's say it has no k components.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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We can call that C. So we'll call that C. And we're going to traverse it in the counterclockwise direction. And let's say that we also have a vector field F that essentially its i component is just going to be a function of x and y. And its j component is only going to be a function of x and y. And let's say it has no k components. So the vector field on this region, it might look something like this. I'm just drawing random things. And then if you go off that region, if you go in the z direction, it's just going to look the same as you go higher and higher.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And let's say it has no k components. So the vector field on this region, it might look something like this. I'm just drawing random things. And then if you go off that region, if you go in the z direction, it's just going to look the same as you go higher and higher. So that vector, it wouldn't change as you change your z component. And all of the vectors would essentially be parallel to, or if z is 0, actually sitting on the xy-plane. Now given this, let's think about what Stokes' theorem would tell us about the value of the line integral over the contour.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And then if you go off that region, if you go in the z direction, it's just going to look the same as you go higher and higher. So that vector, it wouldn't change as you change your z component. And all of the vectors would essentially be parallel to, or if z is 0, actually sitting on the xy-plane. Now given this, let's think about what Stokes' theorem would tell us about the value of the line integral over the contour. Let me draw that a little bit neater. The line integral over the contour C of F dot dr, F dot lowercase dr, where dr is obviously going along the contour. So if we take Stokes' theorem, then this quantity right over here should be equal to this quantity right over here.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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Now given this, let's think about what Stokes' theorem would tell us about the value of the line integral over the contour. Let me draw that a little bit neater. The line integral over the contour C of F dot dr, F dot lowercase dr, where dr is obviously going along the contour. So if we take Stokes' theorem, then this quantity right over here should be equal to this quantity right over here. It should be equal to the double integral over the surface. Well, this region is really just a surface that's sitting in the xy-plane. So it should really just be the double integral.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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So if we take Stokes' theorem, then this quantity right over here should be equal to this quantity right over here. It should be equal to the double integral over the surface. Well, this region is really just a surface that's sitting in the xy-plane. So it should really just be the double integral. Let me write that in that same. It'll be the double integral over our region, which is really just the same thing as our surface of the curl of F dot n. So let's just think about what the curl of F dot n is. And then d of s would just be a little chunk of our region, a little chunk of our flattened surface right over there.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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So it should really just be the double integral. Let me write that in that same. It'll be the double integral over our region, which is really just the same thing as our surface of the curl of F dot n. So let's just think about what the curl of F dot n is. And then d of s would just be a little chunk of our region, a little chunk of our flattened surface right over there. So instead of ds, I'll just write da. But let's think about what curl of F dot n would actually be. So let's work on curl of F first.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And then d of s would just be a little chunk of our region, a little chunk of our flattened surface right over there. So instead of ds, I'll just write da. But let's think about what curl of F dot n would actually be. So let's work on curl of F first. So the curl of F, and the way I always remember it is we're going to take the determinant of this ijk, partial with respect to x, partial with respect to y, partial with respect to z. This is just the definition of taking the curl. We're figuring out how much this vector field would cause something to spin.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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So let's work on curl of F first. So the curl of F, and the way I always remember it is we're going to take the determinant of this ijk, partial with respect to x, partial with respect to y, partial with respect to z. This is just the definition of taking the curl. We're figuring out how much this vector field would cause something to spin. And then we want the i component, which is our function p, which is just a function of x and y, j component, which is just the function q. And there was no z component over here, so 0. And so this is going to be equal to, well, if we look at the i component, it's going to be the partial of y of 0.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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We're figuring out how much this vector field would cause something to spin. And then we want the i component, which is our function p, which is just a function of x and y, j component, which is just the function q. And there was no z component over here, so 0. And so this is going to be equal to, well, if we look at the i component, it's going to be the partial of y of 0. That's just going to be 0 minus the partial of q with respect to z. Well, what's the partial of q with respect to z? Well, q isn't a function of z at all, so that's also going to be 0.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And so this is going to be equal to, well, if we look at the i component, it's going to be the partial of y of 0. That's just going to be 0 minus the partial of q with respect to z. Well, what's the partial of q with respect to z? Well, q isn't a function of z at all, so that's also going to be 0. Let me write this out, just so it's not too confusing. So our i component is going to be partial of 0 with respect to y. Well, that's just going to be 0 minus the partial of q with respect to z.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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Well, q isn't a function of z at all, so that's also going to be 0. Let me write this out, just so it's not too confusing. So our i component is going to be partial of 0 with respect to y. Well, that's just going to be 0 minus the partial of q with respect to z. Well, the partial of q with respect to z is just going to be 0. So we have a 0i component, and then we want to subtract the j component. And then the j component, partial of 0 with respect to x is 0.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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Well, that's just going to be 0 minus the partial of q with respect to z. Well, the partial of q with respect to z is just going to be 0. So we have a 0i component, and then we want to subtract the j component. And then the j component, partial of 0 with respect to x is 0. And then from that, you're going to subtract the partial of p with respect to z. Well, once again, p is not a function of z at all, so that's going to be 0 again. And then you have plus k times the partial of q with respect to x.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And then the j component, partial of 0 with respect to x is 0. And then from that, you're going to subtract the partial of p with respect to z. Well, once again, p is not a function of z at all, so that's going to be 0 again. And then you have plus k times the partial of q with respect to x. Remember, this is just the partial derivative operator. So partial of q with respect to x. And from that, we're going to subtract the partial of p with respect to y.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And then you have plus k times the partial of q with respect to x. Remember, this is just the partial derivative operator. So partial of q with respect to x. And from that, we're going to subtract the partial of p with respect to y. So the curl of f just simplifies to this right over here. Now, what is n? What is the unit normal vector?
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And from that, we're going to subtract the partial of p with respect to y. So the curl of f just simplifies to this right over here. Now, what is n? What is the unit normal vector? Well, we're in the xy plane, so the unit normal vector is just going to be straight up in the z direction. It's going to have a magnitude of 1. So in this case, our unit normal vector is just going to be the k vector.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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What is the unit normal vector? Well, we're in the xy plane, so the unit normal vector is just going to be straight up in the z direction. It's going to have a magnitude of 1. So in this case, our unit normal vector is just going to be the k vector. So we're essentially just going to take curl of f is this, and our unit normal vector is just going to be equal to the k unit vector. It's going to go straight up. So what happens if we take the curl of f dot k?
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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So in this case, our unit normal vector is just going to be the k vector. So we're essentially just going to take curl of f is this, and our unit normal vector is just going to be equal to the k unit vector. It's going to go straight up. So what happens if we take the curl of f dot k? If we just dot this with k? We're just dotting this with this. Well, we're just going to end up with this part right over here.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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So what happens if we take the curl of f dot k? If we just dot this with k? We're just dotting this with this. Well, we're just going to end up with this part right over here. So curl of f dot the unit normal vector is just going to be equal to this business. It's just going to be equal to the partial of q with respect to x minus the partial of p with respect to y. And this is neat, because using Stokes' theorem in this special case where we're dealing with a flattened out surface in the xy plane, in this situation, this just boiled down to Green's theorem.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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Well, we're just going to end up with this part right over here. So curl of f dot the unit normal vector is just going to be equal to this business. It's just going to be equal to the partial of q with respect to x minus the partial of p with respect to y. And this is neat, because using Stokes' theorem in this special case where we're dealing with a flattened out surface in the xy plane, in this situation, this just boiled down to Green's theorem. This thing right over here just boiled down to Green's theorem. So we see that Green's theorem is really just a special case. Let me write theorem a little bit neater.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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