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So these bottom two points here correspond to plus and minus x equals the square root of two. So if we go over here and think about the case where x equals the square root of two, and we plug that into the expression, what are we gonna get? Well, we're gonna get 12 multiplied by, if x equals square root of two, then x squared is equal to two, so that's 12 times two minus eight, so that's 24 minus eight, and we're gonna get 16, which is a positive number, which is why you have positive concavity at each of these points. So as far as the x direction is concerned, it feels like, ah, yes, both of these have positive concavity, so they should look like local minima. And then if you plug in zero, if instead we went over here and we said x equals zero, then when you plug that in, you'd have 12 times zero minus eight, and instead of 16, you would be getting negative eight. So because you have a negative amount, that gives you this negative concavity on the graph, which is why, as far as x is concerned, the origin looks like a local maximum. So let's actually write that down.
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Warm up to the second partial derivative test.mp3
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So as far as the x direction is concerned, it feels like, ah, yes, both of these have positive concavity, so they should look like local minima. And then if you plug in zero, if instead we went over here and we said x equals zero, then when you plug that in, you'd have 12 times zero minus eight, and instead of 16, you would be getting negative eight. So because you have a negative amount, that gives you this negative concavity on the graph, which is why, as far as x is concerned, the origin looks like a local maximum. So let's actually write that down. If we kinda go down here, and we're analyzing each one of these, and we think about what does it look like from the perspective of each variable, as far as x is concerned, that origin should look like a max, and then each of these two points should look like minima. This is kind of what the variable x thinks. And then the variable y, if we do something similar, and we take the second partial derivative with respect to y, yeah, I'll go ahead and write that over here because this'll be pretty quick, second partial derivative with respect to y, we're taking the derivative of this expression with respect to y, and that's just a constant, that's just two.
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Warm up to the second partial derivative test.mp3
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So let's actually write that down. If we kinda go down here, and we're analyzing each one of these, and we think about what does it look like from the perspective of each variable, as far as x is concerned, that origin should look like a max, and then each of these two points should look like minima. This is kind of what the variable x thinks. And then the variable y, if we do something similar, and we take the second partial derivative with respect to y, yeah, I'll go ahead and write that over here because this'll be pretty quick, second partial derivative with respect to y, we're taking the derivative of this expression with respect to y, and that's just a constant, that's just two. And because it's positive, it's telling you that, as far as y is concerned, there's positive concavity everywhere. And on the graph, what that would mean, what that would mean, if you just look at things where you're kind of slicing with a constant x value to see pure movement in the y direction, there's always gonna be positive concavity. And here, I've only drawn the plane where x is constantly equal to zero, but if you imagine kind of sliding that plane around left and right, you're always getting positive concavity.
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Warm up to the second partial derivative test.mp3
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And then the variable y, if we do something similar, and we take the second partial derivative with respect to y, yeah, I'll go ahead and write that over here because this'll be pretty quick, second partial derivative with respect to y, we're taking the derivative of this expression with respect to y, and that's just a constant, that's just two. And because it's positive, it's telling you that, as far as y is concerned, there's positive concavity everywhere. And on the graph, what that would mean, what that would mean, if you just look at things where you're kind of slicing with a constant x value to see pure movement in the y direction, there's always gonna be positive concavity. And here, I've only drawn the plane where x is constantly equal to zero, but if you imagine kind of sliding that plane around left and right, you're always getting positive concavity. So as far as y is concerned, everything looks like a local minimum. So we kind of go down here, and you'd say, everything looks like a local minimum. Minimum, minimum, and minimum.
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Warm up to the second partial derivative test.mp3
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And here, I've only drawn the plane where x is constantly equal to zero, but if you imagine kind of sliding that plane around left and right, you're always getting positive concavity. So as far as y is concerned, everything looks like a local minimum. So we kind of go down here, and you'd say, everything looks like a local minimum. Minimum, minimum, and minimum. So it might be tempting here to think that you're done, to think you found all the information you need to, because you say, well, in the x and y direction, they disagree about whether that origin should be a maximum or a minimum, which is why it looks like a saddle point. And then they agree, they agree on the other two points that both of them should look like a minimum, which is why, you know, which is why you could say, you think you might say, both of these guys look like a minimum. However, that's actually not enough.
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Warm up to the second partial derivative test.mp3
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Minimum, minimum, and minimum. So it might be tempting here to think that you're done, to think you found all the information you need to, because you say, well, in the x and y direction, they disagree about whether that origin should be a maximum or a minimum, which is why it looks like a saddle point. And then they agree, they agree on the other two points that both of them should look like a minimum, which is why, you know, which is why you could say, you think you might say, both of these guys look like a minimum. However, that's actually not enough. There are cases, there are examples that I could draw where doing this kind of analysis would lead you to the wrong conclusion. You would conclude that certain points are, you know, a local minimum, when in fact, they're a saddle point. And the basic reason is that you need to take into account the information given by that other second partial derivative, because in the multivariable world, you can take the partial derivative with respect to one variable, and then with respect to another.
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Warm up to the second partial derivative test.mp3
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However, that's actually not enough. There are cases, there are examples that I could draw where doing this kind of analysis would lead you to the wrong conclusion. You would conclude that certain points are, you know, a local minimum, when in fact, they're a saddle point. And the basic reason is that you need to take into account the information given by that other second partial derivative, because in the multivariable world, you can take the partial derivative with respect to one variable, and then with respect to another. And you have to take into account this mixed partial derivative term in order to make full conclusions. And I'm a little bit afraid that this video might be running long, so I'll cut it short here, and then I will give you the second partial derivative test in its full glory, accounting for this mixed partial derivative term in the next video. And I'll also, you know, give intuition for where this comes in, why it comes in, why this simple analysis that we did in this case is close, and it does give intuition, but it's not quite full, and it won't give you the right conclusion always.
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Warm up to the second partial derivative test.mp3
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And we really need to express it in terms of a double integral with the parameters, in the domain of the parameters. And the first thing I'm going to do is rewrite this part right over here using our parameters. And we already know that n, our normal vector, times our surface differential can also be written as kind of a vector version of our surface differential that points in the same direction as our normal vector. This is going to be the same thing, and we need to make sure that we get the order on the cross product right. As the partial derivative, and I'm going to confirm this in a second, the partial derivative of the parameterization with respect to one of the parameters crossed with the partial derivative of the parameterization with respect to the other parameter. And then that whole thing, I'm not going to take the absolute value because I need a vector right over here, times the differentials of the parameters, d theta, dr. And we can swap these two things around depending on what will make our eventual double integral easier. But we can't swap these two things around because this would actually change the direction of the vector.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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This is going to be the same thing, and we need to make sure that we get the order on the cross product right. As the partial derivative, and I'm going to confirm this in a second, the partial derivative of the parameterization with respect to one of the parameters crossed with the partial derivative of the parameterization with respect to the other parameter. And then that whole thing, I'm not going to take the absolute value because I need a vector right over here, times the differentials of the parameters, d theta, dr. And we can swap these two things around depending on what will make our eventual double integral easier. But we can't swap these two things around because this would actually change the direction of the vector. So we need to make sure that this is popping us out in the right direction. So let's think about the direction that the partial with respect to r will take us. So as r increases, we're going to be moving radially outward from the center of our surface.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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But we can't swap these two things around because this would actually change the direction of the vector. So we need to make sure that this is popping us out in the right direction. So let's think about the direction that the partial with respect to r will take us. So as r increases, we're going to be moving radially outward from the center of our surface. As r increases, let me do this in a different color, as r increases, we'll be moving radially outward. So this quantity will be a vector that looks something like that. And then as theta increases, we'll be going roughly in that direction.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So as r increases, we're going to be moving radially outward from the center of our surface. As r increases, let me do this in a different color, as r increases, we'll be moving radially outward. So this quantity will be a vector that looks something like that. And then as theta increases, we'll be going roughly in that direction. And so if we take the cross product of those two things, and we could use the right-hand rule, you could imagine, take your right hand, point your index finger in the direction of that yellow vector. Let me make it clear, this is the orange vector right over here. Take your index finger in the direction of that yellow vector.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And then as theta increases, we'll be going roughly in that direction. And so if we take the cross product of those two things, and we could use the right-hand rule, you could imagine, take your right hand, point your index finger in the direction of that yellow vector. Let me make it clear, this is the orange vector right over here. Take your index finger in the direction of that yellow vector. So this is my index finger, my shakily drawn yellow index finger. Put your middle finger in the direction of the orange vector. So my middle finger, you bend it and put it in the direction of the orange vector, and we don't care what the other two fingers do.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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Take your index finger in the direction of that yellow vector. So this is my index finger, my shakily drawn yellow index finger. Put your middle finger in the direction of the orange vector. So my middle finger, you bend it and put it in the direction of the orange vector, and we don't care what the other two fingers do. And then your thumb will be in the direction of the cross product. So your thumb will point outward like that. That's my best attempt at drawing it, which is exactly the direction we need it to point in.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So my middle finger, you bend it and put it in the direction of the orange vector, and we don't care what the other two fingers do. And then your thumb will be in the direction of the cross product. So your thumb will point outward like that. That's my best attempt at drawing it, which is exactly the direction we need it to point in. We need it to point upward here in order to be oriented properly with the direction that we are actually traversing the path. This is actually the right order. If when we did this, we got the thumb pointing into it or below the plane, then we would actually have to swap these orders.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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That's my best attempt at drawing it, which is exactly the direction we need it to point in. We need it to point upward here in order to be oriented properly with the direction that we are actually traversing the path. This is actually the right order. If when we did this, we got the thumb pointing into it or below the plane, then we would actually have to swap these orders. So with that out of the way, let's actually evaluate this cross product. Let's evaluate this cross product. So the cross product of the partial of our parameterization with respect to r crossed with the partial of our parameterization with respect to theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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If when we did this, we got the thumb pointing into it or below the plane, then we would actually have to swap these orders. So with that out of the way, let's actually evaluate this cross product. Let's evaluate this cross product. So the cross product of the partial of our parameterization with respect to r crossed with the partial of our parameterization with respect to theta. I like to set up the matrix to take the cross product. So we'll put our i, j, and k components just like that. And then first I will write the partial with respect to r. So the i component, if you take the derivative of this with respect to r, it's just going to be cosine theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So the cross product of the partial of our parameterization with respect to r crossed with the partial of our parameterization with respect to theta. I like to set up the matrix to take the cross product. So we'll put our i, j, and k components just like that. And then first I will write the partial with respect to r. So the i component, if you take the derivative of this with respect to r, it's just going to be cosine theta. The derivative of this with respect to r is just sine theta. And the derivative of this with respect to r is just going to be a negative sine theta. Negative sine theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And then first I will write the partial with respect to r. So the i component, if you take the derivative of this with respect to r, it's just going to be cosine theta. The derivative of this with respect to r is just sine theta. And the derivative of this with respect to r is just going to be a negative sine theta. Negative sine theta. And then if we take, and then we're going to cross it with this. So the derivative of this with respect to theta is going to be negative r sine of theta. So it's negative r sine of theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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Negative sine theta. And then if we take, and then we're going to cross it with this. So the derivative of this with respect to theta is going to be negative r sine of theta. So it's negative r sine of theta. Derivative of our j component with respect to theta will be r cosine theta. r cosine theta. And the derivative of k, of our k component, or our z component with respect to theta is going to be negative r cosine theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So it's negative r sine of theta. Derivative of our j component with respect to theta will be r cosine theta. r cosine theta. And the derivative of k, of our k component, or our z component with respect to theta is going to be negative r cosine theta. Negative r cosine theta. Is that right? Negative derivative of sine theta is cosine theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And the derivative of k, of our k component, or our z component with respect to theta is going to be negative r cosine theta. Negative r cosine theta. Is that right? Negative derivative of sine theta is cosine theta. So it's negative r cosine theta. And now we just evaluate this determinant over here. So this is going to be equal to our i component.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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Negative derivative of sine theta is cosine theta. So it's negative r cosine theta. And now we just evaluate this determinant over here. So this is going to be equal to our i component. It's going to be, ignore that row and that column. And we get sine times negative r cosine theta. So we're going to get, I'll just do this in a new color.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So this is going to be equal to our i component. It's going to be, ignore that row and that column. And we get sine times negative r cosine theta. So we're going to get, I'll just do this in a new color. So we're going to get negative r, that wasn't a new color, I'll do it in purple. We will get negative r cosine theta sine theta. Cosine theta sine theta minus, well this is going to be a negative number, so when you subtract a negative number it's going to be plus.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So we're going to get, I'll just do this in a new color. So we're going to get negative r, that wasn't a new color, I'll do it in purple. We will get negative r cosine theta sine theta. Cosine theta sine theta minus, well this is going to be a negative number, so when you subtract a negative number it's going to be plus. So it's going to be plus r cosine theta sine theta. Plus r cosine theta sine theta. And it's always nice when things cancel out like this.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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Cosine theta sine theta minus, well this is going to be a negative number, so when you subtract a negative number it's going to be plus. So it's going to be plus r cosine theta sine theta. Plus r cosine theta sine theta. And it's always nice when things cancel out like this. This plus this is just zero. Negative of it plus the positive of it. So that all cancels out to zero.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And it's always nice when things cancel out like this. This plus this is just zero. Negative of it plus the positive of it. So that all cancels out to zero. We don't have an i component. Now let's go to the j component. And remember we need to do our little checkerboard pattern.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So that all cancels out to zero. We don't have an i component. Now let's go to the j component. And remember we need to do our little checkerboard pattern. So it's going to be minus j. And it's going to be, ignore this column, that row. Cosine theta times negative r cosine theta is negative r cosine squared theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And remember we need to do our little checkerboard pattern. So it's going to be minus j. And it's going to be, ignore this column, that row. Cosine theta times negative r cosine theta is negative r cosine squared theta. I just multiplied those two. And then from that I'm going to subtract this times that. And so this times that, the negative cancels out, we get r sine squared theta.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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Cosine theta times negative r cosine theta is negative r cosine squared theta. I just multiplied those two. And then from that I'm going to subtract this times that. And so this times that, the negative cancels out, we get r sine squared theta. So this is minus, let me make sure, I'm going to subtract the product of these two. I'm going to get this positive and it's going to be r sine squared theta. This is always the hard part.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And so this times that, the negative cancels out, we get r sine squared theta. So this is minus, let me make sure, I'm going to subtract the product of these two. I'm going to get this positive and it's going to be r sine squared theta. This is always the hard part. You can make a lot of careless mistakes here. And this looks like we might be able to simplify this in a second, but I'll wait. Actually I'll just distribute this negative sign just for fun.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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This is always the hard part. You can make a lot of careless mistakes here. And this looks like we might be able to simplify this in a second, but I'll wait. Actually I'll just distribute this negative sign just for fun. So if we distribute the negative sign, these all become positive. Helps simplify things a little bit. And now let's worry about the k component.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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Actually I'll just distribute this negative sign just for fun. So if we distribute the negative sign, these all become positive. Helps simplify things a little bit. And now let's worry about the k component. K component I will be doing in purple. I'll do it in blue. K component, ignore this row, ignore this column.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And now let's worry about the k component. K component I will be doing in purple. I'll do it in blue. K component, ignore this row, ignore this column. So plus k times cosine theta times r cosine theta is r cosine squared theta. And then from that I'm going to subtract negative r sine theta times sine theta. So that's going to be negative r sine squared theta, but I'm subtracting it.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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K component, ignore this row, ignore this column. So plus k times cosine theta times r cosine theta is r cosine squared theta. And then from that I'm going to subtract negative r sine theta times sine theta. So that's going to be negative r sine squared theta, but I'm subtracting it. So it's going to be plus r sine squared theta. And this looks like we're going to simplify it as well. And so this piece right over here we can factor out, let me just rewrite it.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So that's going to be negative r sine squared theta, but I'm subtracting it. So it's going to be plus r sine squared theta. And this looks like we're going to simplify it as well. And so this piece right over here we can factor out, let me just rewrite it. This can be written as r times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1, so this is just r times j.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And so this piece right over here we can factor out, let me just rewrite it. This can be written as r times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1, so this is just r times j. And this over here simplifies for the exact same reason. This is r times cosine squared theta plus sine squared theta. This also is just 1.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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That just evaluates to 1, so this is just r times j. And this over here simplifies for the exact same reason. This is r times cosine squared theta plus sine squared theta. This also is just 1. So this just simplifies to r times k. And so this whole cross product, all of this business right over here, simplified to, quite luckily, is equal to r times our j unit vector plus r times our k unit vector. And so now we can write our surface integral, our original surface integral, we can write as the double integral. And we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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This also is just 1. So this just simplifies to r times k. And so this whole cross product, all of this business right over here, simplified to, quite luckily, is equal to r times our j unit vector plus r times our k unit vector. And so now we can write our surface integral, our original surface integral, we can write as the double integral. And we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later. Double integral. Now it's going to be over our parameter domain, the r theta domain. So it's the double integral of, we still have the curl of f, and we're going to have to evaluate the curl of f. So I'll just write curl of our vector field f dotted with this business, rj plus rk, and then we have our two parameters.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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And we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later. Double integral. Now it's going to be over our parameter domain, the r theta domain. So it's the double integral of, we still have the curl of f, and we're going to have to evaluate the curl of f. So I'll just write curl of our vector field f dotted with this business, rj plus rk, and then we have our two parameters. And we might want to switch the order. So maybe we could write d theta dr, and then if we do it in this order, theta goes from 0 to 2 pi and r is going from 0 to 1. But if we swap these two, then obviously we're going to have to swap these two as well.
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Stokes example part 3 Surface to double integral Multivariable Calculus Khan Academy.mp3
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So let's do another curvature example. This time, I'll just take a two-dimensional curve. So it'll have two different components, x of t and y of t. And the specific components here will be t minus the sine of t, t minus sine of t, and then one minus cosine of t, one minus cosine of t. And this is actually the curve, if you watched the very first video that I did about curvature, introducing it, this is that curve. This is the curve that I said, imagine that it's a road and you're driving along it, and if your steering wheel gets stuck, you're thinking of the circle that you trace out as a result, and at various different points, you're going to be turning at various different amounts, so the circle that your car ends up tracing out would be of varying sizes. So if the curvature is high, if you're steering a lot, radius of curvature is low, and things like that. Here, let's actually compute it. And in the last example, I walked through thinking in terms of the derivative of the unit tangent vector with respect to arc length, but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double prime minus y prime times x double prime, and then all of that divided by, all of that divided by x prime squared plus y prime squared, and I'm writing x prime and y prime and such, and all of these you should think of as taking in the variable t, I'm just being a little too lazy to write it, and you take that to the 3 halves power.
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Curvature of a cycloid.mp3
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This is the curve that I said, imagine that it's a road and you're driving along it, and if your steering wheel gets stuck, you're thinking of the circle that you trace out as a result, and at various different points, you're going to be turning at various different amounts, so the circle that your car ends up tracing out would be of varying sizes. So if the curvature is high, if you're steering a lot, radius of curvature is low, and things like that. Here, let's actually compute it. And in the last example, I walked through thinking in terms of the derivative of the unit tangent vector with respect to arc length, but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double prime minus y prime times x double prime, and then all of that divided by, all of that divided by x prime squared plus y prime squared, and I'm writing x prime and y prime and such, and all of these you should think of as taking in the variable t, I'm just being a little too lazy to write it, and you take that to the 3 halves power. So this was a formula, and I'm not a huge fan of memorizing formulas and then hoping to apply them later. I really do think the one thing you should take away from curvature is the idea that it's the derivative of the unit tangent vector with respect to arc length, and if you need to, you can just look up a formula like this, but it's worth pointing out that it makes some things easier to compute because finding the tangent vector and everything can be kind of like reinventing the wheel when you already have the results here. So first thing to do is just find x prime, y double prime, y prime, and x prime.
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Curvature of a cycloid.mp3
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And in the last example, I walked through thinking in terms of the derivative of the unit tangent vector with respect to arc length, but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double prime minus y prime times x double prime, and then all of that divided by, all of that divided by x prime squared plus y prime squared, and I'm writing x prime and y prime and such, and all of these you should think of as taking in the variable t, I'm just being a little too lazy to write it, and you take that to the 3 halves power. So this was a formula, and I'm not a huge fan of memorizing formulas and then hoping to apply them later. I really do think the one thing you should take away from curvature is the idea that it's the derivative of the unit tangent vector with respect to arc length, and if you need to, you can just look up a formula like this, but it's worth pointing out that it makes some things easier to compute because finding the tangent vector and everything can be kind of like reinventing the wheel when you already have the results here. So first thing to do is just find x prime, y double prime, y prime, and x prime. So let's go ahead and write those out. So the first derivative of x of t, if we go up here, that's t minus sine of t, so its derivative is 1 minus cosine of t, and the derivative of the y component of 1 minus cosine t, y prime of t, is going to be derivative of cosine is negative sine, so negative derivative of that is sine, and that 1 goes to a constant. And then when we take the second derivatives of those guys, so maybe change the color for the second derivative here, x double prime of t, so now we're taking the derivative of this, which actually we just did because by coincidence, the first derivative of x is also the y component, so that also equals sine of t. And then y double prime is just the derivative of sine here, so that's just going to be cosine, cosine of t. So now, when we just plug those four values in for kappa, for our curvature, what we get is x prime was 1 minus cosine of t, multiplied by y double prime is cosine of t, we subtract off from that y prime, which is sine of t, multiplied by x double prime, x double prime is also sine of t, so I could just say sine of t squared, and the whole thing is divided by x prime squared, so x prime was 1 minus cosine of t squared, plus y prime squared, so y prime was just sine, so that's just going to be sine squared of t, and that whole thing to the power 3 halves.
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Curvature of a cycloid.mp3
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So first thing to do is just find x prime, y double prime, y prime, and x prime. So let's go ahead and write those out. So the first derivative of x of t, if we go up here, that's t minus sine of t, so its derivative is 1 minus cosine of t, and the derivative of the y component of 1 minus cosine t, y prime of t, is going to be derivative of cosine is negative sine, so negative derivative of that is sine, and that 1 goes to a constant. And then when we take the second derivatives of those guys, so maybe change the color for the second derivative here, x double prime of t, so now we're taking the derivative of this, which actually we just did because by coincidence, the first derivative of x is also the y component, so that also equals sine of t. And then y double prime is just the derivative of sine here, so that's just going to be cosine, cosine of t. So now, when we just plug those four values in for kappa, for our curvature, what we get is x prime was 1 minus cosine of t, multiplied by y double prime is cosine of t, we subtract off from that y prime, which is sine of t, multiplied by x double prime, x double prime is also sine of t, so I could just say sine of t squared, and the whole thing is divided by x prime squared, so x prime was 1 minus cosine of t squared, plus y prime squared, so y prime was just sine, so that's just going to be sine squared of t, and that whole thing to the power 3 halves. And that's your answer, right? You apply the formula, you get the answer. So for example, when I was drawing this curve and kind of telling the computer to draw out the appropriate circle, I didn't go through the entire find the unit tangent vector, differentiate it with respect to arc length process, even though that's, you know, decently easy to do in the case of things like circles or helixes, but instead I just went to that formula, I looked it up because I had forgotten, and I found the radius of curvature that way.
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Curvature of a cycloid.mp3
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So let's try to think of a parameterization. And let me just copy and paste this entire drawing just so that I can use it down below as we parameterize it. So let me copy it, and then go all the way down here and let me paste it. OK, that is our shape again, our surface. And then let me go to the layer that I wanted to get on. And then let me start evaluating it. So what we want to care about is the integral over surface 3 of z ds.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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OK, that is our shape again, our surface. And then let me go to the layer that I wanted to get on. And then let me start evaluating it. So what we want to care about is the integral over surface 3 of z ds. And surface 3 here, we see that the x and y values essentially take on all of the possible x and y values inside of the unit circle, including the boundary. And then the z values are going to be a function of the x values. We know that this plane, that this top surface right over here, S3, it is a subset of the plane z. z is equal to 1 minus x.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So what we want to care about is the integral over surface 3 of z ds. And surface 3 here, we see that the x and y values essentially take on all of the possible x and y values inside of the unit circle, including the boundary. And then the z values are going to be a function of the x values. We know that this plane, that this top surface right over here, S3, it is a subset of the plane z. z is equal to 1 minus x. It's a subset that's kind of above the unit circle in the xy plane, or kind of the subset that intersects with our cylinder and kind of chops it. So let's think about x and y's first. So first, so x, so let's think about it in terms of polar coordinates, because that's probably the easiest way to think about it.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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We know that this plane, that this top surface right over here, S3, it is a subset of the plane z. z is equal to 1 minus x. It's a subset that's kind of above the unit circle in the xy plane, or kind of the subset that intersects with our cylinder and kind of chops it. So let's think about x and y's first. So first, so x, so let's think about it in terms of polar coordinates, because that's probably the easiest way to think about it. So I'm going to redraw kind of a top view. So that is my y-axis, and this is my x-axis. And the x's and y's can take on any value.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So first, so x, so let's think about it in terms of polar coordinates, because that's probably the easiest way to think about it. So I'm going to redraw kind of a top view. So that is my y-axis, and this is my x-axis. And the x's and y's can take on any value. They essentially have to fill the unit circle. So if you were to kind of project this top surface down onto the xy plane, you would get this orange surface, that bottom surface, which looked like this. It was essentially the unit circle, just like that.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And the x's and y's can take on any value. They essentially have to fill the unit circle. So if you were to kind of project this top surface down onto the xy plane, you would get this orange surface, that bottom surface, which looked like this. It was essentially the unit circle, just like that. Let me draw it a little bit neater than that. I can do a better job. So let me draw the unit circle as neatly as I can.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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It was essentially the unit circle, just like that. Let me draw it a little bit neater than that. I can do a better job. So let me draw the unit circle as neatly as I can. So there's my unit circle. And so we can have one parameter that essentially says how far around the unit circle we're going. So essentially, that would be our angle.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So let me draw the unit circle as neatly as I can. So there's my unit circle. And so we can have one parameter that essentially says how far around the unit circle we're going. So essentially, that would be our angle. And let's use theta, because that's just for fun. We haven't used theta as a parameter yet. That's theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So essentially, that would be our angle. And let's use theta, because that's just for fun. We haven't used theta as a parameter yet. That's theta. But if we had x's and y's as just a function of theta, and we had a fixed radius, that would essentially just give us the points on the outside of the unit circle. But we need to be able to have all of the xy's that are outside and inside the unit circle. So we actually have to have two parameters.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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That's theta. But if we had x's and y's as just a function of theta, and we had a fixed radius, that would essentially just give us the points on the outside of the unit circle. But we need to be able to have all of the xy's that are outside and inside the unit circle. So we actually have to have two parameters. We need to not only vary this angle, but we also need to vary the radius. So we would want to trace out the outside of that unit circle, and maybe we'd want to shorten it a little bit, and then trace it out again, and then shorten it some more, and then trace it out again. And so you want to actually have a variable radius as well.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So we actually have to have two parameters. We need to not only vary this angle, but we also need to vary the radius. So we would want to trace out the outside of that unit circle, and maybe we'd want to shorten it a little bit, and then trace it out again, and then shorten it some more, and then trace it out again. And so you want to actually have a variable radius as well. And so you could have how far out you're going. You could call that r. So for example, if r is fixed, and you change your ranges of theta, then you would essentially get all of those points right over there. And you would do that for all of the r's, and from r0 all the way to r1, and you would essentially fill up the entire unit circle.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And so you want to actually have a variable radius as well. And so you could have how far out you're going. You could call that r. So for example, if r is fixed, and you change your ranges of theta, then you would essentially get all of those points right over there. And you would do that for all of the r's, and from r0 all the way to r1, and you would essentially fill up the entire unit circle. And so let's do that. So r is going to go between 0 and 1. r is going to be between 0 and 1. And our theta is going to go all the way around.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And you would do that for all of the r's, and from r0 all the way to r1, and you would essentially fill up the entire unit circle. And so let's do that. So r is going to go between 0 and 1. r is going to be between 0 and 1. And our theta is going to go all the way around. So our theta is going to go between 0 and 2 pi. This is, let me write this down. I wrote 0 instead of theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And our theta is going to go all the way around. So our theta is going to go between 0 and 2 pi. This is, let me write this down. I wrote 0 instead of theta. Our theta is going to be greater than or equal to 0, less than or equal to 2 pi. And now we're ready to parameterize it. x of r and theta is going to be equal to, so whatever r is, it's going to be r cosine theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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I wrote 0 instead of theta. Our theta is going to be greater than or equal to 0, less than or equal to 2 pi. And now we're ready to parameterize it. x of r and theta is going to be equal to, so whatever r is, it's going to be r cosine theta. So x is going to be r cosine theta. y is going to be r sine theta. That's going to be the y value, r sine theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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x of r and theta is going to be equal to, so whatever r is, it's going to be r cosine theta. So x is going to be r cosine theta. y is going to be r sine theta. That's going to be the y value, r sine theta. And now z is essentially just a function of x. z is going to be equal to 1 minus x, but x is just r cosine theta. So there you have it. We have our parameterization of this surface right over here.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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That's going to be the y value, r sine theta. And now z is essentially just a function of x. z is going to be equal to 1 minus x, but x is just r cosine theta. So there you have it. We have our parameterization of this surface right over here. The x's and y's can take all the values of the unit circle, but then the z is up here based on a function of, well, based really on a function of x. It's 1 minus x. And so that will give us all of the possible points right over here on the surface.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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We have our parameterization of this surface right over here. The x's and y's can take all the values of the unit circle, but then the z is up here based on a function of, well, based really on a function of x. It's 1 minus x. And so that will give us all of the possible points right over here on the surface. You pick an x and y, and then the z is going to pop us right here someplace on that surface. And we can write it as a position vector function. Instead of calling that position vector function r, since we've already used r for radius, I will call it, I don't know, let's call it, I'm just going to pick a random letter here.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And so that will give us all of the possible points right over here on the surface. You pick an x and y, and then the z is going to pop us right here someplace on that surface. And we can write it as a position vector function. Instead of calling that position vector function r, since we've already used r for radius, I will call it, I don't know, let's call it, I'm just going to pick a random letter here. Let's call it p for position vector function. And so p, our surface p, we can write it as, actually, I just call it surface 3. So surface 3, I'll do it in that same purple color too, so we know we're talking about this.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Instead of calling that position vector function r, since we've already used r for radius, I will call it, I don't know, let's call it, I'm just going to pick a random letter here. Let's call it p for position vector function. And so p, our surface p, we can write it as, actually, I just call it surface 3. So surface 3, I'll do it in that same purple color too, so we know we're talking about this. Surface 3 as a position vector function, as a function of theta and r. Maybe I'll write r and theta, because that's how I think of things. r and theta is going to be equal to r cosine theta i plus r sine of theta j plus 1 minus r cosine theta. Need to get some real estate here.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So surface 3, I'll do it in that same purple color too, so we know we're talking about this. Surface 3 as a position vector function, as a function of theta and r. Maybe I'll write r and theta, because that's how I think of things. r and theta is going to be equal to r cosine theta i plus r sine of theta j plus 1 minus r cosine theta. Need to get some real estate here. 1 minus r cosine of theta k. And now we are ready to start doing all of the business of evaluating the actual surface integral. So the first thing we do is take the cross product of this, the partial of this with respect to r, and the partial of this with respect to theta. And so let's just get down to business.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Need to get some real estate here. 1 minus r cosine of theta k. And now we are ready to start doing all of the business of evaluating the actual surface integral. So the first thing we do is take the cross product of this, the partial of this with respect to r, and the partial of this with respect to theta. And so let's just get down to business. Let's take the cross product. And so we have our i unit vector, we have our j unit vector, and we have our k unit vector. And this might get a little bit involved, but we'll try our best to just work through it, give myself a little bit more space.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And so let's just get down to business. Let's take the cross product. And so we have our i unit vector, we have our j unit vector, and we have our k unit vector. And this might get a little bit involved, but we'll try our best to just work through it, give myself a little bit more space. And so the partial of this with respect to r. So let's take the partial of this with respect to r. I'll do it in blue. The partial of this with respect to r is just the cosine theta i so this is just cosine theta. I said I was going to do it in blue, and that's not blue.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And this might get a little bit involved, but we'll try our best to just work through it, give myself a little bit more space. And so the partial of this with respect to r. So let's take the partial of this with respect to r. I'll do it in blue. The partial of this with respect to r is just the cosine theta i so this is just cosine theta. I said I was going to do it in blue, and that's not blue. So this is going to be cosine theta. The partial of this with respect to r is sine theta and the partial of this This term right over here with respect to r is negative cosine theta. Now let's take the partial with respect to theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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I said I was going to do it in blue, and that's not blue. So this is going to be cosine theta. The partial of this with respect to r is sine theta and the partial of this This term right over here with respect to r is negative cosine theta. Now let's take the partial with respect to theta. The partial of this with respect to theta is negative r sine of theta. The partial of this with respect to theta is r cosine theta. And the partial here, this is 0, and then this would be negative r sine theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Now let's take the partial with respect to theta. The partial of this with respect to theta is negative r sine of theta. The partial of this with respect to theta is r cosine theta. And the partial here, this is 0, and then this would be negative r sine theta. Oh, no, let me be careful. This is going to be, you have a negative r. So the derivative of cosine theta with respect to theta is negative sine theta. So the negatives are going to cancel out, and so you're going to have r sine theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And the partial here, this is 0, and then this would be negative r sine theta. Oh, no, let me be careful. This is going to be, you have a negative r. So the derivative of cosine theta with respect to theta is negative sine theta. So the negatives are going to cancel out, and so you're going to have r sine theta. And now we can actually evaluate this determinant to figure out the cross product of the partial of this with respect to r and the partial of this with respect to theta. I'm not writing it down just to kind of save some real estate here. And so we have, actually, maybe I will write it down just to be clear what we're doing.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So the negatives are going to cancel out, and so you're going to have r sine theta. And now we can actually evaluate this determinant to figure out the cross product of the partial of this with respect to r and the partial of this with respect to theta. I'm not writing it down just to kind of save some real estate here. And so we have, actually, maybe I will write it down just to be clear what we're doing. The partial of S3 with respect to r crossed with the partial of S3 with respect to theta is equal to, now our i component is going to be sine theta times r sine theta. So that's going to be r sine squared theta minus r cosine theta times negative cosine theta. So that's plus r cosine squared theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And so we have, actually, maybe I will write it down just to be clear what we're doing. The partial of S3 with respect to r crossed with the partial of S3 with respect to theta is equal to, now our i component is going to be sine theta times r sine theta. So that's going to be r sine squared theta minus r cosine theta times negative cosine theta. So that's plus r cosine squared theta. All of that times i. And a simplification might be popping out here at you. And then you have minus the j component.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So that's plus r cosine squared theta. All of that times i. And a simplification might be popping out here at you. And then you have minus the j component. The j component is going to be cosine theta times r sine theta. So it's r cosine theta sine theta. And then we're going to subtract from that.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And then you have minus the j component. The j component is going to be cosine theta times r sine theta. So it's r cosine theta sine theta. And then we're going to subtract from that. See, the negative sines cancel out. So you're going to subtract r sine theta cosine theta, or r cosine theta sine theta. Well, this is interesting because these are the negatives of each other.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And then we're going to subtract from that. See, the negative sines cancel out. So you're going to subtract r sine theta cosine theta, or r cosine theta sine theta. Well, this is interesting because these are the negatives of each other. r cosine theta sine theta minus r cosine theta sine theta. This just evaluates to 0. So we have no j component.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Well, this is interesting because these are the negatives of each other. r cosine theta sine theta minus r cosine theta sine theta. This just evaluates to 0. So we have no j component. And then finally, for our k component, we have cosine theta times r cosine theta. So we have r cosine squared theta minus r sine theta times sine theta. Or minus negative r sine theta times sine theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So we have no j component. And then finally, for our k component, we have cosine theta times r cosine theta. So we have r cosine squared theta minus r sine theta times sine theta. Or minus negative r sine theta times sine theta. So this would give you a negative, but we're going to have to subtract it, so it gives you a positive. So plus r sine squared theta k. And so this simplifies quite nicely because this is going to be equal to this term up here. You can factor out an r. This is r times sine squared theta plus cosine squared theta, which is just that part simplifies to 1.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Or minus negative r sine theta times sine theta. So this would give you a negative, but we're going to have to subtract it, so it gives you a positive. So plus r sine squared theta k. And so this simplifies quite nicely because this is going to be equal to this term up here. You can factor out an r. This is r times sine squared theta plus cosine squared theta, which is just that part simplifies to 1. So that's just r times i. So this is equal to r times i. And we do the same thing over here.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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You can factor out an r. This is r times sine squared theta plus cosine squared theta, which is just that part simplifies to 1. So that's just r times i. So this is equal to r times i. And we do the same thing over here. This also simplifies. This is actually the same thing. This also simplifies to r. So this whole thing simplifies.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And we do the same thing over here. This also simplifies. This is actually the same thing. This also simplifies to r. So this whole thing simplifies. Let me write it this way. This is also r sine squared theta plus cosine squared theta also simplifies to r. So you have r times k. And so if you want the magnitude of this business, so let me make it clear. So the magnitude, we'll go back to the magenta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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This also simplifies to r. So this whole thing simplifies. Let me write it this way. This is also r sine squared theta plus cosine squared theta also simplifies to r. So you have r times k. And so if you want the magnitude of this business, so let me make it clear. So the magnitude, we'll go back to the magenta. The magnitude of, I don't feel like rewriting it all. I'll just copy and paste it. Edit, copy, and paste.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So the magnitude, we'll go back to the magenta. The magnitude of, I don't feel like rewriting it all. I'll just copy and paste it. Edit, copy, and paste. The magnitude of all of this business is going to be equal to the square root of this squared, which is r squared, plus this squared, which is r squared, which simplifies to this is 2r squared. So you take the square root of both of those. You get the square root of 2 times r. So this is equal to the square root of 2 times, it would be the absolute value of r, but we know that r only takes on positive values.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Edit, copy, and paste. The magnitude of all of this business is going to be equal to the square root of this squared, which is r squared, plus this squared, which is r squared, which simplifies to this is 2r squared. So you take the square root of both of those. You get the square root of 2 times r. So this is equal to the square root of 2 times, it would be the absolute value of r, but we know that r only takes on positive values. r only takes on positive values. So it's the square root of 2 times r, which is very nice because now we can evaluate ds. ds is going to be this business times dr d theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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You get the square root of 2 times r. So this is equal to the square root of 2 times, it would be the absolute value of r, but we know that r only takes on positive values. r only takes on positive values. So it's the square root of 2 times r, which is very nice because now we can evaluate ds. ds is going to be this business times dr d theta. So let's do it. So our surface integral, the thing that we were dealing with from the beginning, that thing right over there. So our surface integral, the surface S3 of z ds is now equal to, so I'm going to use different colors for the different variables of integration.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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ds is going to be this business times dr d theta. So let's do it. So our surface integral, the thing that we were dealing with from the beginning, that thing right over there. So our surface integral, the surface S3 of z ds is now equal to, so I'm going to use different colors for the different variables of integration. So one on the outside, and then I'll do one on the inside. I'll do the inside one in pink. z is equal to 1 minus r cosine theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So our surface integral, the surface S3 of z ds is now equal to, so I'm going to use different colors for the different variables of integration. So one on the outside, and then I'll do one on the inside. I'll do the inside one in pink. z is equal to 1 minus r cosine theta. So z is equal to 1 minus r cosine theta. It involves both, so I'll use a different color. Minus r cosine theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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z is equal to 1 minus r cosine theta. So z is equal to 1 minus r cosine theta. It involves both, so I'll use a different color. Minus r cosine theta. And then I just have to integrate relative to both of the variables. 1 minus r cosine theta. Oh, no, I have to do the times ds.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Minus r cosine theta. And then I just have to integrate relative to both of the variables. 1 minus r cosine theta. Oh, no, I have to do the times ds. ds is this thing. It's this thing times d theta d dr. So let's see.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Oh, no, I have to do the times ds. ds is this thing. It's this thing times d theta d dr. So let's see. Let's write this down. Times square root of 2r. So let's write that down.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So let's see. Let's write this down. Times square root of 2r. So let's write that down. So this times the square root of 2r. Times the square root of 2. And we can write the square root of 2 out front since it's a constant.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So let's write that down. So this times the square root of 2r. Times the square root of 2. And we can write the square root of 2 out front since it's a constant. So let me just do that. Simplify this thing. Square root of 2 times r. And now d theta dr. Or we could write dr d theta either way.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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And we can write the square root of 2 out front since it's a constant. So let me just do that. Simplify this thing. Square root of 2 times r. And now d theta dr. Or we could write dr d theta either way. So let's do that. Let's do dr d theta. So dr d theta.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Square root of 2 times r. And now d theta dr. Or we could write dr d theta either way. So let's do that. Let's do dr d theta. So dr d theta. We could do it either way. It's going to be about the same level of complexity. And so first we're going to integrate with respect to r. And let me do the colors the same way, actually.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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So dr d theta. We could do it either way. It's going to be about the same level of complexity. And so first we're going to integrate with respect to r. And let me do the colors the same way, actually. So dr d theta. If I've got colors, I might as well use them. And actually, I just realized that I'm way out of time.
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Surface integral ex3 part 3 Top surface Multivariable Calculus Khan Academy.mp3
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Hello everyone. So I have here the graph of a two variable function and I'd like to talk about how you can interpret the partial derivative of that function. So specifically, the function that you're looking at is f of x, y is equal to x squared times y plus sine of y. And the question is, if I take the partial derivative of this function, so maybe I'm looking at the partial derivative of f with respect to x, and let's say I want to do this at negative one, one. So I'll be looking at the partial derivative at a specific point. How do you interpret that on this whole graph? So first, let's consider where the point negative one, one is if we're looking above, this is our x-axis, this is our y-axis, the point negative one, one is sitting right there.
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Partial derivatives and graphs.mp3
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And the question is, if I take the partial derivative of this function, so maybe I'm looking at the partial derivative of f with respect to x, and let's say I want to do this at negative one, one. So I'll be looking at the partial derivative at a specific point. How do you interpret that on this whole graph? So first, let's consider where the point negative one, one is if we're looking above, this is our x-axis, this is our y-axis, the point negative one, one is sitting right there. So negative one, move up one, and it's the point that's sitting on the graph. And the first thing you might do is you say, well, when we're taking the partial derivative with respect to x, we're gonna pretend that y is a constant. So let's actually just go ahead and evaluate that.
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Partial derivatives and graphs.mp3
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So first, let's consider where the point negative one, one is if we're looking above, this is our x-axis, this is our y-axis, the point negative one, one is sitting right there. So negative one, move up one, and it's the point that's sitting on the graph. And the first thing you might do is you say, well, when we're taking the partial derivative with respect to x, we're gonna pretend that y is a constant. So let's actually just go ahead and evaluate that. So when you're doing this, it looks, it says x squared looks like a variable, y looks like a constant, sine of y also looks like a constant. So this is gonna be, we differentiate x squared, and that's two times x times y, which is like a constant, and then the derivative of a constant there is zero. And we're evaluating this whole thing at x is equal to negative one, and y is equal to one.
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Partial derivatives and graphs.mp3
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So let's actually just go ahead and evaluate that. So when you're doing this, it looks, it says x squared looks like a variable, y looks like a constant, sine of y also looks like a constant. So this is gonna be, we differentiate x squared, and that's two times x times y, which is like a constant, and then the derivative of a constant there is zero. And we're evaluating this whole thing at x is equal to negative one, and y is equal to one. So when we actually plug that in, it'll be two times negative one, multiplied by one, which is two, negative two, excuse me. But what does that mean, right? We evaluate this, and maybe you're thinking this is kind of a slight nudge in the x direction, this is the resulting nudge of f. What does that mean for the graph?
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Partial derivatives and graphs.mp3
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And we're evaluating this whole thing at x is equal to negative one, and y is equal to one. So when we actually plug that in, it'll be two times negative one, multiplied by one, which is two, negative two, excuse me. But what does that mean, right? We evaluate this, and maybe you're thinking this is kind of a slight nudge in the x direction, this is the resulting nudge of f. What does that mean for the graph? Well, first of all, treating y as a constant is basically like slicing the whole graph with a plane that represents a constant y value. So this is the y-axis, and the plane that cuts it perpendicularly, that represents a constant y value. This one represents the constant y value one, but you could imagine it, you could imagine sliding the plane back and forth, and that would represent various different y values.
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Partial derivatives and graphs.mp3
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We evaluate this, and maybe you're thinking this is kind of a slight nudge in the x direction, this is the resulting nudge of f. What does that mean for the graph? Well, first of all, treating y as a constant is basically like slicing the whole graph with a plane that represents a constant y value. So this is the y-axis, and the plane that cuts it perpendicularly, that represents a constant y value. This one represents the constant y value one, but you could imagine it, you could imagine sliding the plane back and forth, and that would represent various different y values. So for the general partial derivative, you can imagine whichever one you want, but this one is y equals one. And I'll go ahead and slice the actual graph at that point, and draw a red line. And this red line is basically all the points on the graph where y is equal to one.
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Partial derivatives and graphs.mp3
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