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So this thing that we were viewing sideways, now we're viewing it from the top. And so now our unit sphere is going to look something like this, viewed from above. And this, what I just drew, this dotted circle right over here, this is going to be where our unit sphere intersects what I labeled that y. That should be x. I don't want to confuse you already. Let me clear that. So this is our x-axis. This is our x-axis.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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That should be x. I don't want to confuse you already. Let me clear that. So this is our x-axis. This is our x-axis. So this little dotted blue circle, this is where our unit sphere intersects the xy-plane. And so using this, we can start to think about how to parameterize at least our x and y values, our x and y coordinates, as a function of a first parameter. So the first parameter, we can think of something that is, so this is the z-axis popping straight out at us.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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This is our x-axis. So this little dotted blue circle, this is where our unit sphere intersects the xy-plane. And so using this, we can start to think about how to parameterize at least our x and y values, our x and y coordinates, as a function of a first parameter. So the first parameter, we can think of something that is, so this is the z-axis popping straight out at us. So essentially if we're rotating around that z-axis, viewed from above, we can imagine an angle, and I'll call that angle, I will call that angle s, which is essentially saying how much we're rotating from the x-axis towards the y-axis. You could think about it in the xy-plane or in a plane that is parallel to the xy-plane, or you could say going around the z-axis, the z-axis popping straight up at us. And the radius here is always 1.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So the first parameter, we can think of something that is, so this is the z-axis popping straight out at us. So essentially if we're rotating around that z-axis, viewed from above, we can imagine an angle, and I'll call that angle, I will call that angle s, which is essentially saying how much we're rotating from the x-axis towards the y-axis. You could think about it in the xy-plane or in a plane that is parallel to the xy-plane, or you could say going around the z-axis, the z-axis popping straight up at us. And the radius here is always 1. It's a unit sphere. So given this parameter s, what would be your x and y coordinates? And now we're thinking about it right if we're sitting in the xy-plane.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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And the radius here is always 1. It's a unit sphere. So given this parameter s, what would be your x and y coordinates? And now we're thinking about it right if we're sitting in the xy-plane. Well, the x coordinate, this goes back to the unit circle definition of our trig functions, the x coordinate is going to be cosine of s. It would be the radius, which is 1, times the cosine of s. And the y coordinate would be 1 times the sine of s. That's actually where we get our definitions for cosine and sine from. So that's pretty straightforward. And in this case, z is obviously equal to 0.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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And now we're thinking about it right if we're sitting in the xy-plane. Well, the x coordinate, this goes back to the unit circle definition of our trig functions, the x coordinate is going to be cosine of s. It would be the radius, which is 1, times the cosine of s. And the y coordinate would be 1 times the sine of s. That's actually where we get our definitions for cosine and sine from. So that's pretty straightforward. And in this case, z is obviously equal to 0. So if we wanted to add our z coordinate here, z is 0. We are sitting in the xy-plane. But now let's think about what happens if we go above and below the xy-plane.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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And in this case, z is obviously equal to 0. So if we wanted to add our z coordinate here, z is 0. We are sitting in the xy-plane. But now let's think about what happens if we go above and below the xy-plane. Remember, this is in any plane that is parallel to the xy-plane. This is saying how we are rotated around the z-axis. Now let's think about if we go above and below it.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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But now let's think about what happens if we go above and below the xy-plane. Remember, this is in any plane that is parallel to the xy-plane. This is saying how we are rotated around the z-axis. Now let's think about if we go above and below it. And to figure out how far above or below it, I'm going to introduce another parameter. And this new parameter I'm going to introduce is t. t is how much we have rotated above and below the xy-plane. Now, what's interesting about that is if we take any other cross-section that is parallel to the xy-plane now, we are going to have a smaller radius.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Now let's think about if we go above and below it. And to figure out how far above or below it, I'm going to introduce another parameter. And this new parameter I'm going to introduce is t. t is how much we have rotated above and below the xy-plane. Now, what's interesting about that is if we take any other cross-section that is parallel to the xy-plane now, we are going to have a smaller radius. Let me make that clear. So if we're right over there, now where this plane intersects our unit sphere, the radius is smaller. The radius is smaller than it was before.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Now, what's interesting about that is if we take any other cross-section that is parallel to the xy-plane now, we are going to have a smaller radius. Let me make that clear. So if we're right over there, now where this plane intersects our unit sphere, the radius is smaller. The radius is smaller than it was before. Well, what would be this new radius? Well, a little bit of trigonometry. It's the same as this length right over here, which is going to be cosine of t. So the radius is going to be cosine of t. And it still works over here because if t goes all the way to 0, cosine of 0 is 1, and then that works right over there when we're in the xy-plane.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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The radius is smaller than it was before. Well, what would be this new radius? Well, a little bit of trigonometry. It's the same as this length right over here, which is going to be cosine of t. So the radius is going to be cosine of t. And it still works over here because if t goes all the way to 0, cosine of 0 is 1, and then that works right over there when we're in the xy-plane. So the radius over here is going to be cosine of 0. So this is when t is equal to 0, and we haven't rotated above or below the xy-plane. But if we have rotated above the xy-plane, the radius has changed.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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It's the same as this length right over here, which is going to be cosine of t. So the radius is going to be cosine of t. And it still works over here because if t goes all the way to 0, cosine of 0 is 1, and then that works right over there when we're in the xy-plane. So the radius over here is going to be cosine of 0. So this is when t is equal to 0, and we haven't rotated above or below the xy-plane. But if we have rotated above the xy-plane, the radius has changed. It is now cosine of t. And now we can use that to truly parameterize x and y anywhere. So now let's look at this cross-section. So we're not necessarily in the xy-plane, we're in something that's parallel to the xy-plane.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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But if we have rotated above the xy-plane, the radius has changed. It is now cosine of t. And now we can use that to truly parameterize x and y anywhere. So now let's look at this cross-section. So we're not necessarily in the xy-plane, we're in something that's parallel to the xy-plane. And so if we're up here, now all of a sudden the cross-section, if we view it from above, might look something like this. We're viewing it from above, this cross-section right over here. Our radius right over here is cosine of t. And so given that, I guess, altitude that we're at, what would now be the parameterization using s of x and y?
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So we're not necessarily in the xy-plane, we're in something that's parallel to the xy-plane. And so if we're up here, now all of a sudden the cross-section, if we view it from above, might look something like this. We're viewing it from above, this cross-section right over here. Our radius right over here is cosine of t. And so given that, I guess, altitude that we're at, what would now be the parameterization using s of x and y? Well, it's the exact same thing, except now our radius isn't a fixed one. It is now a function of t. So we're now a little bit higher. So now our x-coordinate is going to be our radius, which is cosine of t. That's just our radius, times cosine of s. How much we've angled around.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Our radius right over here is cosine of t. And so given that, I guess, altitude that we're at, what would now be the parameterization using s of x and y? Well, it's the exact same thing, except now our radius isn't a fixed one. It is now a function of t. So we're now a little bit higher. So now our x-coordinate is going to be our radius, which is cosine of t. That's just our radius, times cosine of s. How much we've angled around. And in this case, s has gone all the way around here. So it's going to be cosine of t times cosine of s. And then our y-coordinate is going to be our radius, which is cosine of t, times sine of s. Same exact logic here, except now we have a different radius. Our radius is no longer 1, times sine of s. Running out of space, let me scroll to the right a little bit.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So now our x-coordinate is going to be our radius, which is cosine of t. That's just our radius, times cosine of s. How much we've angled around. And in this case, s has gone all the way around here. So it's going to be cosine of t times cosine of s. And then our y-coordinate is going to be our radius, which is cosine of t, times sine of s. Same exact logic here, except now we have a different radius. Our radius is no longer 1, times sine of s. Running out of space, let me scroll to the right a little bit. And this looks very confusing, but you just have to say, at any given level we are, we're parallel to the x-axis. We're kind of tracing out another circle where another plane intersects our unit sphere. We're now then rotating around with s. And so our radius will change.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Our radius is no longer 1, times sine of s. Running out of space, let me scroll to the right a little bit. And this looks very confusing, but you just have to say, at any given level we are, we're parallel to the x-axis. We're kind of tracing out another circle where another plane intersects our unit sphere. We're now then rotating around with s. And so our radius will change. It's a function of how much above or below the xy-plane we've rotated. So this is just our radius instead of 1. And then s is how much we've rotated around the z-axis.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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We're now then rotating around with s. And so our radius will change. It's a function of how much above or below the xy-plane we've rotated. So this is just our radius instead of 1. And then s is how much we've rotated around the z-axis. Same there for the y-coordinate. And then the z-coordinate is pretty straightforward. It's going to be completely a function of t. It's not dependent on how much we've rotated around here at any given altitude.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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And then s is how much we've rotated around the z-axis. Same there for the y-coordinate. And then the z-coordinate is pretty straightforward. It's going to be completely a function of t. It's not dependent on how much we've rotated around here at any given altitude. It is what our altitude actually is. And we can go straight to this diagram right over here. Our z-coordinate is just going to be the sine of t. So our z is equal to sine of t. So let me write that down.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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It's going to be completely a function of t. It's not dependent on how much we've rotated around here at any given altitude. It is what our altitude actually is. And we can go straight to this diagram right over here. Our z-coordinate is just going to be the sine of t. So our z is equal to sine of t. So let me write that down. So z is going to be equal to sine of t. So now every point on this sphere can be described as a function of t and s. And we have to think about over what range will they be defined. Well, s is going to go at any given level, you could think. For any given t, s is going to go all the way around.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Our z-coordinate is just going to be the sine of t. So our z is equal to sine of t. So let me write that down. So z is going to be equal to sine of t. So now every point on this sphere can be described as a function of t and s. And we have to think about over what range will they be defined. Well, s is going to go at any given level, you could think. For any given t, s is going to go all the way around. We see that right over here. At any given level viewed from above, s is going to go all the way around. So thinking about it in radians, s is going to be between 0 and 2 pi.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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For any given t, s is going to go all the way around. We see that right over here. At any given level viewed from above, s is going to go all the way around. So thinking about it in radians, s is going to be between 0 and 2 pi. And t is essentially our altitude in the z direction. So t can go all the way down here, which would be negative pi over 2. So t can be between negative pi over 2.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So thinking about it in radians, s is going to be between 0 and 2 pi. And t is essentially our altitude in the z direction. So t can go all the way down here, which would be negative pi over 2. So t can be between negative pi over 2. And it can go all the way up to pi over 2. It doesn't need to go all the way back down again. And so it goes all the way back.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So t can be between negative pi over 2. And it can go all the way up to pi over 2. It doesn't need to go all the way back down again. And so it goes all the way back. It always goes only up to pi over 2. And then we have our parameterization. Let me write this down in a form that you might recognize even more.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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And so it goes all the way back. It always goes only up to pi over 2. And then we have our parameterization. Let me write this down in a form that you might recognize even more. If we wanted to write our surface as a position vector function, we could write it like this. We could write it r is a function of s and t. And it is equal to our x component, our i component, is going to be cosine of t, cosine of s, i, and then plus our y component is cosine of t, sine of s, plus our z component, which is the sine, which is just, oh, I forgot our j vector. j plus the z component, which is just sine of t, sine of t, k. And we're done.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Let me write this down in a form that you might recognize even more. If we wanted to write our surface as a position vector function, we could write it like this. We could write it r is a function of s and t. And it is equal to our x component, our i component, is going to be cosine of t, cosine of s, i, and then plus our y component is cosine of t, sine of s, plus our z component, which is the sine, which is just, oh, I forgot our j vector. j plus the z component, which is just sine of t, sine of t, k. And we're done. And these are the ranges that those parameters will take on. So that's just the first step. We've parameterized this surface.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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j plus the z component, which is just sine of t, sine of t, k. And we're done. And these are the ranges that those parameters will take on. So that's just the first step. We've parameterized this surface. Now we're going to have to actually set up the surface integral. It's going to involve a little bit of taking a cross product, which can get hairy. And then we can actually evaluate the integral itself.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So F dot dr. Well, we remember what F is. F is all the way up here. It's i, j, and k components. We're just p, q, and r, so that's kind of easy to remember. But let's think about what dr is equal to. Let's think about what dr is. And we're going to have to break out a little bit of our three-dimensional, our multivariable chain rule right over here.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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We're just p, q, and r, so that's kind of easy to remember. But let's think about what dr is equal to. Let's think about what dr is. And we're going to have to break out a little bit of our three-dimensional, our multivariable chain rule right over here. So dr is the same thing as dr dt times dt. So we really just have to figure out what the derivative of r with respect to t, and this is where we're going to have to break out a little bit of the chain rule. So let me write this down.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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And we're going to have to break out a little bit of our three-dimensional, our multivariable chain rule right over here. So dr is the same thing as dr dt times dt. So we really just have to figure out what the derivative of r with respect to t, and this is where we're going to have to break out a little bit of the chain rule. So let me write this down. So dr dt is going to be equal to the derivative of x with respect to t times i, plus the derivative of y with respect to t. I'm just taking the derivative with respect to t, j. And now we're going to have to break the, because z is a function of x, which is a function of t, and z is also a function of y, which is a function of t, we're going to have to break out our multivariable chain rule. So if we want to take the derivative of z with respect to t, I'll do it separately here, and then I'll write it down down here.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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So let me write this down. So dr dt is going to be equal to the derivative of x with respect to t times i, plus the derivative of y with respect to t. I'm just taking the derivative with respect to t, j. And now we're going to have to break the, because z is a function of x, which is a function of t, and z is also a function of y, which is a function of t, we're going to have to break out our multivariable chain rule. So if we want to take the derivative of z with respect to t, I'll do it separately here, and then I'll write it down down here. The derivative of z with respect to t, you really, the way I conceptualize this, what's all the different ways that z can change from a change in t? Well, it could change because x is changing due to a change in t, so we could have, so z could change due to x, the partial of z with respect to x, when x changes with respect to t. But then that's not the only way that z can change. We have to add to that how z can change with respect to y.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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So if we want to take the derivative of z with respect to t, I'll do it separately here, and then I'll write it down down here. The derivative of z with respect to t, you really, the way I conceptualize this, what's all the different ways that z can change from a change in t? Well, it could change because x is changing due to a change in t, so we could have, so z could change due to x, the partial of z with respect to x, when x changes with respect to t. But then that's not the only way that z can change. We have to add to that how z can change with respect to y. So partial of z with respect to y times how fast or how y is changing with respect to t. And this is just our multivariable chain rule. And so this is dz dt, so I'll just write it right over here. And I'll use slightly different notation that's consistent with what we were doing before, and it'll help make things a little bit clearer.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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We have to add to that how z can change with respect to y. So partial of z with respect to y times how fast or how y is changing with respect to t. And this is just our multivariable chain rule. And so this is dz dt, so I'll just write it right over here. And I'll use slightly different notation that's consistent with what we were doing before, and it'll help make things a little bit clearer. So it's going to be the partial of z with respect to x, dx dt, plus the, actually let me write it this way, plus the partial of z with respect to y, dy dt, and then we're going to multiply everything times our k. We are going to multiply everything times our k. So with this out of the way, so if we wanted what dr is, dr is just going to be this whole thing times dt. So let's do that. So now we can rewrite our line integral right over here.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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And I'll use slightly different notation that's consistent with what we were doing before, and it'll help make things a little bit clearer. So it's going to be the partial of z with respect to x, dx dt, plus the, actually let me write it this way, plus the partial of z with respect to y, dy dt, and then we're going to multiply everything times our k. We are going to multiply everything times our k. So with this out of the way, so if we wanted what dr is, dr is just going to be this whole thing times dt. So let's do that. So now we can rewrite our line integral right over here. And we're going to now go into the t domain. And so t is going to go between a and b, and f dot dr, remember f's components were just the functions p, q, and r. And each of those were functions of x, y, and z. And z is a function of x and y, so we'll have to think about all of that in a little bit.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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So now we can rewrite our line integral right over here. And we're going to now go into the t domain. And so t is going to go between a and b, and f dot dr, remember f's components were just the functions p, q, and r. And each of those were functions of x, y, and z. And z is a function of x and y, so we'll have to think about all of that in a little bit. Use a little more of our multivariable chain rule. But when we take the dot products, we're just going to take the corresponding components and multiply them. So it is going to be, actually let me just copy and paste.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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And z is a function of x and y, so we'll have to think about all of that in a little bit. Use a little more of our multivariable chain rule. But when we take the dot products, we're just going to take the corresponding components and multiply them. So it is going to be, actually let me just copy and paste. So f is, let me rewrite it down here. Our vector field f, and I'm going to write it a little bit shorter, our vector field f is p times i plus q times j plus r times k. So when we take our dot product of f dot dr, we're essentially taking the dot product of this and this, and we have to throw a dt at the end. So we're going to get p times dx dt plus q times dy dt plus r times all of this business over here, which is the partial of z with respect to x dx dt plus the partial of z with respect to y dy dt.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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So it is going to be, actually let me just copy and paste. So f is, let me rewrite it down here. Our vector field f, and I'm going to write it a little bit shorter, our vector field f is p times i plus q times j plus r times k. So when we take our dot product of f dot dr, we're essentially taking the dot product of this and this, and we have to throw a dt at the end. So we're going to get p times dx dt plus q times dy dt plus r times all of this business over here, which is the partial of z with respect to x dx dt plus the partial of z with respect to y dy dt. And then we have to multiply all of this times dt. We can't forget that part right over there. So we're going to multiply all of this right here times dt.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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So we're going to get p times dx dt plus q times dy dt plus r times all of this business over here, which is the partial of z with respect to x dx dt plus the partial of z with respect to y dy dt. And then we have to multiply all of this times dt. We can't forget that part right over there. So we're going to multiply all of this right here times dt. Now I'm going to leave you there in this video just because I'm afraid of making careless mistakes. What we're going to do now is we're going to rearrange this whole thing. Recognize that this is the same thing as this right over here.
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Stokes' theorem proof part 5 Multivariable Calculus Khan Academy.mp3
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Let's say I have a path in the xy-plane that's essentially the unit circle. This is essentially the unit circle, so it's my y-axis. This is my x-axis. And our path is going to be the unit circle. It's going to be the unit circle, and we're going to traverse it just like that. We're going to traverse it clockwise. I think you get the idea.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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And our path is going to be the unit circle. It's going to be the unit circle, and we're going to traverse it just like that. We're going to traverse it clockwise. I think you get the idea. And so you can see its equation is the unit circle. So the equation of this is x squared plus y squared is equal to 1, has a radius of 1, unit circle. And what we're concerned with is the line integral.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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I think you get the idea. And so you can see its equation is the unit circle. So the equation of this is x squared plus y squared is equal to 1, has a radius of 1, unit circle. And what we're concerned with is the line integral. The line integral over this curve C. It's a closed curve C. It's actually going in that direction. Of 2y dx minus 3x dy. So we are probably tempted to use Green's Theorem and why not?
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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And what we're concerned with is the line integral. The line integral over this curve C. It's a closed curve C. It's actually going in that direction. Of 2y dx minus 3x dy. So we are probably tempted to use Green's Theorem and why not? So let's try. This is our path. So Green's Theorem tells us that the integral of some curve F dot dr, F dot dr over some path where F is equal to, where F is, let me write it a little bit neater, where F of xy is equal to P of xyi plus Q of xyj, that this integral is equal to the double integral over the region, this would be the region under question in this example, over the region of the partial of Q with respect to x, minus the partial of P with respect to y, all of that da, the differential of area.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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So we are probably tempted to use Green's Theorem and why not? So let's try. This is our path. So Green's Theorem tells us that the integral of some curve F dot dr, F dot dr over some path where F is equal to, where F is, let me write it a little bit neater, where F of xy is equal to P of xyi plus Q of xyj, that this integral is equal to the double integral over the region, this would be the region under question in this example, over the region of the partial of Q with respect to x, minus the partial of P with respect to y, all of that da, the differential of area. And of course the region is that, what I just showed you. Now, you may or may not remember, I made, well, there's a slight subtle thing in this which would give you the wrong answer. In the last video we said that Green's Theorem applies when we're going counterclockwise.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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So Green's Theorem tells us that the integral of some curve F dot dr, F dot dr over some path where F is equal to, where F is, let me write it a little bit neater, where F of xy is equal to P of xyi plus Q of xyj, that this integral is equal to the double integral over the region, this would be the region under question in this example, over the region of the partial of Q with respect to x, minus the partial of P with respect to y, all of that da, the differential of area. And of course the region is that, what I just showed you. Now, you may or may not remember, I made, well, there's a slight subtle thing in this which would give you the wrong answer. In the last video we said that Green's Theorem applies when we're going counterclockwise. Notice even on this little thing on the integral, I made it go counterclockwise. In our example, the curve goes clockwise. The region is to our right.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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In the last video we said that Green's Theorem applies when we're going counterclockwise. Notice even on this little thing on the integral, I made it go counterclockwise. In our example, the curve goes clockwise. The region is to our right. Green's Theorem, this applies when the region is to our left. So in this situation, when the region is to our right and we're going clockwise, so this is counterclockwise. So in our example, where we're going clockwise, so this is to our right, Green's Theorem is going to be the negative of this.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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The region is to our right. Green's Theorem, this applies when the region is to our left. So in this situation, when the region is to our right and we're going clockwise, so this is counterclockwise. So in our example, where we're going clockwise, so this is to our right, Green's Theorem is going to be the negative of this. So in our example, we're going to have the integral of C and we're going to go in the clockwise direction. So maybe I'll draw it like that, of F dot dr. This is going to be equal to the double integral over the region.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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So in our example, where we're going clockwise, so this is to our right, Green's Theorem is going to be the negative of this. So in our example, we're going to have the integral of C and we're going to go in the clockwise direction. So maybe I'll draw it like that, of F dot dr. This is going to be equal to the double integral over the region. And we could just swap these two, the partial of P with respect to y, minus the partial of Q with respect to x, da. So let's do that. So this is going to be equal to, in this example, the integral over the region.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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This is going to be equal to the double integral over the region. And we could just swap these two, the partial of P with respect to y, minus the partial of Q with respect to x, da. So let's do that. So this is going to be equal to, in this example, the integral over the region. Let's just keep it abstract for now. We could start setting the boundaries, but let's just keep the region abstract. And what is the partial of P with respect to, let's just remember, this right here is our, I think we could recognize right now that this, if we take F dot dr, we're going to get this.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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So this is going to be equal to, in this example, the integral over the region. Let's just keep it abstract for now. We could start setting the boundaries, but let's just keep the region abstract. And what is the partial of P with respect to, let's just remember, this right here is our, I think we could recognize right now that this, if we take F dot dr, we're going to get this. The dr contributes those components. The F contributes these two components. So this is P of xy, that is P of xy, and then this is Q of xy.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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And what is the partial of P with respect to, let's just remember, this right here is our, I think we could recognize right now that this, if we take F dot dr, we're going to get this. The dr contributes those components. The F contributes these two components. So this is P of xy, that is P of xy, and then this is Q of xy. And we've seen it, I don't want to go into the whole dot dr and take the dot product over and over again. I think you can see that this is the dot product of two vectors. This is the x component of F, y component of F. This is the x component of dr, y component of dr.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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So this is P of xy, that is P of xy, and then this is Q of xy. And we've seen it, I don't want to go into the whole dot dr and take the dot product over and over again. I think you can see that this is the dot product of two vectors. This is the x component of F, y component of F. This is the x component of dr, y component of dr. So let's take the partial of P with respect to y. Take the derivative of this with respect to y, you get 2. The derivative of 2y is just 2.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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This is the x component of F, y component of F. This is the x component of dr, y component of dr. So let's take the partial of P with respect to y. Take the derivative of this with respect to y, you get 2. The derivative of 2y is just 2. So you get 2, and then minus the derivative of Q with respect to x. The derivative of this with respect to x is minus 3. So we're going to get minus 3, and then all of that dA.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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The derivative of 2y is just 2. So you get 2, and then minus the derivative of Q with respect to x. The derivative of this with respect to x is minus 3. So we're going to get minus 3, and then all of that dA. And this is equal to the integral over the region. What's this? This is 2 minus minus 3.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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So we're going to get minus 3, and then all of that dA. And this is equal to the integral over the region. What's this? This is 2 minus minus 3. That's the same thing as 2 plus 3. So it's the integral over the region of 5 dA. 5 is just a constant, so we can take it out of the integral.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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This is 2 minus minus 3. That's the same thing as 2 plus 3. So it's the integral over the region of 5 dA. 5 is just a constant, so we can take it out of the integral. So this is going to turn out to be quite a simple problem. So this is going to be equal to 5 times the double integral over the region r dA. Now what is this thing?
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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5 is just a constant, so we can take it out of the integral. So this is going to turn out to be quite a simple problem. So this is going to be equal to 5 times the double integral over the region r dA. Now what is this thing? What is this thing right here? It looks very abstract, but we can solve this. This is just the area of the region.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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Now what is this thing? What is this thing right here? It looks very abstract, but we can solve this. This is just the area of the region. That's what that double integral represents. You just sum up all the little dAs. That's a dA.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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This is just the area of the region. That's what that double integral represents. You just sum up all the little dAs. That's a dA. You sum up the infinite sums of those little dAs over the region. Well, what's the area of this unit circle? Here we just break out a little bit of 9th grade, actually even earlier than that, pre-algebra or middle school geometry.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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That's a dA. You sum up the infinite sums of those little dAs over the region. Well, what's the area of this unit circle? Here we just break out a little bit of 9th grade, actually even earlier than that, pre-algebra or middle school geometry. Area is equal to pi r squared. What's our radius? It's a unit circle.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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Here we just break out a little bit of 9th grade, actually even earlier than that, pre-algebra or middle school geometry. Area is equal to pi r squared. What's our radius? It's a unit circle. Our radius is 1. Length is 1. So the area here is pi.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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It's a unit circle. Our radius is 1. Length is 1. So the area here is pi. So this thing over here, that whole thing is just equal to pi. So the answer to our line integral is just 5 pi, which is pretty straightforward. We could have taken the trouble of setting up a double integral where we take the antiderivative with respect to y first and write y is equal to the negative square root of 1 minus x squared, y is equal to the positive square root, x goes from minus 1 to 1.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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So the area here is pi. So this thing over here, that whole thing is just equal to pi. So the answer to our line integral is just 5 pi, which is pretty straightforward. We could have taken the trouble of setting up a double integral where we take the antiderivative with respect to y first and write y is equal to the negative square root of 1 minus x squared, y is equal to the positive square root, x goes from minus 1 to 1. But that would have been super hairy and a huge pain. We just have to realize, no, this is just the area. The other interesting thing is I challenge you to solve the same integral without using Green's Theorem.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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We could have taken the trouble of setting up a double integral where we take the antiderivative with respect to y first and write y is equal to the negative square root of 1 minus x squared, y is equal to the positive square root, x goes from minus 1 to 1. But that would have been super hairy and a huge pain. We just have to realize, no, this is just the area. The other interesting thing is I challenge you to solve the same integral without using Green's Theorem. The way we actually take generating a parameterization for this curve, going in that direction, taking the derivatives of x of t and y of t, multiplying by the appropriate thing, and then taking the antiderivative, way hairier than what we just did using Green's Theorem to get 5 pi. And remember, the reason why it wasn't minus 5 pi here is because we're going in a clockwise direction. If we were going in a counterclockwise direction, we could have applied the straight-up Green's Theorem and we would have gotten minus 5 pi.
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Green's theorem example 2 Multivariable Calculus Khan Academy.mp3
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If I try to draw that, let's see if I can have a good attempt at it. That is my y axis, I'm going to do a little perspective here. This is my x axis, I could make it do the negative x and y axis, you could do it in that direction. This is my x axis here. And if I were to graph this, when y is 0, it's going to be just a, let me draw it in yellow, it's going to be just a straight line that looks something like that. And then for any given x, you're going to have a parabola in y. Y is going to look something like that. I'm just going to do it in the positive quadrant, it's going to look something like that.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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This is my x axis here. And if I were to graph this, when y is 0, it's going to be just a, let me draw it in yellow, it's going to be just a straight line that looks something like that. And then for any given x, you're going to have a parabola in y. Y is going to look something like that. I'm just going to do it in the positive quadrant, it's going to look something like that. When you go into the negative y, you're going to see the other half of the parabola, but I'm not going to worry about it too much. So you're going to have this surface that looks something like that. Maybe I'll do another attempt at drawing it.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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I'm just going to do it in the positive quadrant, it's going to look something like that. When you go into the negative y, you're going to see the other half of the parabola, but I'm not going to worry about it too much. So you're going to have this surface that looks something like that. Maybe I'll do another attempt at drawing it. But this is our ceiling we're going to deal with again. And then I'm going to have a path in the xy plane. I'm going to start at the point 2,0, x is equal to 2, y is 0.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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Maybe I'll do another attempt at drawing it. But this is our ceiling we're going to deal with again. And then I'm going to have a path in the xy plane. I'm going to start at the point 2,0, x is equal to 2, y is 0. And I'm going to travel, just like we did in the last video, I'm going to travel along a circle, but this time the circle is going to have a radius 2. I'm going to go counterclockwise in that circle. This is on the xy plane, just to be able to visualize it properly.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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I'm going to start at the point 2,0, x is equal to 2, y is 0. And I'm going to travel, just like we did in the last video, I'm going to travel along a circle, but this time the circle is going to have a radius 2. I'm going to go counterclockwise in that circle. This is on the xy plane, just to be able to visualize it properly. So this right here is a point 0,2. Then I'm going to come back along the y-axis. This is my path.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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This is on the xy plane, just to be able to visualize it properly. So this right here is a point 0,2. Then I'm going to come back along the y-axis. This is my path. I'm going to come back along the y-axis, just like that. And then I'm going to take a left here, and then I'm going to take another left here, and then come back along the x-axis. So what I drew in these two shades of green, that is my contour.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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This is my path. I'm going to come back along the y-axis, just like that. And then I'm going to take a left here, and then I'm going to take another left here, and then come back along the x-axis. So what I drew in these two shades of green, that is my contour. And what I want to do is I want to evaluate the surface area of essentially this little building that has the roof of f of x, y is equal to x plus y squared. And I want to find the surface area of its walls. So you'll have this wall right here, whose base is the x-axis.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So what I drew in these two shades of green, that is my contour. And what I want to do is I want to evaluate the surface area of essentially this little building that has the roof of f of x, y is equal to x plus y squared. And I want to find the surface area of its walls. So you'll have this wall right here, whose base is the x-axis. Then you're going to have this wall, which is along the curve. It's going to look something like a funky wall on that curved side right there. I'll try my best effort to try to... Actually, it's going to be curving way up like that.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So you'll have this wall right here, whose base is the x-axis. Then you're going to have this wall, which is along the curve. It's going to look something like a funky wall on that curved side right there. I'll try my best effort to try to... Actually, it's going to be curving way up like that. It's going to be curved up like that. And then along the y-axis, along the y-axis when x is equal to 0, it's going to have like a half of a parabolic wall right there. I'll do that back wall along the y-axis.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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I'll try my best effort to try to... Actually, it's going to be curving way up like that. It's going to be curved up like that. And then along the y-axis, along the y-axis when x is equal to 0, it's going to have like a half of a parabolic wall right there. I'll do that back wall along the y-axis. I'll do that in orange. Actually, I was already using orange. I'll use magenta.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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I'll do that back wall along the y-axis. I'll do that in orange. Actually, I was already using orange. I'll use magenta. That is the back wall along the y-axis. Then you have this front wall along the x-axis. And then you have this weird, curvy, curvy curtain or wall.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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I'll use magenta. That is the back wall along the y-axis. Then you have this front wall along the x-axis. And then you have this weird, curvy, curvy curtain or wall. I'll do that maybe in blue. That goes along this curve right here, this part of a circle of radius 2. Hopefully you get that visualization.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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And then you have this weird, curvy, curvy curtain or wall. I'll do that maybe in blue. That goes along this curve right here, this part of a circle of radius 2. Hopefully you get that visualization. It's a little harder. I'm not using any graphic program this time. But I want to figure out the surface area, the combined surface area of these three walls.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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Hopefully you get that visualization. It's a little harder. I'm not using any graphic program this time. But I want to figure out the surface area, the combined surface area of these three walls. In very simple notation, we could say, well, the surface area of those walls, of this wall plus that wall plus that wall, is going to be equal to the line integral along this curve or along this contour, however you want to call it, of f of x, y, so that's x plus y squared, ds, where ds is just a little length along our contour. And since this is a closed loop, we'll call this a closed line integral. And you'll sometimes see this notation right here.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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But I want to figure out the surface area, the combined surface area of these three walls. In very simple notation, we could say, well, the surface area of those walls, of this wall plus that wall plus that wall, is going to be equal to the line integral along this curve or along this contour, however you want to call it, of f of x, y, so that's x plus y squared, ds, where ds is just a little length along our contour. And since this is a closed loop, we'll call this a closed line integral. And you'll sometimes see this notation right here. That notation, often you'll see that in physics books. You'll be dealing with a lot more. You put a circle on that integral sign.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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And you'll sometimes see this notation right here. That notation, often you'll see that in physics books. You'll be dealing with a lot more. You put a circle on that integral sign. And all that means is that the contour we're dealing with is a closed contour. We get back to where we started from. But how do we solve this thing?
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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You put a circle on that integral sign. And all that means is that the contour we're dealing with is a closed contour. We get back to where we started from. But how do we solve this thing? A good place to start is to just define the contour itself. And just to simplify it, we're going to divide it into three pieces and essentially just do three separate line integrals because this isn't a very continuous contour. So the first part, let's do this first part of the curve as we're going along a circle of radius 2.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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But how do we solve this thing? A good place to start is to just define the contour itself. And just to simplify it, we're going to divide it into three pieces and essentially just do three separate line integrals because this isn't a very continuous contour. So the first part, let's do this first part of the curve as we're going along a circle of radius 2. And that's pretty easy to construct. If we have x is equal to... Let me do that in each part of the contour in a different color. So if I do an orange, this part of the contour, if we say that x is equal to 2 cosine of t and y is equal to 2 sine of t, if we say that t... and this is really just building off just what we saw in the last video.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So the first part, let's do this first part of the curve as we're going along a circle of radius 2. And that's pretty easy to construct. If we have x is equal to... Let me do that in each part of the contour in a different color. So if I do an orange, this part of the contour, if we say that x is equal to 2 cosine of t and y is equal to 2 sine of t, if we say that t... and this is really just building off just what we saw in the last video. If we say that t is greater than or equal to 0 and is less than or equal to pi over 2, t is essentially going to be the angle that we're going along this circle right here. And this will actually describe this path. And how I constructed this is a little confusing.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So if I do an orange, this part of the contour, if we say that x is equal to 2 cosine of t and y is equal to 2 sine of t, if we say that t... and this is really just building off just what we saw in the last video. If we say that t is greater than or equal to 0 and is less than or equal to pi over 2, t is essentially going to be the angle that we're going along this circle right here. And this will actually describe this path. And how I constructed this is a little confusing. You might want to review the video on parametric equations. So this is the first part of our path. So if we just wanted to find the surface area of that wall right there, we know we're going to have to find dx, dt, and dy, dt.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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And how I constructed this is a little confusing. You might want to review the video on parametric equations. So this is the first part of our path. So if we just wanted to find the surface area of that wall right there, we know we're going to have to find dx, dt, and dy, dt. So let's get that out of the way right now. So if we say dx, dt is going to be equal to minus 2 sine of t. dy, dt is going to be equal to 2 cosine of t. Just took the derivatives of these. We've seen that many times before.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So if we just wanted to find the surface area of that wall right there, we know we're going to have to find dx, dt, and dy, dt. So let's get that out of the way right now. So if we say dx, dt is going to be equal to minus 2 sine of t. dy, dt is going to be equal to 2 cosine of t. Just took the derivatives of these. We've seen that many times before. So if we want this orange wall's surface area, we can take the integral. And if any of this is confusing, there are two videos before this where we kind of derive this formula. But we could take the integral from t is equal to 0 to pi over 2 of our function of x plus y squared.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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We've seen that many times before. So if we want this orange wall's surface area, we can take the integral. And if any of this is confusing, there are two videos before this where we kind of derive this formula. But we could take the integral from t is equal to 0 to pi over 2 of our function of x plus y squared. And then times the ds. So x plus y squared will give the height of each little block. And then we want to get the width of each little block, which is ds.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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But we could take the integral from t is equal to 0 to pi over 2 of our function of x plus y squared. And then times the ds. So x plus y squared will give the height of each little block. And then we want to get the width of each little block, which is ds. But we know that we can rewrite ds as the square root of dx of the derivative of x with respect to t squared. So that is minus 2 sine of t squared plus the derivative of y with respect to t squared. So plus 2 cosine of t squared.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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And then we want to get the width of each little block, which is ds. But we know that we can rewrite ds as the square root of dx of the derivative of x with respect to t squared. So that is minus 2 sine of t squared plus the derivative of y with respect to t squared. So plus 2 cosine of t squared. dt. This will give us the orange section. And then we can worry about the other two walls.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So plus 2 cosine of t squared. dt. This will give us the orange section. And then we can worry about the other two walls. And so how can we simplify this? Well, this is going to be equal to the integral from 0 to pi over 2 of x plus y squared. And actually, let me write everything in terms of t. So x is equal to 2 cosine of t. So let me write that down.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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And then we can worry about the other two walls. And so how can we simplify this? Well, this is going to be equal to the integral from 0 to pi over 2 of x plus y squared. And actually, let me write everything in terms of t. So x is equal to 2 cosine of t. So let me write that down. So it's 2 cosine of t plus y, which is 2 sine of t. And we're going to square everything. And then all of that times this crazy radical. Right now it looks like a hard antiderivative or integral to solve, but I think we'll find out it's not too bad.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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And actually, let me write everything in terms of t. So x is equal to 2 cosine of t. So let me write that down. So it's 2 cosine of t plus y, which is 2 sine of t. And we're going to square everything. And then all of that times this crazy radical. Right now it looks like a hard antiderivative or integral to solve, but I think we'll find out it's not too bad. This is going to be equal to 4 sine squared of t plus 4 cosine squared of t. We can factor a 4 out. Now we don't want to forget the dt. This over here, let me just simplify this expression so I don't have to keep rewriting it.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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Right now it looks like a hard antiderivative or integral to solve, but I think we'll find out it's not too bad. This is going to be equal to 4 sine squared of t plus 4 cosine squared of t. We can factor a 4 out. Now we don't want to forget the dt. This over here, let me just simplify this expression so I don't have to keep rewriting it. That is the same thing as the square root of 4 times sine squared of t plus cosine squared of t. We know what that is. That's just 1. So this whole thing just simplifies to the square root of 4, which is just 2.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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This over here, let me just simplify this expression so I don't have to keep rewriting it. That is the same thing as the square root of 4 times sine squared of t plus cosine squared of t. We know what that is. That's just 1. So this whole thing just simplifies to the square root of 4, which is just 2. So this whole thing simplifies to 2, which is nice for us solving our antiderivative. It simplifies things a lot. So this whole thing simplifies down to, I'll do it over here.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So this whole thing just simplifies to the square root of 4, which is just 2. So this whole thing simplifies to 2, which is nice for us solving our antiderivative. It simplifies things a lot. So this whole thing simplifies down to, I'll do it over here. I don't want to waste too much space. I have two more walls to figure out. The integral from t is equal to 0 to pi over 2.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So this whole thing simplifies down to, I'll do it over here. I don't want to waste too much space. I have two more walls to figure out. The integral from t is equal to 0 to pi over 2. And I want to make it very clear. I just chose the simplest parametrization I could for x and y, but I could have picked other parametrizations and I would have had to change t accordingly. So as long as you're consistent with how you do it, it should all work out.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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The integral from t is equal to 0 to pi over 2. And I want to make it very clear. I just chose the simplest parametrization I could for x and y, but I could have picked other parametrizations and I would have had to change t accordingly. So as long as you're consistent with how you do it, it should all work out. There isn't just one parametrization for this curve. It's kind of depending on how fast you want to go along the curve. Watch the parametric functions videos if you want a little bit more depth on that.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So as long as you're consistent with how you do it, it should all work out. There isn't just one parametrization for this curve. It's kind of depending on how fast you want to go along the curve. Watch the parametric functions videos if you want a little bit more depth on that. But anyway, this thing simplifies. We have a 2 here, 2 times cosine of t. That's 4 cosine of t. And then here we have 2 sine squared sine of t squared. So that's 4 sine squared of t. And then we have to multiply it times this 2 again.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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Watch the parametric functions videos if you want a little bit more depth on that. But anyway, this thing simplifies. We have a 2 here, 2 times cosine of t. That's 4 cosine of t. And then here we have 2 sine squared sine of t squared. So that's 4 sine squared of t. And then we have to multiply it times this 2 again. So that gives us an 8. 8 times sine squared of t dt. And then sine squared of t, that looks like a tough thing to find the antiderivative for, but we can remember that sine squared of really anything, we could say sine squared of u, is equal to 1 half times 1 minus cosine of 2u.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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So that's 4 sine squared of t. And then we have to multiply it times this 2 again. So that gives us an 8. 8 times sine squared of t dt. And then sine squared of t, that looks like a tough thing to find the antiderivative for, but we can remember that sine squared of really anything, we could say sine squared of u, is equal to 1 half times 1 minus cosine of 2u. So we can reuse this identity. I could write a t here. Sine squared of t is equal to 1 half times 1 minus cosine of 2t.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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And then sine squared of t, that looks like a tough thing to find the antiderivative for, but we can remember that sine squared of really anything, we could say sine squared of u, is equal to 1 half times 1 minus cosine of 2u. So we can reuse this identity. I could write a t here. Sine squared of t is equal to 1 half times 1 minus cosine of 2t. Let me rewrite it that way, because that will make the integral a lot easier to solve. So we get the integral from 0 to pi over 2. And actually I could break up, well, I won't break it up.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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