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This is a pretty neat thing. All of a sudden, this vector, and q of xy, I didn't draw it out like I did the last time. This q of xy only has things in the j direction. If I were to draw it, it's a vector field. The vectors only go up and down. They have no horizontal component to them. What we saw, when you start with a vector field like this, you take the line integral around this closed loop. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
If I were to draw it, it's a vector field. The vectors only go up and down. They have no horizontal component to them. What we saw, when you start with a vector field like this, you take the line integral around this closed loop. I'll rewrite it right here. You take this line integral around this closed loop of q dot dr, which is equal to the integral around the closed loop of q of xy dy. We just figured out that that's equivalent to the double integral over the region. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
What we saw, when you start with a vector field like this, you take the line integral around this closed loop. I'll rewrite it right here. You take this line integral around this closed loop of q dot dr, which is equal to the integral around the closed loop of q of xy dy. We just figured out that that's equivalent to the double integral over the region. This is the region. That's exactly what we're doing over here. If I just gave you the region and you had to define it, you might say, x is going from this function to that function, and y is going from a to b. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
We just figured out that that's equivalent to the double integral over the region. This is the region. That's exactly what we're doing over here. If I just gave you the region and you had to define it, you might say, x is going from this function to that function, and y is going from a to b. You might want to review the double integral videos if that confuses you. We're taking the double integral over the region of the partial of q with respect to x. You could write dx dy, or you could even write a little da, the differential of area. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
If I just gave you the region and you had to define it, you might say, x is going from this function to that function, and y is going from a to b. You might want to review the double integral videos if that confuses you. We're taking the double integral over the region of the partial of q with respect to x. You could write dx dy, or you could even write a little da, the differential of area. That's what we can imagine as a da, which is the same thing as a dx dy. If we combine that with the last video, this is kind of the neat bringing it all together part, the result of the last video was this. If I had a function that's defined completely in terms of x, we had this right here. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
You could write dx dy, or you could even write a little da, the differential of area. That's what we can imagine as a da, which is the same thing as a dx dy. If we combine that with the last video, this is kind of the neat bringing it all together part, the result of the last video was this. If I had a function that's defined completely in terms of x, we had this right here. We had that result. Actually, let me copy and paste both of these to a nice, clean part of my whiteboard, and then we can do the exciting conclusion. Let me copy and paste that. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
If I had a function that's defined completely in terms of x, we had this right here. We had that result. Actually, let me copy and paste both of these to a nice, clean part of my whiteboard, and then we can do the exciting conclusion. Let me copy and paste that. That's what we got in the last video. In this video, we got this result. I'll actually just copy and paste that part right there. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
Let me copy and paste that. That's what we got in the last video. In this video, we got this result. I'll actually just copy and paste that part right there. You might already predict where this is going. Then let me paste it over here. This is the result from this video. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
I'll actually just copy and paste that part right there. You might already predict where this is going. Then let me paste it over here. This is the result from this video. Now, let's think about an arbitrary vector field that is defined as, let's say I have an arbitrary vector field. I'll do it in pink. Let's say F is a vector field defined over the xy-plane, and F is equal to P of xyi plus Q of xyj. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
This is the result from this video. Now, let's think about an arbitrary vector field that is defined as, let's say I have an arbitrary vector field. I'll do it in pink. Let's say F is a vector field defined over the xy-plane, and F is equal to P of xyi plus Q of xyj. You can almost imagine F being the addition of our vector fields P and Q that we did in the last two videos. Q was this video, and we did P in the video before that. This is really any arbitrary vector field. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
Let's say F is a vector field defined over the xy-plane, and F is equal to P of xyi plus Q of xyj. You can almost imagine F being the addition of our vector fields P and Q that we did in the last two videos. Q was this video, and we did P in the video before that. This is really any arbitrary vector field. Let's say we wanted to take the line integral of this vector field along some path. It could be the same one we've done, which has been a very arbitrary one. It's really any arbitrary path. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
This is really any arbitrary vector field. Let's say we wanted to take the line integral of this vector field along some path. It could be the same one we've done, which has been a very arbitrary one. It's really any arbitrary path. Let me draw some arbitrary path over here. Let's say that is my arbitrary path, my arbitrary curve. Let's say it goes in that counterclockwise direction, just like that. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
It's really any arbitrary path. Let me draw some arbitrary path over here. Let's say that is my arbitrary path, my arbitrary curve. Let's say it goes in that counterclockwise direction, just like that. I'm interested in what the closed line integral around that path of F dot dr is. We've seen it multiple times. dr is equal to dx times i plus dy times j. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
Let's say it goes in that counterclockwise direction, just like that. I'm interested in what the closed line integral around that path of F dot dr is. We've seen it multiple times. dr is equal to dx times i plus dy times j. This line integral can be rewritten as this is equal to the line integral around the path C. F dot dr, that's going to be this term times dx. That's P of xy times dx plus this term, Q of xy times dy. This whole thing, this is the same thing as the line integral of P of xy dx plus the line integral of Q of xy dy. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
dr is equal to dx times i plus dy times j. This line integral can be rewritten as this is equal to the line integral around the path C. F dot dr, that's going to be this term times dx. That's P of xy times dx plus this term, Q of xy times dy. This whole thing, this is the same thing as the line integral of P of xy dx plus the line integral of Q of xy dy. What are these things? This is what we figured out in the first video. This is what we figured out just now in this video. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
This whole thing, this is the same thing as the line integral of P of xy dx plus the line integral of Q of xy dy. What are these things? This is what we figured out in the first video. This is what we figured out just now in this video. This thing right here is the exact same thing as that over there. This is going to be equal to the double integral over this region of the minus partial of P with respect to y. Instead of a dy dx, we can say over the differential of area. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
This is what we figured out just now in this video. This thing right here is the exact same thing as that over there. This is going to be equal to the double integral over this region of the minus partial of P with respect to y. Instead of a dy dx, we can say over the differential of area. Plus this result, this thing right here is exactly what we just proved. It's exactly what we just showed in this video. That's plus, I'll leave it up there, maybe I'll do it in the yellow. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
Instead of a dy dx, we can say over the differential of area. Plus this result, this thing right here is exactly what we just proved. It's exactly what we just showed in this video. That's plus, I'll leave it up there, maybe I'll do it in the yellow. Plus the double integral over the same region of the partial of Q with respect to x, da. That's just dy dx or dx dy, you can switch the order, it's the differential of area. Now we can add these two integrals. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
That's plus, I'll leave it up there, maybe I'll do it in the yellow. Plus the double integral over the same region of the partial of Q with respect to x, da. That's just dy dx or dx dy, you can switch the order, it's the differential of area. Now we can add these two integrals. What do we get? This is equal to, and this is kind of our big grand conclusion, maybe magenta is called for here. The double integral of over the region of, I'll write this one first because it's positive, that one's negative. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
Now we can add these two integrals. What do we get? This is equal to, and this is kind of our big grand conclusion, maybe magenta is called for here. The double integral of over the region of, I'll write this one first because it's positive, that one's negative. Over the region of the partial of Q with respect to x minus the partial of P with respect to y, d, the differential of area. This is our big takeaway. The line integral of the closed loop of F dot dr is equal to the double integral of this expression. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
The double integral of over the region of, I'll write this one first because it's positive, that one's negative. Over the region of the partial of Q with respect to x minus the partial of P with respect to y, d, the differential of area. This is our big takeaway. The line integral of the closed loop of F dot dr is equal to the double integral of this expression. Just remember, we're taking the function that was associated with the x component, or the i component, we're taking the partial with respect to y. The function that was associated with the y component, we're taking the partial with respect to x. That first one we're taking the negative of, that's a good way to remember it. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
The line integral of the closed loop of F dot dr is equal to the double integral of this expression. Just remember, we're taking the function that was associated with the x component, or the i component, we're taking the partial with respect to y. The function that was associated with the y component, we're taking the partial with respect to x. That first one we're taking the negative of, that's a good way to remember it. This result right here, this is, maybe I should write it in green, this is Green's Theorem. It's a neat way to relate a line integral of a vector field that has these partial derivatives, assuming it has these partial derivatives, to the region, to a double integral of the region. Just as a little bit of a side note, we've seen in several videos before, if F is conservative, we've learned that if F is conservative, which means it's the gradient of some function, that it's path independent, that the closed integral around any path is equal to zero. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
That first one we're taking the negative of, that's a good way to remember it. This result right here, this is, maybe I should write it in green, this is Green's Theorem. It's a neat way to relate a line integral of a vector field that has these partial derivatives, assuming it has these partial derivatives, to the region, to a double integral of the region. Just as a little bit of a side note, we've seen in several videos before, if F is conservative, we've learned that if F is conservative, which means it's the gradient of some function, that it's path independent, that the closed integral around any path is equal to zero. That's still true. That tells us that if F is conservative, this thing right here must be equal to zero. That's the only way that you're always going to enforce that this whole integral is going to be equal to zero over any region. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
Just as a little bit of a side note, we've seen in several videos before, if F is conservative, we've learned that if F is conservative, which means it's the gradient of some function, that it's path independent, that the closed integral around any path is equal to zero. That's still true. That tells us that if F is conservative, this thing right here must be equal to zero. That's the only way that you're always going to enforce that this whole integral is going to be equal to zero over any region. You could think of situations where they cancel each other out, but really over any region, that's the only way that this is going to be true, that these two things are going to be equal to zero. Then you could say partial of Q with respect to X minus the partial of P with respect to Y has to be equal to zero, or these two things have to equal each other. Or, this is kind of a corollary to Green's Theorem, kind of a low-hanging fruit you could have figured out. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
That's the only way that you're always going to enforce that this whole integral is going to be equal to zero over any region. You could think of situations where they cancel each other out, but really over any region, that's the only way that this is going to be true, that these two things are going to be equal to zero. Then you could say partial of Q with respect to X minus the partial of P with respect to Y has to be equal to zero, or these two things have to equal each other. Or, this is kind of a corollary to Green's Theorem, kind of a low-hanging fruit you could have figured out. The partial of Q with respect to X is equal to the partial of P with respect to Y. When you study exact equations in differential equation, you'll see this a lot more. I won't go into too much, but conservative fields, the differential form of what you see in the line integral, if it's conservative, it would be an exact equation. | Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3 |
Welcome back. In the last video, we were just figuring out the volume under the surface. And we had set up these integral bounds. So let's see how to evaluate it now. And look at this. I actually realize that I can scroll things, which is quite useful, because now I have a lot more board space. So how do I evaluate this integral? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So let's see how to evaluate it now. And look at this. I actually realize that I can scroll things, which is quite useful, because now I have a lot more board space. So how do I evaluate this integral? Well, the first integral, I'm integrating with respect to x. I'm adding up the little x sums. So I'm forming this rectangle right here. Or you could kind of view it. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So how do I evaluate this integral? Well, the first integral, I'm integrating with respect to x. I'm adding up the little x sums. So I'm forming this rectangle right here. Or you could kind of view it. I'm holding y constant and integrating along the x-axis. I should switch colors. So what's the antiderivative of xy squared with respect to x? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Or you could kind of view it. I'm holding y constant and integrating along the x-axis. I should switch colors. So what's the antiderivative of xy squared with respect to x? Well, it's just x squared over 2. And then I have the y squared. That's just a constant. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So what's the antiderivative of xy squared with respect to x? Well, it's just x squared over 2. And then I have the y squared. That's just a constant. All over 2. And I'm going to evaluate that from x is equal to 1 to x is equal to the square root of y, which you might be daunted by. But you'll see that it's actually not that bad once you evaluate them. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
That's just a constant. All over 2. And I'm going to evaluate that from x is equal to 1 to x is equal to the square root of y, which you might be daunted by. But you'll see that it's actually not that bad once you evaluate them. And then let me draw the outside of the integral. This is y is equal to 0 to y is equal to 1 dy. Now, if x is equal to 1, this expression becomes y squared over 2 minus. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
But you'll see that it's actually not that bad once you evaluate them. And then let me draw the outside of the integral. This is y is equal to 0 to y is equal to 1 dy. Now, if x is equal to 1, this expression becomes y squared over 2 minus. Now, if x is equal to square root of y, what does this expression become? If x is equal to the square root of y, then x squared is just y, and then y times y squared is y to the third. So it's y to the third over 3. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Now, if x is equal to 1, this expression becomes y squared over 2 minus. Now, if x is equal to square root of y, what does this expression become? If x is equal to the square root of y, then x squared is just y, and then y times y squared is y to the third. So it's y to the third over 3. Fair enough. And now I take the integral with respect to y. So now I sum up all of these rectangles in the y direction. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So it's y to the third over 3. Fair enough. And now I take the integral with respect to y. So now I sum up all of these rectangles in the y direction. 0, 1. This is with respect to y. And that's cool, right? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So now I sum up all of these rectangles in the y direction. 0, 1. This is with respect to y. And that's cool, right? Because when you take the first integral with respect to x, you end up with a function of y anyway. So you might as well have your bounds as functions of y's. It really doesn't make it any more difficult. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And that's cool, right? Because when you take the first integral with respect to x, you end up with a function of y anyway. So you might as well have your bounds as functions of y's. It really doesn't make it any more difficult. But anyway, back to the problem. What is the antiderivative of y squared over 2 minus y to the third over 3? Well, the antiderivative of y squared, it's y cubed. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
It really doesn't make it any more difficult. But anyway, back to the problem. What is the antiderivative of y squared over 2 minus y to the third over 3? Well, the antiderivative of y squared, it's y cubed. And you have to divide by 3, so it's y cubed over 6 minus y to the fourth, you have to divide by 4, minus y to the fourth over, did I mess up someplace? No, I think this is correct. y to the fourth over 12. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Well, the antiderivative of y squared, it's y cubed. And you have to divide by 3, so it's y cubed over 6 minus y to the fourth, you have to divide by 4, minus y to the fourth over, did I mess up someplace? No, I think this is correct. y to the fourth over 12. Oh wait, how did I get a 3 here? That's where I messed up. This is a 2, right? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
y to the fourth over 12. Oh wait, how did I get a 3 here? That's where I messed up. This is a 2, right? Let's see. x is square root of y. Yeah, this is a 2. I don't know how I ended up, right? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
This is a 2, right? Let's see. x is square root of y. Yeah, this is a 2. I don't know how I ended up, right? Square root of y, squared is y, times y squared, y to the third over 2. Right. And then when I take the integral of this, it's 4 times 2, 8. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
I don't know how I ended up, right? Square root of y, squared is y, times y squared, y to the third over 2. Right. And then when I take the integral of this, it's 4 times 2, 8. Got to make sure I don't make those careless mistakes. That's the tough part. I just want to make sure that you got that too. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And then when I take the integral of this, it's 4 times 2, 8. Got to make sure I don't make those careless mistakes. That's the tough part. I just want to make sure that you got that too. I hate it when I do that. But I don't want to re-record the whole video. So when I evaluated this, this is right, and then I take the antiderivative of y to the third over 2, I get y to the fourth over 8. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
I just want to make sure that you got that too. I hate it when I do that. But I don't want to re-record the whole video. So when I evaluated this, this is right, and then I take the antiderivative of y to the third over 2, I get y to the fourth over 8. And now I evaluate this at 1 and 0. And that gives us what? 1 sixth minus 1 eighth minus, well both of these when you evaluate them at 0 are going to be 0. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So when I evaluated this, this is right, and then I take the antiderivative of y to the third over 2, I get y to the fourth over 8. And now I evaluate this at 1 and 0. And that gives us what? 1 sixth minus 1 eighth minus, well both of these when you evaluate them at 0 are going to be 0. So it's going to be 0 minus 0. So you don't have to worry about that. So what's 1 sixth minus 1 eighth? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
1 sixth minus 1 eighth minus, well both of these when you evaluate them at 0 are going to be 0. So it's going to be 0 minus 0. So you don't have to worry about that. So what's 1 sixth minus 1 eighth? Let's see. 24, that's 4 minus 3 over 24, which is equal to 1 24th, is the volume of our figure. Is the volume of this figure. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So what's 1 sixth minus 1 eighth? Let's see. 24, that's 4 minus 3 over 24, which is equal to 1 24th, is the volume of our figure. Is the volume of this figure. So this time, the way we just did it, we took the integral with respect to x first, and then we did it with respect to y. Let's do it the other way around. So let me erase some things, and hopefully I won't make these careless mistakes again. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Is the volume of this figure. So this time, the way we just did it, we took the integral with respect to x first, and then we did it with respect to y. Let's do it the other way around. So let me erase some things, and hopefully I won't make these careless mistakes again. I'll keep this figure, but I'll even erase this one. Just because, let me erase all of this stuff. We have room to work with. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So let me erase some things, and hopefully I won't make these careless mistakes again. I'll keep this figure, but I'll even erase this one. Just because, let me erase all of this stuff. We have room to work with. So I kept that figure, but let me redraw just the xy-plane, just so we get the visualization right. It's more important to visualize the xy-plane in these problems than it is to visualize the whole thing in 3D. That's the y-axis, that's the x-axis, y, x. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
We have room to work with. So I kept that figure, but let me redraw just the xy-plane, just so we get the visualization right. It's more important to visualize the xy-plane in these problems than it is to visualize the whole thing in 3D. That's the y-axis, that's the x-axis, y, x. Our upper bound, you can say, is the graph y is equal to x squared, or you can view it as the bound x is equal to square root of y. That is, x is equal to 1, that's y is equal to 1. And we care about the volume above the shaded region. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
That's the y-axis, that's the x-axis, y, x. Our upper bound, you can say, is the graph y is equal to x squared, or you can view it as the bound x is equal to square root of y. That is, x is equal to 1, that's y is equal to 1. And we care about the volume above the shaded region. That shaded region is this yellow region right here. And let's draw our dA, I'll draw a little square actually, and I'll do it in magenta. So that's our little dA. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And we care about the volume above the shaded region. That shaded region is this yellow region right here. And let's draw our dA, I'll draw a little square actually, and I'll do it in magenta. So that's our little dA. And the height is dy, that's a y, and its width is dx. So the volume above this little square, that's the same thing as this little square, just like we said before, the volume above it is equal to the value of the function, the height is the value of the function, which is xy squared. And then we multiply it times the area of the base. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So that's our little dA. And the height is dy, that's a y, and its width is dx. So the volume above this little square, that's the same thing as this little square, just like we said before, the volume above it is equal to the value of the function, the height is the value of the function, which is xy squared. And then we multiply it times the area of the base. Well, the area of the base, you could say it's dA, but we know it's really dy times dx. I didn't have to write that times there, you could ignore it. And I wrote the y first just because we're going to integrate with respect to y first. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And then we multiply it times the area of the base. Well, the area of the base, you could say it's dA, but we know it's really dy times dx. I didn't have to write that times there, you could ignore it. And I wrote the y first just because we're going to integrate with respect to y first. We're going to sum in the y direction. So what does summing in the y direction mean? It means we're going to add that square to that square to that square. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And I wrote the y first just because we're going to integrate with respect to y first. We're going to sum in the y direction. So what does summing in the y direction mean? It means we're going to add that square to that square to that square. Excuse me. So we're adding all the dy's together, right? So my question to you is, what is the upper bound on the y? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
It means we're going to add that square to that square to that square. Excuse me. So we're adding all the dy's together, right? So my question to you is, what is the upper bound on the y? Well, once again, we bump into the curve, right? So the curve is the upper bound when we go upwards. And what is the upper bound on the curve? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So my question to you is, what is the upper bound on the y? Well, once again, we bump into the curve, right? So the curve is the upper bound when we go upwards. And what is the upper bound on the curve? Well, we're holding x fixed. So for any given x, what is this point? Well, that's going to be x squared, right? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And what is the upper bound on the curve? Well, we're holding x fixed. So for any given x, what is this point? Well, that's going to be x squared, right? Because this is the graph of y equals x squared. So our upper bound is y is equal to x squared. And what's our lower bound? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Well, that's going to be x squared, right? Because this is the graph of y equals x squared. So our upper bound is y is equal to x squared. And what's our lower bound? We can keep adding these squares down here. We're adding all the little changes in y. So what's our lower bound? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And what's our lower bound? We can keep adding these squares down here. We're adding all the little changes in y. So what's our lower bound? Well, our lower bound is just 0. That was pretty straightforward. So this expression, as it is written right now, is the volume above this rectangle. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So what's our lower bound? Well, our lower bound is just 0. That was pretty straightforward. So this expression, as it is written right now, is the volume above this rectangle. Let me draw it. It's the volume above this rectangle. These are the same rectangles. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So this expression, as it is written right now, is the volume above this rectangle. Let me draw it. It's the volume above this rectangle. These are the same rectangles. Now what we want to do is add up all the dx's together, and we'll get the volume above the entire surface. So that rectangle, now we'll add it to another dx1, dx1, like that. So what's the upper and lower bounds on the x's? | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
These are the same rectangles. Now what we want to do is add up all the dx's together, and we'll get the volume above the entire surface. So that rectangle, now we'll add it to another dx1, dx1, like that. So what's the upper and lower bounds on the x's? We're going from x is equal to 0, right? We don't bump into the graph when we go all the way down. So we go for x is equal to 0. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So what's the upper and lower bounds on the x's? We're going from x is equal to 0, right? We don't bump into the graph when we go all the way down. So we go for x is equal to 0. And then our upper bound is x is equal to 1. x is equal to 0, x is equal to 1. And in general, one way to think about it is, when you're doing this outer, kind of the last sum or the last integral, you really shouldn't have variable boundaries at this point, right? Because our final answer has to be a number, assuming that we're not dealing with something very, very abstract. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So we go for x is equal to 0. And then our upper bound is x is equal to 1. x is equal to 0, x is equal to 1. And in general, one way to think about it is, when you're doing this outer, kind of the last sum or the last integral, you really shouldn't have variable boundaries at this point, right? Because our final answer has to be a number, assuming that we're not dealing with something very, very abstract. But our final answer is going to have a number. So if you had some variables here, you probably did something wrong. It's always useful, I think, to draw that little da and then figure out, OK, I'm summing to dy first. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Because our final answer has to be a number, assuming that we're not dealing with something very, very abstract. But our final answer is going to have a number. So if you had some variables here, you probably did something wrong. It's always useful, I think, to draw that little da and then figure out, OK, I'm summing to dy first. When I go upwards, I bump into the curve. What is that upward bound if x is constant? Oh, it's x squared. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
It's always useful, I think, to draw that little da and then figure out, OK, I'm summing to dy first. When I go upwards, I bump into the curve. What is that upward bound if x is constant? Oh, it's x squared. y is equal to x squared. If I go down, I bump into the x-axis where y is equal to 0. And so forth and so on. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Oh, it's x squared. y is equal to x squared. If I go down, I bump into the x-axis where y is equal to 0. And so forth and so on. So now let's just evaluate this and confirm that we get the same answer. So we're integrating with respect to y first. So that's xy to the third over 3 at x squared and 0. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And so forth and so on. So now let's just evaluate this and confirm that we get the same answer. So we're integrating with respect to y first. So that's xy to the third over 3 at x squared and 0. And then we have our outer integral. x goes from 0 to 1 dx. If we substitute x squared in for y, x squared to the third power is x to the sixth. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So that's xy to the third over 3 at x squared and 0. And then we have our outer integral. x goes from 0 to 1 dx. If we substitute x squared in for y, x squared to the third power is x to the sixth. x to the sixth times x. Let me write that. So we have x times x squared to the third power over 3. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
If we substitute x squared in for y, x squared to the third power is x to the sixth. x to the sixth times x. Let me write that. So we have x times x squared to the third power over 3. That equals x to the seventh. x to the squared to the third, you multiply the exponents and then you add these. x to the seventh over 3 minus this evaluated when y is 0. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So we have x times x squared to the third power over 3. That equals x to the seventh. x to the squared to the third, you multiply the exponents and then you add these. x to the seventh over 3 minus this evaluated when y is 0. But that's just going to be 0, right? And then we evaluate that from 0 to 1 dx. We're almost there. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
x to the seventh over 3 minus this evaluated when y is 0. But that's just going to be 0, right? And then we evaluate that from 0 to 1 dx. We're almost there. Increment the exponent. You get x to the eighth over 8. And we already have a 3 down there, so it's over 24. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
We're almost there. Increment the exponent. You get x to the eighth over 8. And we already have a 3 down there, so it's over 24. And you evaluate that from 0 to 1. And I think we get the same answer. When you evaluate it at 1, you get 1 24th minus 0. | Double integrals 6 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
What I want to do in the next few videos is try to see what happens to a line integral, either a line integral over a scalar field or a vector field. But what happens to that line integral when we change the direction of our path? So let's say when I say change direction, let's say that I have some curve C that looks something like this. We draw the x and y axis. That's my y axis. That is my x axis. And let's say my parametrization starts there and then as t increases, ends up over there, just like that. | Parametrization of a reverse path Khan Academy.mp3 |
We draw the x and y axis. That's my y axis. That is my x axis. And let's say my parametrization starts there and then as t increases, ends up over there, just like that. So it's moving in that direction. And when I say I reverse the path, we could define another curve, let's call it minus C, that looks something like this. That is my y axis. | Parametrization of a reverse path Khan Academy.mp3 |
And let's say my parametrization starts there and then as t increases, ends up over there, just like that. So it's moving in that direction. And when I say I reverse the path, we could define another curve, let's call it minus C, that looks something like this. That is my y axis. That is my x axis. And it looks exactly the same, but it starts up here. And then as t increases, it goes down to the starting point of the other curve. | Parametrization of a reverse path Khan Academy.mp3 |
That is my y axis. That is my x axis. And it looks exactly the same, but it starts up here. And then as t increases, it goes down to the starting point of the other curve. So it's the exact same shape of a curve, but it goes in the opposite direction. So what I'm going to do in this video is just understand how we can construct a parametrization like this and hopefully understand it pretty well. And in the next two videos after this, we'll try to see what this actually does to the line integral, one for a scalar field and then one for a vector field. | Parametrization of a reverse path Khan Academy.mp3 |
And then as t increases, it goes down to the starting point of the other curve. So it's the exact same shape of a curve, but it goes in the opposite direction. So what I'm going to do in this video is just understand how we can construct a parametrization like this and hopefully understand it pretty well. And in the next two videos after this, we'll try to see what this actually does to the line integral, one for a scalar field and then one for a vector field. So let's just say this parametrization right here, this is defined in the basic way that we've always defined them. Let's say that this is x is equal to x of t, y is equal to y of t. And let's say this is from t is equal, or t, let me write it this way, t starts at a, so t is greater than or equal to a, and it goes up to b. So in this example, this was when t is equal to a, and the point right here is the coordinate x of a, y of a. | Parametrization of a reverse path Khan Academy.mp3 |
And in the next two videos after this, we'll try to see what this actually does to the line integral, one for a scalar field and then one for a vector field. So let's just say this parametrization right here, this is defined in the basic way that we've always defined them. Let's say that this is x is equal to x of t, y is equal to y of t. And let's say this is from t is equal, or t, let me write it this way, t starts at a, so t is greater than or equal to a, and it goes up to b. So in this example, this was when t is equal to a, and the point right here is the coordinate x of a, y of a. And then when t is equal to b up here, this is really just a review of what we've seen before, really just a review of parametrization. When t is equal to b up here, this is the point x of b, y of b. Nothing new there. | Parametrization of a reverse path Khan Academy.mp3 |
So in this example, this was when t is equal to a, and the point right here is the coordinate x of a, y of a. And then when t is equal to b up here, this is really just a review of what we've seen before, really just a review of parametrization. When t is equal to b up here, this is the point x of b, y of b. Nothing new there. Now given these functions, how can we construct another parametrization here that has the same shape but starts here? So I want this to be t is equal to a. Let me switch colors. | Parametrization of a reverse path Khan Academy.mp3 |
Nothing new there. Now given these functions, how can we construct another parametrization here that has the same shape but starts here? So I want this to be t is equal to a. Let me switch colors. Let me switch to maybe magenta. I want this to be t is equal to a, and as t increases, I want this to be t equals b. So I want to move in the opposite direction. | Parametrization of a reverse path Khan Academy.mp3 |
Let me switch colors. Let me switch to maybe magenta. I want this to be t is equal to a, and as t increases, I want this to be t equals b. So I want to move in the opposite direction. So when t is equal to a, I want my coordinate to still be x of b, y of b. When t is equal to a, I want a b in each of these functions. And when t is equal to b, I want the coordinate to be x of a, y of a. | Parametrization of a reverse path Khan Academy.mp3 |
So I want to move in the opposite direction. So when t is equal to a, I want my coordinate to still be x of b, y of b. When t is equal to a, I want a b in each of these functions. And when t is equal to b, I want the coordinate to be x of a, y of a. Notice they're opposites now. Here t is equal to a, x of a, y of a. Here t is equal to b, our end point. | Parametrization of a reverse path Khan Academy.mp3 |
And when t is equal to b, I want the coordinate to be x of a, y of a. Notice they're opposites now. Here t is equal to a, x of a, y of a. Here t is equal to b, our end point. Now I'm at this coordinate x of a, y of a. So how do I construct that? Well, if you think about it, when t is equal to a, we want both of these functions to evaluate it at b. | Parametrization of a reverse path Khan Academy.mp3 |
Here t is equal to b, our end point. Now I'm at this coordinate x of a, y of a. So how do I construct that? Well, if you think about it, when t is equal to a, we want both of these functions to evaluate it at b. So what if we define our x in this case for our minus c curve. What if we say x is equal to x of, I'm going to say x of, I'm talking about the same exact function. Actually, maybe I should write it in that same exact color. | Parametrization of a reverse path Khan Academy.mp3 |
Well, if you think about it, when t is equal to a, we want both of these functions to evaluate it at b. So what if we define our x in this case for our minus c curve. What if we say x is equal to x of, I'm going to say x of, I'm talking about the same exact function. Actually, maybe I should write it in that same exact color. x of, but instead of putting t in there, instead of putting a straight up t in there, what if I put an a plus b minus t in there? What happens? Well, let me do it for the y as well. | Parametrization of a reverse path Khan Academy.mp3 |
Actually, maybe I should write it in that same exact color. x of, but instead of putting t in there, instead of putting a straight up t in there, what if I put an a plus b minus t in there? What happens? Well, let me do it for the y as well. y is equal to y of a plus b minus t. I'm using slightly different shades of yellow. It might be a little disconcerting. Anyway, what happens when we define this? | Parametrization of a reverse path Khan Academy.mp3 |
Well, let me do it for the y as well. y is equal to y of a plus b minus t. I'm using slightly different shades of yellow. It might be a little disconcerting. Anyway, what happens when we define this? When t is equal to a, and let's say that this parametrization is also for t starts at a and then goes up to b. So let's just experiment and confirm that this parametrization really is the same thing as this thing, but it goes in an opposite direction, or at least confirm in our minds intuitively. So when t is equal to a, x will be equal to x of a plus b minus a. | Parametrization of a reverse path Khan Academy.mp3 |
Anyway, what happens when we define this? When t is equal to a, and let's say that this parametrization is also for t starts at a and then goes up to b. So let's just experiment and confirm that this parametrization really is the same thing as this thing, but it goes in an opposite direction, or at least confirm in our minds intuitively. So when t is equal to a, x will be equal to x of a plus b minus a. This is when t is equal to a. So minus t or minus a, which is equal to what? Well, a minus a cancel out. | Parametrization of a reverse path Khan Academy.mp3 |
So when t is equal to a, x will be equal to x of a plus b minus a. This is when t is equal to a. So minus t or minus a, which is equal to what? Well, a minus a cancel out. That's equal to x of b. Similarly, when t is equal to a, y will be equal to y of a plus b minus a. The a's cancel out, so it's equal to y of b. | Parametrization of a reverse path Khan Academy.mp3 |
Well, a minus a cancel out. That's equal to x of b. Similarly, when t is equal to a, y will be equal to y of a plus b minus a. The a's cancel out, so it's equal to y of b. So that worked. When t is equal to a, my parametrization evaluates to the coordinate x of b, y of b. When t is equal to a, x of b, y of b. | Parametrization of a reverse path Khan Academy.mp3 |
The a's cancel out, so it's equal to y of b. So that worked. When t is equal to a, my parametrization evaluates to the coordinate x of b, y of b. When t is equal to a, x of b, y of b. Then we can do the exact same thing when t is equal to b. I'll do it over here because I don't want to lose this. Let me just draw a line here. I'm still dealing with this parametrization over here. | Parametrization of a reverse path Khan Academy.mp3 |
When t is equal to a, x of b, y of b. Then we can do the exact same thing when t is equal to b. I'll do it over here because I don't want to lose this. Let me just draw a line here. I'm still dealing with this parametrization over here. Actually, let me scroll over to the right just so that I don't get confused. When t is equal to b, what is x equal? x is equal to x of a plus b minus b. a plus b minus b when t is equal to b. | Parametrization of a reverse path Khan Academy.mp3 |
I'm still dealing with this parametrization over here. Actually, let me scroll over to the right just so that I don't get confused. When t is equal to b, what is x equal? x is equal to x of a plus b minus b. a plus b minus b when t is equal to b. So that's equal to x of a. Then when t is equal to b, y is equal to y of a plus b minus b. Of course, that's going to be equal to y of a. | Parametrization of a reverse path Khan Academy.mp3 |
x is equal to x of a plus b minus b. a plus b minus b when t is equal to b. So that's equal to x of a. Then when t is equal to b, y is equal to y of a plus b minus b. Of course, that's going to be equal to y of a. The endpoints work, and if you think about it intuitively, as t increases, so when t is at a, this thing is going to be x of b, y of b. We saw that down here. As t increases, this value is going to decrease. | Parametrization of a reverse path Khan Academy.mp3 |
Of course, that's going to be equal to y of a. The endpoints work, and if you think about it intuitively, as t increases, so when t is at a, this thing is going to be x of b, y of b. We saw that down here. As t increases, this value is going to decrease. We start at x of b, y of b, and as t increases, this value is going to decrease to a. It starts from b and it goes to a. This one obviously starts at a and it goes to b. Hopefully that should give you the intuition why this is the exact same curve as that. | Parametrization of a reverse path Khan Academy.mp3 |
As t increases, this value is going to decrease. We start at x of b, y of b, and as t increases, this value is going to decrease to a. It starts from b and it goes to a. This one obviously starts at a and it goes to b. Hopefully that should give you the intuition why this is the exact same curve as that. It just goes in a completely opposite direction. Now, with that out of the way, if you accept what I've told you, that these are really the same parametrizations, just opposite directions. I shouldn't say same parametrizations. | Parametrization of a reverse path Khan Academy.mp3 |
This one obviously starts at a and it goes to b. Hopefully that should give you the intuition why this is the exact same curve as that. It just goes in a completely opposite direction. Now, with that out of the way, if you accept what I've told you, that these are really the same parametrizations, just opposite directions. I shouldn't say same parametrizations. Same curve going in an opposite direction or the same path going in the opposite direction. In the next video, I'm going to see what happens when we evaluate this line integral, f of x, ds, versus this line integral. This is a line integral of a scalar field using this curve or this path. | Parametrization of a reverse path Khan Academy.mp3 |
I shouldn't say same parametrizations. Same curve going in an opposite direction or the same path going in the opposite direction. In the next video, I'm going to see what happens when we evaluate this line integral, f of x, ds, versus this line integral. This is a line integral of a scalar field using this curve or this path. What happens if we take a line integral over the same scalar field but we do it over this reverse path? That's what we're going to do in the next video. The video after that, we'll do it for vector fields. | Parametrization of a reverse path Khan Academy.mp3 |
There's S, the tons of steel that you're using, H, the hours of labor, and then lambda, this Lagrange multiplier we introduced that's basically a proportionality constant between the gradient vectors of the revenue function and the constraint function. And always the third equation that we're dealing with here to solve this is the constraint itself. That gives us another equation that'll help solve for H and S, and ultimately lambda, if that's something you want as well. So as kind of a first pass here, I'm gonna do a little more simplifying, but I'm gonna make a substitution that'll make this easier for us. So I see S over H here, and they're both to the same power, so I feel it might be a little bit easier if I just substitute U in for S divided by H. And what that'll let me do is rewrite this first equation here as 200 thirds, 200 thirds times U to the power 1 third, and that's equal to 20 times lambda. And then likewise, what that means for this guy is, well, this is H over S, not S over H, so that one's gonna be 100 thirds, not times U to the power 2 thirds, but times U to the negative 2 thirds because we swapped the H and S here. So that's U to the negative 2 thirds, and this is just to make it a little bit cleaner, I think. | Lagrange multiplier example, part 2.mp3 |
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