00:03
All right, for this problem, we are given the equation for a force, 2xy cubed and 4x squared y squared, and we're told that c is the boundary of the region enclosed by the x -axis, x equals 1, and y equals x cubed.
00:17
I have that drawn in the diagram here.
00:19
And the idea is we want to traverse this curve once counterclockwise and find the work done on the particle.
00:28
So to do that, the work equation is we integrate our force along the closed curve c.
00:43
Now, we can do this line integral directly.
00:47
However, i would recommend against it because you'll notice that we have three different curves that we would have to follow, which would translate into three separate line integrals.
00:59
So that's no good.
01:00
But what we can do instead is we can rewrite the line integral as pdx plus qdy, and the p and q are the x and y coordinates of the vector field.
01:19
And so long as we're integrating along a closed curve, we can apply green's theorem.
01:28
Green's theorem allows us to rewrite this line integral as a single double integral, which will be much easier for us to compute.
01:41
So it'll be a double integral over the region d.
01:43
The region d is the one that's enclosed by our curve.
01:50
And then you're going to have the z component to the curl, which is the partial of q with respect to x minus partial of p with respect to y.
02:09
Okay.
02:11
So again, this greatly simplifies the problem that we're working with.
02:14
So let's go ahead and find each of these partial derivatives.
02:17
So up top, our x partial of q is going to be 8 x, y squared, and then our y partial of the p component is going to be 6 x y squared.
02:41
Okay, so substituting into our double integral, qx minus p y, so we have 8x, x, y squared minus six yep x y squared d a couple things we can do with this first we can combine like terms oops sorry one second okay so we can combine like terms and what we also need to do is get the bounds of integration so that's not too hard we have our diagram here so what i would do is integrate with respect to y first.
03:39
So let's go ahead and write this out.
03:40
So i'm going to pull out the constant.
03:45
We're going to do this, d -y -d -x.
03:49
So if we integrate with respect to y first, then we're going from one curve to the other curve.
03:54
So the bottom boundary will be y equals zero, and then the upper boundary is going to be the y equals x cubed...