00:01
Okay, so for this question we have this interesting region here in the first quadrant, bounded by the following two curves.
00:08
So i have x squared plus y squared equals 4, and then x squared plus y squared equals 2x.
00:13
And the curves on the actual graph are color -coded to match the equations.
00:18
So we want to find a double integral over this region for the function x.
00:23
Now, because our two regions are circular, it would make the most sense for us to use polar coordinates to figure this out.
00:33
So how do we do polar coordinates? we're going to use the following change of variables.
00:38
We will let x equal r cosine theta, and y will equal r sine theta.
00:47
So that then our double integral that we'll be calculating will be some double integral of r cosine theta.
00:58
Now, whenever you convert from cartesian to polar coordinates, you need to include an additional factor of r, and then we'll have dr d theta.
01:11
So we need to figure out the radius and theta bounds.
01:16
So let's see how we can do this.
01:19
Well, we need to get polar forms of our two equations for the boundaries.
01:27
So we have x squared plus y squared equals 4.
01:29
The nice part about polar coordinates is we get this really nice pythagorean relationship where x squared plus y squared equals r squared.
01:39
So that gives us this equation, which becomes r equals 2.
01:45
This will be the upper bound for the radius.
01:48
Now, the other one's a little more complicated, but you should get r squared equals 2r cosine theta.
01:56
Take out a factor of r, and that will give you the other bound for the radius.
02:07
So when we write out our integral, we should go from 2 cosine theta up until 2.
02:18
On the inside of the integral, we should get r squared cosine theta.
02:27
And then the theta bounds, we're in the first quadrant, so we should go from 0 to pi over 2.
02:34
Make sure, just to double check, that if you plug in both 0 and pi over 2 into the red function, you should end up getting the two points on this half circle.
02:45
Plug in 0, you should get 2.
02:47
Plug in pi over 2, you should get 0.
02:50
Okay, so here's our double integral.
02:54
Let's go ahead and solve it.
02:57
So first we're going to integrate the radius.
03:00
So the integral of r squared is r cubed over 3.
03:03
We'll pull out that factor of 1 third.
03:09
We'll have the cosine theta, and then we have r cubed from 2 cosine theta to 2.
03:22
Okay? let's see.
03:25
So we have the 1 third integral, 0 over 2, cosine, and then what do we have? 2 cubed, which is 8, minus 8 cosine cubed of theta.
03:46
There we go.
03:48
So we can pull out the 8, and not only that, but we can distribute in the cosine as well.
04:02
So you should get cosine minus cosine to the fourth...