00:01
For this problem, we are asked to use a line integral to find the area of the region bounded by the graph of x squared plus y squared equals a squared.
00:08
Now to begin, we know that we can find the area of a region by a line integral, by writing a equals one -half of the line integral over c of x -d -y -minus y -d -x.
00:23
You can parameterize that region or that curve by writing that x equals a cos theta.
00:32
And y equals a sine theta, which then means that we'd have that dx will be equal to negative a sine theta, d theta, and d y will be equal to a cos theta, d theta.
00:53
So we'd then have that a will be equal to one -half, the integral from, now theta will be between zero and two pi, of x, d -y -y, so we can see that would be a -cost theta times a -cose -theta, so we get a squared, co -squared theta, minus y -d -x...