00:01
In this problem we are asked to use green's theorem to evaluate the line integral, integral over c, x e -raise to negative 2x, dx, plus x -raised 4 plus 2x2x2x2 -y square -d -y, where c is the positively oriented curve that forms the boundary of the region between the circles, x square plus y square is equal to 4 and x square plus y square is equal to 25.
00:37
Now by green's theorem we can evaluate an integral, integral over c, fdx plus g, d ,y as double integral over the region e, do g by da x minus do f by do y d x d x where e is the domain whose boundary is c so here comparing with this formula we have f of x y is x e raised to negative 2x and g of x y is x rise to 4 plus 2x2 y square and d o x x x x x x plus 2 x square and do f by do y is 0 since f is a function of x only.
01:26
And we have do f by do x is do do g by da x is 4 x cube plus 4 x y square.
01:41
And e is the region between the circles x square plus y square is equal to 4 and x square plus y square is equal to 25 so that we can use polar coordinates to get x is equal to r cost theta and y is equal to r sine theta therefore the region has boundary r ranges between two and five while theta ranges between zero and two pi therefore we can evaluate the integral integral over c x erase to negative 2x dx plus x raised to 4 plus 2x square y square d y using green's theorem as integral 0 to 2 pi integral 2 to 5 the indignant here is this term in polar coordinates it is 4 times r cube cos cube theta plus r cose x theta times r square sine square theta and d xdy becomes r d r d theta now this simplifies to four times integral to two pi integral two to five r raised to four cos theta times cos square theta plus sine square theta, dr, d theta.
03:29
Since cos square theta plus sine square theta is one, we have this is 4 times integral 0 to 2 pi, integral 2 to 5, r raised to 4, cos theta, dr d theta.
03:42
Now this can be evaluated...