
help with greens theorem
just trying figure this example out using greens theorem:
$\displaystyle
I = \oint_R {9y^3 dx  9x^3 dy}
$
mapped counterclockwise, is the circle
$\displaystyle
x^2 + y^2 = 81
$
enter as a multiple of pi
so using substitution,
$\displaystyle
\int {udx + vdy = \int_{} {\int_R {(\frac{{\delta y}}
{{\delta x}}  \frac{{\delta u}}
{{\delta x}}} } } )dxdy
$
so
$\displaystyle
u = 9y^3
$
$\displaystyle
v =  9x^3
$
and
$\displaystyle
 27\int {\int_R {x^2 + y^2 dxdy} }
$
$\displaystyle
 27\int\limits_0^{2\pi } {\int\limits_0^1 {r^2 rdrd\theta } }
$
so i got this far how does this end bit work?? do i integrate? where do i go from here...

Doesn't r vary from 0 to 9?.
$\displaystyle 27\int_{0}^{2\pi}\int_{0}^{9}r^{3}drd{\theta}$

oops yeah my mistake, so how would i proceed, from here, i know im supposed to change polars, but i dont know exactly how to solve this

Just integrate. You should know how to do this simple integration since you're in Calc III.
$\displaystyle \int_{0}^{9}r^{3}dr=\frac{1}{4}r^{4}$
$\displaystyle \frac{1}{4}(9)^{4}\frac{1}{4}(0)^{4}=\frac{6561}{4}$
Now, integrate wrt to theta:
$\displaystyle \int_{0}^{2\pi}\frac{6561}{4}d{\theta}=\frac{6561} {4}(2{\pi})\frac{6561}{4}(0)=\frac{6561\pi}{2}$
Now, multiply by 27: $\displaystyle \frac{177147\pi}{2}$

any help with this?
help...(Doh)
int {[xy4x^4yysqrt(x^2+y^2+1/2y^2sinx]dx
+ [4xy^4+8/3x^3y^2+sqrtx^2+y^2ycosx]dy}
where r is the path mapped counterclockwise: along y=o,
0<x<1; then along x^2+y^2=1, from (1,0)to (1/sqrt2,1/sqrt2);
finally y=x from (1/sqrt2,1sqrt2) to (0,0)