Given that R is the region bounded by the curve C defined by 4x^2+9y^2=36, then by using Green's Theorem, the double integral int int x^2 dA is equal to...?
Any idea to solve it?
Green's theorem states that
$\displaystyle \int_{C} Pdx+Qdy=\int \int_D \left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) dA$
so to use the theorem backwards we need to find a function that P or Q that would gives the proper integrand.
There is more than one, for example
let $\displaystyle P=0 \mbox{ and } Q=\frac{1}{3}x^3$
Then by greens theorem we get
$\displaystyle \int_C \frac{1}{3}x^3dy=\int \int_Dx^2dA$
note that simple closed path is$\displaystyle r(t)=3\cos(t) \vec i +2\sin(t) \vec j \mbox{ for } 0 \le t \le 2\pi$