$$\iint_R x^2y^2\,dy\,dx$$
I have to solve the above integral over the region $R$ defined by:
$R$ is the region enclosed by the circle $x^2+y^2=1$ and below the line $y=x+1$.
These give me the regions of:
$$\left\{(x,y) \mid -1 \le x \le 0, -\sqrt{1-x^2} \le y \le x + 1 ∪ 0 \le x \le 1, -\sqrt{1-x^2} \le y \le \sqrt{1-x^2}\right\}$$
Doing the integration for the above problem leave me with an absurdly large second integral, that's not impossible to solve but extremely lengthy.
$$\frac{1}{3}\int x^2(x+1)^3+x^2(1-x^2)^{3/2} + \frac{2}{3}\int x^2(1-x^2)^{3/2}$$
Surely there must be a simpler way?
Alternatively, I've completely botched this and none of this is nearly as complicated as I've made it which is also welcome.
Split the area into the triangle in the second quadrant and the rest of the circle. Use the transform to polar coordinates on the second area. $$ \iint_Rx^2y^2dydx=\int_{-1}^0\int_0^{x+1}x^2y^2dydx+\int_0^1\int_{-\pi}^{\frac\pi2}\rho^5\cos^2\phi\sin^2\phi d\phi d\rho $$ And now it's straightforward.