TITLE: Integration – what should I do next?
QUESTION [5 upvotes]: $$\iint_\Omega (x^2+3y^3)\mathrm dx \mathrm dy $$
$$\Omega: 0\le x^2+y^2\le 1$$
I integrated it twice and then got this             
$$Cy+\frac{x^2y^2}{4}+xy+c  $$                          
What should I do next?

REPLY [7 votes]: Alternatively:
$$\int \int_{\Omega} x^2+3y^3 dxdy=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2+3y^3dydx=\int_{-1}^1 (x^2y+\frac{3}{4}y^4)\big|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx=\int_{-1}^12x^2\sqrt{1-x^2}dx \stackrel{x=\sin t}=\int_{-\pi/2}^{\pi/2}2\sin^2t\cdot\cos^2tdt=\int_{-\pi/2}^{\pi/2}\frac12\sin^2 2tdt=\int_{-\pi/2}^{\pi/2}\frac{1}{4}(1-\cos4t)dt=\frac{\pi}{4}.$$