Integration – what should I do next? $$\iint_\Omega (x^2+3y^3)\mathrm dx \mathrm dy $$
$$\Omega: 0\le x^2+y^2\le 1$$
I integrated it twice and then got this             
$$Cy+\frac{x^2y^2}{4}+xy+c  $$                          
What should I do next?
 A: Alternatively:
$$\int \int_{\Omega} x^2+3y^3 dxdy=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2+3y^3dydx=\int_{-1}^1 (x^2y+\frac{3}{4}y^4)\big|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx=\int_{-1}^12x^2\sqrt{1-x^2}dx \stackrel{x=\sin t}=\int_{-\pi/2}^{\pi/2}2\sin^2t\cdot\cos^2tdt=\int_{-\pi/2}^{\pi/2}\frac12\sin^2 2tdt=\int_{-\pi/2}^{\pi/2}\frac{1}{4}(1-\cos4t)dt=\frac{\pi}{4}.$$
A: \begin{align}
&\int \int_{\Omega} x^2+3y^3 dxdy
\\&= \int\int_{\Omega} x^2dxdy
\\&=4 \int_0^{\frac{\pi}2}\int_0^{1}r^3\cos^2(\theta)  drd\theta
\\&=4 \int_0^{\frac{\pi}2}\cos^2(\theta)d\theta\int_0^{1}r^3  dr \\
&= \int_0^{\frac{\pi}2}\cos^2(\theta) d\theta \\
&= \int_0^{\frac{\pi}2} \frac{\cos(2\theta)+1}{2}d\theta \\
&= \left[ \frac{\sin(2\theta)}4+\frac{\theta}2\right]_0^{\frac{\pi}2} \\
&=\frac{\pi}4
\end{align}
Remark:
You might want to check your integration steps in your working, it is counter-intuitive to me that you obtain low order terms.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\iint_{x^{2} + y^{2} < 1}\pars{x^{2} + y^{3}}\,\dd x\,\dd y & =
{1 \over 2}\iint_{x^{2} + y^{2} < 1}r^{2}\,\dd x\,\dd y =
{1 \over 2}\int_{0}^{2\pi}\int_{0}^{1}r^{2}\,r\,\dd r\,\dd\theta =
\pi\int_{0}^{1}r^{3}\,\dd r\,\dd\theta
\\[5mm] & = \bbx{\pi \over 4}
\end{align}

$$\mbox{Note that}\quad
\left\{\substack{\ds{\iint_{x^{2} + y^{2} < 1}y^{3}\,\dd x\,\dd y\,\,\,
                     \stackrel{y\ \mapsto\ -y}{=}\,\,\,
                     -\iint_{x^{2} + y^{2} < 1}y^{3}\,\dd x\,\dd y
                     \implies
                     \iint_{x^{2} + y^{2} < 1}y^{3}\,\dd x\,\dd y =
                     \color{#f00}{0}}
                     \\[1cm]
                  \ds{\iint_{x^{2} + y^{2} < 1}x^{2}\,\dd x\,\dd y\,\,\,
                     \stackrel{x\ \leftrightarrow\ y}{=}\,\,\,
                     \iint_{x^{2} + y^{2} < 1}y^{2}\,\dd x\,\dd y}
                     \\[3mm] \ds{\implies
                     \iint_{x^{2} + y^{2} < 1}x^{2}\,\dd x\,\dd y
                     =
                     {1 \over 2}\iint_{x^{2} + y^{2} < 1}
                     \pars{x^{2} + y^{2}}\,\dd x\,\dd y =
                     {1 \over 2}\iint_{x^{2} + y^{2} < 1}r^{2}
                     \,\dd x\,\dd y}}\right.
$$
