Calculate the integral $ \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3)\, dy dx .$ 
Calculate the integral $$ \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3) dy dx .$$

My attempt: Notice that $ x^2y^3 $ is an odd function with respect to $ y $, so \begin{align*} \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3)\, dy dx &= \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}( x^2 )\, dy dx \\
&=\int_{0}^{1} 2x^2\sqrt{1-x^2}\, dx\\
&= \int_{0}^{\frac{\pi}{2}}2\sin^2\theta\cos^2\theta \,d\theta\\
&= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin^22\theta \,d\theta\\
&= \frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-\cos 4\theta) \,d\theta\\
&= \frac{\pi}{8}.\end{align*}
Am I right? I think it's a little bit too complicated than what it should be?

It's fine.
I would perhaps compute it using polar coordinates:$$\int_{-\frac\pi2}^\frac\pi2\int_0^1r^3\cos^2(\theta)+r^6\cos^2(\theta)\sin^3(\theta)\,\mathrm dr\,\mathrm d\theta.$$Using the same argument as yours, this is just$$\int_{-\frac\pi2}^\frac\pi2\int_0^1r^3\cos^2(\theta)\,\mathrm dr\,\mathrm d\theta,$$which is equal to$$\left(\int_0^1r^3\,\mathrm dr\right)\left(\int_{-\frac\pi2}^\frac\pi2\cos^2(\theta)\,\mathrm d\theta\right)=\frac\pi8.$$
