Evaluating $\iint_R\big(x^2+y^2\big)\,dA$ 
Evaluate the following double integral:
  $$\iint_R\big(x^2+y^2\big)\,dA,$$
  where $R$ is the region given by plane $x^2+y^2\leq a^2$.

My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA 
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\ 
&=\int_{-a}^{a}\left(x^2y+\dfrac{y^3}{3}\right)_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dx\\ 
&=\dfrac{2}{3}\int_{-a}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx.
\end{align}
I can't go further from here, please help.

As has been hinted, polar coordinates are the way to go. But if you really need to continue, the first thing I would note is that your integrand is an even function of $x$ over a symmetric integral. Therefore, you can simplify a tad to
$$\dfrac{4}{3}\int_{0}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx. $$
Now, whenever I see something like $\sqrt{a^2-x^2},$ I think trig substitution. In this case, you draw a triangle with $a$ as the hypotenuse, and one of the other sides as $x$. The third side will be $\sqrt{a^2-x^2}.$ I'll just put $\theta$ as the angle opposite $x,$ so that we have the following:
\begin{align*}
\frac{x}{a}&=\sin(\theta)\\
x&=a\sin(\theta)\\
dx&=a\cos(\theta)\,d\theta\\
\frac{\sqrt{a^2-x^2}}{a}&=\cos(\theta).
\end{align*}
For $x$ going from $0$ to $a,$ we can write that $0\le\theta\le\pi/2.$ Our integral becomes
\begin{align*}
\dfrac{4}{3}\int_{0}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx
&=\dfrac{4}{3}\int_{0}^{\pi/2}a\cos(\theta)\cdot\left(2a^2\sin^2(\theta)+a^2\right)a\cos(\theta)\,d\theta\\
&=\frac{4a^4}{3}\int_0^{\pi/2}\cos^2(\theta)\left(2\sin^2(\theta)+1\right)d\theta.
\end{align*}
This integral breaks up into two pieces, each of which succumbs to the usual methods for integrating products of $\sin(\theta)$ and $\cos(\theta).$ Can you take it from here? The result should be 
$$\frac{4a^4}{3}\cdot\frac{3\pi}{8}=\frac{\pi a^4}{2}.$$
Check with polar version:
\begin{align*}
\iint_R\big(x^2+y^2\big)\,dA
&=\int_0^{2\pi}\int_0^a r^2\cdot r\,dr\,d\theta\\
&=2\pi\,\frac{r^4}{4}\bigg|_0^a\\
&=\frac{\pi a^4}{2},
\end{align*}
as required.
