What is the value of this double integral? Let $C$ be the subset of the plane given by 
$$ C \colon= \{ \ (x,y) \in \mathbb{R}^2 \ | \ 0 \leq x^2 + y^2 \leq 1 \}.$$ 
Then what is the value of the double integral 
$$ \int_{C} \int (x^2 + y^2) \ dx \  dy?$$ 
My work: 
In $C$, we have $-1 \leq x \leq 1$ and $-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$. So we have 
$$ \int_C \int (x^2 + y^2) \ dx \  dy = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \ (x^2 + y^2) \ dy \ dx = 2 \int_{-1}^1 (x^2 \sqrt{1-x^2} + \frac{ (\sqrt{1-x^2})^3}{3} ) \ dx = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx.  $$ Now let $x= \sin t$. Then we obtain
$$ \int_C \int (x^2 + y^2) \ dx \  dy = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx = \frac{4}{3} \int_{t = 0}^{\pi/2} (2\sin^2 t + 1) \cos^2 t \ dt \ \ = \frac{1}{3} \int_{t=0}^{\pi/2} (2\sin^2 2t + 4 \cos^2 t) \ dt = \frac{4}{3} \int_{t=0}^{\pi/2}  ( 1 - \cos 4t + 2(1+ \cos 2t) ) \ dt = \frac{1}{3} \int_{t=0}^{\pi/2} (3 + 2 \cos 2t - \cos 4t) \ dt = \frac{\pi}{2}. $$ Am I right? 

If you use polar coordinates you will get the following

$$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \  d\theta $$

