Computing a double integral over $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$ 
Compute the integral $\int_{D} f \ dx \ dy$, when $f : D \to \mathbf{R}, f(x,y)= y^2x+x^3$ and $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$

It seems that $D$ defines a semicircle with radius $5$ in the $xy$-plane. So converting to polar coordinates I have $x=r\cos\theta, y=r\sin\theta$ and the limits for integration would be $r \in [0,5]$ and $\theta \in [0, \pi]$, thus the integral would become $$\int_{0}^{\pi} \int_{0}^{5} r^2\sin^2\theta \cdot r\cos\theta +r^3\cos^3\theta \ dr \ d\theta.$$
However I doubt that this is correct since seems to evaluate to zero?

Note that
$$
\begin{split}
 I
 &= \int_0^\pi \int_0^5
      \left(r^2\sin^2\theta \cdot r\cos\theta +r^3\cos^3\theta\right) r\ dr \ d\theta \\
 &= \int_0^5 r^4 dr
    \int_0^\pi \left(\sin^2\theta +\cos^2 \theta\right) \cos\theta  d\theta \\
 &= \frac{5^5}{5} \int_0^\pi \cos\theta  d\theta
\end{split}
$$
which does evaluate to zero since cosine is symmetric around $\pi/2$...

A different way to see it without going polar is the boundary curve (since $y>0$) is given by $y < \sqrt{25-x^2}$ and $f(x,y) = x(x^2+y^2)$, which is odd in $x$, and since we will integrate an odd function over a symmetric interval, the integral will be zero.
