Solution
= Solution
The integrand separates into a radial and an angular factor, so
$$
\int_0^{\pi/2}\int_0^R r^2\cos\theta\sin\theta\,dr\,d\theta
=\left[\frac{r^3}{3}\right]_0^R
\left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2}
=\boxed{\frac{R^3}{6}}.
$$
The differential printed in the PDF is already $dr\,d\theta$. No further polar Jacobian is to be inserted into this given iterated <integral>.