I have the following question:
Let $A$ be the upper half of the disk centred at the origin with radius $\pi/2$. Use polar coordinates to calculate the double integral $I = \iint_A y\cos(x) dxdy$.
I have worked out the following limits: $0 \leq R \leq \pi/2$ and $0 \leq \phi \leq \pi$ where $x = R\cos(\phi)$ and $y = R\sin(\phi)$.
But how do I convert the $\cos(x)$ part to polar coordinates? It can't be $\cos(R\cos(\phi))$ surely? If it is, how would I integrate this?
When you convert you get $y = r \sin \phi$ and $\cos x = \cos (r \cos \phi)$ as you indicated. Now you have $$ \int r \sin(\phi) \cos(r \cos \phi) r\ dr\ d\phi $$ and you can substitute $u = r \cos \phi$ with $du = -r \sin \phi \ d\phi$ in the internal ($d\phi$) integral
UPDATE
So $u(0) = r$ and $u(\pi) = -r$ and you get $$ \begin{split} I &= \int_0^{\pi/2} \int_0^\pi r \sin(\phi) \cos(r \cos \phi) r\ d\phi\ dr\\ &= \int_0^{\pi/2} \int_{-r}^r \cos(u) r\ du dr \\ &= \int_0^{\pi/2} r \left[\int_{-r}^r \cos(u) du\right] dr \\ &= \int_0^{\pi/2} r \left[\sin(-r) - \sin(r)\right] dr \\ &= -2\int_0^{\pi/2} r \sin(r) dr \end{split} $$ which you can take by parts, differentiating $r$.