# How to solve a double integral with cos(x) using polar coordinates?

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
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 $$r$$.
• How can you set $u$ to have to variables ($R$ and $\phi$) in it? Does a single substitution with more than one variable work and if so how does the change in variable work? Dec 16, 2019 at 18:51
• @abcd12112144wdwd you are in the inner integral, and while integrating $d \phi$, $r$ is constant. Dec 16, 2019 at 18:59
• Of course - makes sense. Even with this substitution I'm stuck with $\frac{-Ru\cos(u)}{\sqrt{R^2 - u ^2}}$ inside the inner integral. Does a further substitution need to be made? Dec 16, 2019 at 19:11