Polar coordinates integration Compute the following integrals  over $R$
$f(x,y)\,dx\,dy$ over the area $R$ where:
$f(x, y) = x$
 and
$R$ is given by $0 ≤ r ≤ \cos θ$ and $f(x, y) = x$.
I understand polar coordinates is probably the most suitable.
We can convert $f$ into  $r^2\cos(\theta) \,dr\,d\theta$.
The bounds for the $r$ variable is $0$ to $\cos(\theta)$.
I'm not too sure how to get the bounds for the theta variable. 
My first guess is that it's from $0$ to $2\pi$ (simply because whenever I did integration with polar coordinates, it was always integrating a whole circle, so from $0$ to $2\pi$. I'm not too sure what to do in this case). 
But in the solutions, they are integrating theta from $\frac{-\pi}{2}$ to $\frac{\pi}{2}$.
I'm having a hard time understanding why. 
 A: Anytime $r$ is given as a function of $\theta$ (or bounded by a function of $\theta$) you have a polar plot. You can plot a lot of shapes with simple equations relating $r$ and $\theta$. If you're doing problems like this, but haven't seen polar plots before, ask your professor; this isn't something you should "just know".
Here's the region (disk of radius $\frac{1}{2}$ sitting half a unit along the $x$-axis. This is not strictly necessary in order to do the integration, but I always find it helpful.

Our bounds are $0 \le r \le \cos(\theta)$, obviously, and also $-\pi/2 \le \theta \le \pi/2$, since $\theta$ needs to swing from the bottom to the top to trace out the region.
$$
\begin{aligned}
\iint_R f(x, y)\,dx\,dy &= \int_{-\pi/2}^{\pi/2} \int_0^{\cos(\theta)} r^2\cos(\theta)\,dr\,d\theta\\
&= \int_{-\pi/2}^{\pi/2} \cos(\theta)\left(\frac{r^3}{3} \bigg|_{0}^{\cos(\theta)}\right) \,d\theta\\
&= \frac{1}{3} \int_{-\pi/2}^{\pi/2} \cos^4(\theta)
\end{aligned}
$$
Here I appeal to Wolfram again, to integrate $\cos^4(\theta)$ on those bounds. It gives $\frac{3\pi}{8}$, so we have:
$$
\frac{\pi}{8}
$$
