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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.

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    $\begingroup$ So the problem is to integrate $f$ over $R$? As it stands, I'm not sure what $R$ is -- I assume it's a set of points, like $R = \{(r, \theta) \mid 0 \le r \le \cos(\theta), ? \le \theta \le ?\}$, but the bounds on $\theta$ could be anything, as you've realized. Is there any more info in the problem? $\endgroup$ – Eli Rose -- REINSTATE MONICA Jan 2 '16 at 3:05
  • $\begingroup$ Sorry. I've added a bit more to the body. R is simply give as 0 ≤ r ≤ cos θ and the function f(x,y) is given as x. The question has the integral set up as the double integral over the area R, and the integrand as f(x,y) dx dy $\endgroup$ – user1801048 Jan 2 '16 at 3:11
  • $\begingroup$ I still don't understand, unfortunately. Do you see the problem I'm talking about? To specify a 2D region, you need to put bounds on both coordinates. $\endgroup$ – Eli Rose -- REINSTATE MONICA Jan 2 '16 at 3:14
  • $\begingroup$ According to the solutions: R is the region: disk of radius 1/2 centered at (1/2, 0). However, this information isn't given in the problem itself. The final answer is pi/8. $\endgroup$ – user1801048 Jan 2 '16 at 3:18
  • $\begingroup$ Ah, it's a polar plot! $\endgroup$ – Eli Rose -- REINSTATE MONICA Jan 2 '16 at 3:23
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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.

disk to be integrated over

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:

$$ \frac{\pi}{8} $$

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