# How to integrate over polar coordinates

Evaluate the following double integral by rewriting it in polar coordinates:

$\displaystyle\iint\limits_Dxy\,dA$, where $D$ is the disc with center at the origin and radius 5

I have very little understanding about how to do this. The most I know right now is the following:

1. $x=r\cos(\theta)$
2. $y=r\sin(\theta)$
3. $dA=r\,dr\,d\theta$
4. $D=\{(x,y)\mid x^2+y^2\leq 25\}$ or $D=\{(r,\theta)\mid r\leq 5\}$

It's given in the problem that $r=5$, so that's a start. I'm assuming then that my limits for $r$ is $0\leq r\leq 5$. But I have no idea how to define the limits for $\theta$. My guess would be $0\leq\theta\leq 2\pi$, but several examples with different regions seem to use $0\leq\theta\leq\pi$.

So here's part of the integral with missing limits on $\theta$:

$$\int\limits_{\alpha}^{\beta}\int\limits_{0}^{5}r^3\sin{\theta}\cos{\theta}\,dr\,d\theta$$

Is my limited understanding correct so far? How do I fill in the holes of this problem? I know how to integrate after I have the proper limits; I just don't know how to define the limits given the information I have.

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You are off to a good start. Limits of integration for $\theta$ are a full revolution $0$ to $2\pi$ for the disk of radius 5 centered at the origin. – hardmath Aug 1 '13 at 2:56
This looks really good! It seems like what you're having trouble with is the polar parametrization of the circle. You've got that the limits on $r$ should be $0$ to $5$, which is good. For the limits on $\theta$, intuitively, what angle does a circle sweep out? If our disc is parameterized by $x = r\cos\theta, y = r\sin\theta$, what must be the range on $\theta$? – Alex Wertheim Aug 1 '13 at 2:56
Well, as I had stated, my guess is $0\leq\theta\leq 2\pi$... that seems logical to me since that's one full revolution of a circle. But in one problem I saw on youtube, the domain was defined by $x^2+(y-1)^2=1$, and the limits on $\theta$ there was $0\leq\theta\leq\pi$... I don't see why that one was different. – agent154 Aug 1 '13 at 3:00
@agent154, If you draw the circle centered at $(0,1)$, you will see that $\theta$ does indeed lie between $0$ and $\pi$ unlike the original problem where the disk is centered at the origin. – Doctor Dan Aug 1 '13 at 3:04
@DoctorDan why is that though? – agent154 Aug 1 '13 at 3:08

That's exactly right! As far as the limits on $\theta$:

Imagine the radius of a circle "sweeping" out the upper limit on $\theta$. (The starting place for most problems is $\theta = 0$.) I'll call this upper limit $\beta$. See these animated GIFs on various upper limits:

$\beta = \frac{\pi}{2}$:

$\beta = \pi$:

$\beta = 2\pi$:

Since you want the whole disk of a given radius, we want $\beta$ to be $2\pi$. That is, we want to "sweep out" the whole circle.

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