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:
- $x=r\cos(\theta)$
- $y=r\sin(\theta)$
- $dA=r\,dr\,d\theta$
- $D=\{(x,y)\mid x^2+y^2\leq 25\}$ 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.
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.