I'm new to double integral and doing this problem: (from the Supplemental Problems, MIT)
Express each double integral over the given region R as an iterated integral in polar
coordinates. Use the method described in Notes I to supply the limits of integration. For
some of them, it may be necessary to break the integral up into two parts. In each case,
begin by sketching the region.
b) The circle of radius 1, and center at (0, 1).
What I have done so far:
Since I know the area of a radius 1 circle is $\pi$, the result must also be $\pi$.
I know the bounds must be from 0 to $\pi$. Then, I do the following double integral:
$$\int_0^{\pi}\int_0^{2sin\theta} 1 drd\theta $$
But the result is 4, which is a wrong answer.
Therefore, my question is, where is the mistake? I do try to replace the $drd\theta$ term to $rdrd\theta$ and it gives me the correct answer, which is $\pi$.
But there are some problems, which do not use the $rdrd\theta$ term. For example, this problem:
a) The region lying inside the circle with center at the origin and radius 2. and to the left of the vertical line through (−1, 0).
The answer of this problem is:
$$\int_{2\pi/3}^{4\pi/3} \int_{-sec\theta}^2 1 drd\theta$$
which does not include a $rdrd\theta$ term
Furthermore, I will appreciate if anyone can help me to point out when to use the $drd\theta$ term and when to use $rdrd\theta$ term.
P/S: Sorry for my bad English. I'm not a native speaker.
Best Answer
As I mentioned in comments, when you convert area element from cartesian into polar, it should have Jacobian $r$ for the change of variable.
In both examples of yours, you should use the Jacobian $r$.
The first integral is,
$ \displaystyle \int_0^{\pi} \int_0^{2\sin\theta} r \ dr \ d\theta = \pi$
The second integral is,
$\displaystyle \int_{2\pi/3}^{4\pi/3} \int_{-sec\theta}^2 r \ dr \ d\theta = \frac{4\pi}{3} - \sqrt3$