<s>[INST] Computing double integral $\ \iint \sqrt{4-r^2}r \ dr d\theta $ I am trying to solve the following integral:
$$\ 2 \int_0^\pi \int_0^{2\sin\theta} \sqrt{4-r^2} \ r \ dr d\theta $$
My attempt: 
$$\ 2 \int_0^\pi \int_0^{2\sin\theta} \sqrt{4-r^2} \ r \ dr \ d\theta \stackrel{t = 4-r^2}{=} -\int_0^\pi \int t^{\frac{1}{2}} \ dt d\theta = -\frac{2}{3} \int_0^\pi t^{\frac{3}{2}} \ d\theta = -\frac{2}{3} \int_0^\pi \left[(4-r^2)^{3/2}\right]_0^{2\sin\theta} \ d \theta= -\frac{2}{3} \int_0^\pi (4 - 4 \sin^2\theta)^{3/2} - 8 \ d\theta = - \frac{2}{3}\int_0^\pi 8(1-\sin^2\theta)^{3/2} - 8 \ d \theta $$
because $\ (1-\sin^2\theta) = \cos^2\theta$ and $\ (\cos^2\theta)^{3/2} = \cos^3\theta $
$$\ = -\frac{16}{3} \int_0^\pi \cos^3\theta - 1 \ d\theta = \frac{16\pi}{3}  $$ 
but I'm wrong somewhere along the way because the answer should be $\ \frac{16}{9} (3\pi -4) $ and with the limits I set I get the right answer using Wolfram integral calculator.
 [/INST] The error is due to assuming $(x^2)^{3/2}=x^3$ as opposed to $|x^3|$. This can be solved by noticing that $\cos^3x=|\cos^3x|$ for $x\in[0,\pi/2]$ and $-\cos^3x=|\cos^3x|$ for $x\in[\pi/2,\pi]$. Thus the integral should become $$2\times\left(-\frac{16}{3} \int_0^{\pi/2} \cos^3\theta - 1 \ d\theta\right)=\frac{16(3\pi-4)}9.$$
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