I'm asked to solve the following
$$ \int^2_0 \int^\sqrt{4-y²}_0 \sqrt{4-x^2-y^2} dxdy $$
I thought about using polar coordinates:
(1) $0 \le x \le \sqrt{4-y^2}$ is the upper half of a circumference with radius 2 and centered on the origin.
(2) $ 0 \le y \le 2 $ is the region between y=0 and y=2, which contains (1)
So
$-2 \le r \le 2$ because of (1)
$0 \le \theta \le \pi$ because of (1)
And since $dxdy \rightarrow rdrd\theta$ and $r^2 = x^2 + y^2$
$$\int^\pi_0 \int^2_{-2} \sqrt{4-r^2} r drd\theta$$
But $ \int^2_{-2} \sqrt{4-r^2} r dr = 0$ (using $u=4-r^2, du=-2r$ for the indefinite) so the whole thing is zero.
Which seems wrong - shouldn't the result be the area of the region limited by (1) and (2)? I then input the original integral on Wolfram|Alpha to check and it did not return zero. Any hints on what I'm doing wrong or what I should do?
Write your intergral as double integral $$\iint_X\sqrt{4-x^2-y^2}dxdy$$ and draw the picture of $X.$ Then you will get $$ X=\{(x,y)\mid x\geq 0,\ y\geq 0, x^2+y^2\leq 4\}=\{(r,\theta)\mid 0\leq r\leq 2,\ 0\leq\theta\leq\pi/2\} $$