Is this double integral correct?

Question

The question is: what is $$\iint_D \sqrt{1-x^2-y^2}dxdy$$ Where the domain $D$ is defined by $xy<0$ and $1 \geq x^2+y^2$.

So, we can convert everything to polar coordinates and get:

$$\int_{-\pi/2}^0 \int_0^1 r\sqrt{1-r^2}drd\theta + \int_{\pi/2}^\pi \int_0^1 r\sqrt{1-r^2}drd\theta$$

Which gives us $\pi/3$... Correct? Or have I done something wrong? Because the answer I'm given is $\pi/6$.

Answer 1

It looks fine. I did the computations and I also got $\frac\pi3$.

Answer 2

You are correct in your evaluation, as $$\int_0^1 r\sqrt{1-r^2}dr=-\frac{1}{2} (2/3)(1-r^2)^{3/2}\bigg|_0^1=\frac{1}{3}$$ I suggest you look at the problem again, for if $D$ was over $x,y<0$ instead of $xy<0$, we'd multiply our answer to the above integral by $\pi/2$ instead of by $\pi$, thus getting $\pi/6$