Calculating a nasty integral with polar coordinates

Question

I have a p.d.f., $\frac{3}{2\pi}\sqrt{1-x^2-y^2}, \text{ for all } x^2 + y^2 \le 0$. The question asks to find $P(x^2 + y^2 \le \frac{1}{2}).$

I cannot for the life of me figure this one out. I know the solution involves converting to polar coordinates (at least without sacrificing my sanity) but that is giving me fits. It also seems like the answer should be 1/8, but of course that won't get me any points (probability of landing in a sphere with radius 1/2 when selecting from a sphere with radius 1), but I'm not sure how to get there.

EDIT: The correct support is $x^2 + y^2 \le 1$. My apologies.

Answer 1

The integral is $$ \frac{3}{2\pi}\int_0^{2\pi}\int_0^{\frac{\sqrt{2}}{2}} r\sqrt{1-r^2}\mathrm dr \mathrm d\theta\\ =3\int_0^{\frac{\sqrt{2}}{2}} r\sqrt{1-r^2}\mathrm dr $$ where a u substitution will do.

Answer 2

$$I=\frac{3}{2\pi}\int_0^{2 \pi } \int_0^{\frac{1}{\sqrt{2}}} r \sqrt{1-r^2} \, dr \, d\theta$$

Integral $\int r\sqrt{1-r^2}\,dr=-\dfrac{1}{2}\int \left(-2r(1-r^2)^{1/2}\right)\,dr=-\dfrac{1}{2}\cdot \dfrac{1}{1+\frac12}(1-r^2)^{1+1/2}+C=\\=-\dfrac{1}{3} \left(1-r^2\right)^{3/2}+C$

and then $$\int_0^{\frac{1}{\sqrt{2}}} r \sqrt{1-r^2} \, dr=\frac{1}{12} \left(4-\sqrt{2}\right)$$

Finally $$I=\frac{3}{2\pi} \frac{1}{12} \left(4-\sqrt{2}\right)\int_0^{2\pi}\,d\theta=\frac{3}{2\pi} \frac{1}{12} \left(4-\sqrt{2}\right)(2\pi)=\frac{4-\sqrt 2}{4}=1-\frac{\sqrt 2}{4}$$