$$I=\int_0^1\int_0^\sqrt{1-x^2}\sqrt{1-y^2}\space dydx$$ After switching to polar coordinates, it becomes: $$I=\int_0^{\pi/2}\int_0^1r^2\cos\theta \space drd\theta=\int_0^{\pi/2}\frac{r^3\cos\theta}{3}\Bigg|_0^1 \space d\theta=\int_0^{\pi/2}\frac{\cos\theta}{3} \space d\theta=\frac{\sin\theta}{3}\Bigg|_0^{\pi/2}=\frac13(?)$$ since the region of integration is the first quadrant of the unit circle.
However, I believe that the integrand wasn't transformed properly. How $f(y)=\sqrt{1-y^2}$ can be expressed in polar?
Best Answer
Plug in $y=r\sin\theta$ to get
$$ \int_0^{\pi/2}\int_0^1 \sqrt{1-r^2\sin^2\theta}\ \ r dr d\theta $$
The inner integral evaluates to
$$ \int_0^1 r\sqrt{1-r^2\sin^2\theta}\ dr = -\frac13 \left(1-r^2\sin^2\theta\right)^{3/2}\Bigg|_{r=0}^{r=1} = \frac13 (1 - \cos^3\theta) $$
You can use the triple-angle formula to evaluate the remaining integral.