calculus - Evaluating $int _0^{pi }int _0^xsqrt{1-x^2}:dydx$

Evaluate:

$\int _0^{\pi }\int _0^x\sqrt{1-x^2}\:dydx$

I've gotten it done to:

$\int _0^{\pi }x\sqrt{1-x^2}dx$

Should I now change to polar coordinates because of the pi, or how should I proceed?

Answer

I think you made a typo, the upper limit of the integral looks more like $1$ instead of $\pi$ (otherwise the square root is not well defined), I will give my answer based on this correction.

---

Try to make the change of variable $x = \sin \theta$, the $dx = \cos\theta d\theta$, and the integral becomes
\begin{align}
& \int_0^1 x\sqrt{1 - x^2} dx \\
= & \int_0^{\pi/2} \sin\theta \cos \theta \cos\theta d\theta \\
= & -\int_0^{\pi/2} \cos^2(\theta) d(\cos\theta) \\
= & -\left.\frac{1}{3}\cos^3(\theta)\right|_0^{\pi/2} \\
= & \frac{1}{3}.
\end{align}