How do i solve this double integral $$\int_0^1\int_{-\pi}^\pi x\sqrt{1-x^2\sin^2(y)}\mathrm{d}y\mathrm{d}x$$
How do I solve this question here? 

Switch the order of integration.
Integrating first over $x$, we obtain ($u=1-x^2 \sin^2y$)
$$\int_0^1\!dx\, x \sqrt{1-x^2 \sin^2 y}
= \frac{1}{2 \sin^2 y} \int_{\cos^2y}^1\!du\,\sqrt{u}
= \frac{1}{3\sin^2 y} ( 1 - |\cos^3 y|).$$
What is missing is the integral over $y$. Using one of the standard method to integral rational functions of trigonometric function over a full period, we obtain finally
$$\int_0^1\!dx\int_{-\pi}^\pi \!dy\,x\sqrt{1-x^2\sin^2(y)} = \frac{8}{3}.$$
