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? 
 A: $$
   \int_0^1 \int_{-\pi}^\pi x\sqrt{1-x^2\sin^2(y)}\mathrm{d} y \mathrm{d} x = \int_0^1 x \left(\int_{-\pi}^\pi \sqrt{1-x^2\sin^2(y)} \mathrm{d} y \right) \mathrm{d}x = \int_0^1 x \left(4 \underbrace{\int_{0}^{\pi/2} \sqrt{1-x^2\sin^2(y)} \mathrm{d} y}_{\mathrm{E}(x^2)} \right) \mathrm{d}x = 4 \int_0^1 x \mathrm{E}(x^2) \mathrm{d}x = 2 \int_0^1 \mathrm{E}(u) \mathrm{d} u = \frac{4}{3}\left.\left((1+u) \mathrm{E}(u) + (u-1) \mathrm{K}(u)\right)\right|_{0}^1 = \frac{8}{3}
$$
where $\mathrm{E}(m)$ is the complete elliptic integral of the second kind and $\mathrm{K}(m)$ is the complete elliptic integral of the first kind.
A: 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
$$\int_0^1\!dx\int_{-\pi}^\pi \!dy\,x\sqrt{1-x^2\sin^2(y)} = \frac{8}{3}.$$
A: Integrate with respect to $y$ first $\displaystyle \int_0^1 \left (\int_{-\pi}^\pi x\sqrt{1-x^2sin^2(y)}dy\right)dx$, This $\displaystyle \int_{-\pi}^\pi x\sqrt{1-x^2sin^2(y)}d y = f(x)$  will give you a function $f(x)$, then you can compute $\displaystyle \int_0^1f(x)dx$.
