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$$\int_0^1\int_{-\pi}^\pi x\sqrt{1-x^2\sin^2(y)}\mathrm{d}y\mathrm{d}x$$

How do I solve this question here?

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Try switching the order of integration – Jean-Sébastien Oct 25 '12 at 18:10
For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Oct 25 '12 at 18:10
@Ting, Have you tried anything that did not work? What specifically is the part where you are stuck? – Justin Oct 25 '12 at 18:42
@Justin I'm actually not sure how to integrate the trigo functions. I've changed the order of integration, integrated it with respect to x, but got stuck at the integration with respect to y because of the trigo functions. – Ting Oct 26 '12 at 3:35

$$ \int_0^1 \int_{-\pi}^\pi x\sqrt{1-x^2\sin^2(y)}\mathrm{d} y \mathrm{d} x = \int_0^1 x \left(\int_{-\pi}^\pi \sqrt{1-x^2\sin^2(y)} \mathrm{d} y \right) \mathrm{d}x = \int_0^1 x \left(4 \underbrace{\int_{0}^{\pi/2} \sqrt{1-x^2\sin^2(y)} \mathrm{d} y}_{\mathrm{E}(x^2)} \right) \mathrm{d}x = 4 \int_0^1 x \mathrm{E}(x^2) \mathrm{d}x = 2 \int_0^1 \mathrm{E}(u) \mathrm{d} u = \frac{4}{3}\left.\left((1+u) \mathrm{E}(u) + (u-1) \mathrm{K}(u)\right)\right|_{0}^1 = \frac{8}{3} $$ 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.

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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 $$\int_0^1\!dx\int_{-\pi}^\pi \!dy\,x\sqrt{1-x^2\sin^2(y)} = \frac{8}{3}.$$

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i've actually tried this method... but i got stuck at the last part where i had to integrate the whole trigo expression over the full period – Ting Oct 26 '12 at 3:27

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 $\displaystyle \int_0^1f(x)dx$.

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