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By performing a suitable scaling and rotation of the coordinates, or otherwise, evaluate $\iiint_V \sin^2{(x + y + z)}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}x$ where $V$ is the region $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} < 1$ and $a,b,c$ are positive constants.

This is a question in a past exam paper that I am revising for.

My attempt at the question:

First scale the coordinates by making the change $(x,y,z) = (au,bv,cw)$, which means we are now integrating over a sphere, as opposed to an ellipsoid. The Jacobian for this transformation is abc and thus our new integral is $$abc\iiint_{\text{Unit Sphere}}\sin^2(au + bv + cw) \,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w$$

Now I need to rotate the coordinates so that our integral will be easier to evaluate once we eventually transform to spherical polar coordinates. We consider this as $$abc\iiint_{\text{Unit Sphere}}\sin^2((a,b,c)\cdot\mathbf{r}) \,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w$$ where $\mathbf{r}$ represents our coordinate system. Then by rotating the coordinates to form new coordinates $X,Y,Z$ in such a way that the $Z$ axis points in the direction of the vector $(a,b,c)$ the integral becomes $$abc\iiint_{\text{Unit Sphere}}\sin^2(\sqrt{a^2+b^2+c^2}\mathbf{e_3}\cdot\mathbf{r})\, \mathrm{d}V = abc\iiint_{\text{Unit Sphere}}\sin^2(Z\sqrt{a^2+b^2+c^2}) \,\mathrm{d}V$$

Now we change to spherical polar coordinates $(X,Y,Z) = (r\sin\theta\cos\phi, r\sin\theta\sin\phi,r\cos\theta)$ and our integral becomes $$abc\int_{r=0}^1\int_{\theta=0}^\pi\int_{\phi=0}^{2\pi} \left( \sin^2(r\cos\theta\sqrt{a^2+b^2+c^2})\cdot r^2\sin\theta\right)\,\mathrm{d}\phi\,\mathrm{d}\theta\,\mathrm{d}r$$

And it is at this point I become stumped. Does anybody have any pointers about where to go from here?

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    $\begingroup$ Perhaps try $u=\cos\theta$? (Or, which is more or less the same, use cylindrical coordinates instead of spherical.) $\endgroup$ – Hans Lundmark Apr 12 '15 at 13:40
  • $\begingroup$ I fail to see how using that substitution will help... And how could cylindrical coordinates help me integrate over a sphere? @HansLundmark $\endgroup$ – elDin0 Apr 12 '15 at 13:54
  • $\begingroup$ If $u=\cos \theta$, then $du = \sin \theta \, d\theta$ (which is part of the expression that you have). $\endgroup$ – Hans Lundmark Apr 12 '15 at 14:56
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Integrate with respect to the $x$ and $y$ coordinates first. You're integrating a constant, so you get a factor which is the area of the cross section for a fixed $z$, in other words $\pi (1 - z^2)$. So you are evaluating $$\pi abc\int_{-1}^1(1 - z^2)\sin^2(z\sqrt{a^2+b^2+c^2}) \,dz$$ This can now be done by standard calculus methods.

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