I understand we can approach this problem under the change of variables:
$$x=au; y= bv; z=cw$$
Thus we get:
$$V= \iiint_R \,dxdydz = abc\iiint_S \,dudvdw$$
At this point the ellipsoid has become a sphere. Thus we could use spherical coordinates to compute the volume.
My issue is with the extremes of the integral; concretely with the $\theta$ angle. I would set up the integral like this:
$$\int_{0}^{\pi / 4} d\theta \int_{0}^{\pi / 2} d\phi \int_{0}^{1} dr$$
But the stated solution is:
$$\int_{0}^{\tan^{-1} (a/b)} d\theta \int_{0}^{\pi / 2} d\phi \int_{0}^{1} dr$$
My extremes make sense to me; it is just about visualizing a sphere and two intersecting planes. But $\tan^{-1} (a/b)$ confuses me.
What's wrong and why?
Best Answer
When you change variables, the plane $y=x$ becomes $bv=au.$ So your upper limit for $\theta$ isn't $\arctan 1=\pi/4$ but $\arctan(a/b). $