I was given the following problem:

$$\iiint\limits_D (4x^2+9y^2+36z^2)\,dV,$$

where $V$ is the interior of the ellipsoid $$\frac{x^2}{9}+\frac{y^2}{4}+z^2=1.$$

The problem gives what the new coordinate system will be: \begin{align} x&=3\rho\sin\theta\cos\phi,\\ y&=2\rho\sin\theta\sin\phi,\\ z&=\rho\cos\theta. \end{align}

I don't really know why that would work. Let's take the ellipse on the $xy$ plane and polar coordinates: $$\frac{x^2}{9}+\frac{y^2}{4}=1;~~~~~~~~~~~~x=3r\cos\theta,~~y=2r\sin\theta.$$ How do I know that for every $\theta$ I will end up with a point on the ellipse?

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Moreover, how do I know that with the change of variables given by the problem I will end up with a point on the ellipsoid? I appreciate your thoughts.


If you take $r=1$ in your case (where you made the change of variables to polar coordinates), you know for every $\theta$ you will be on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ simply because for every $\theta$, the equation for that ellipse will be true.

$$ \begin{align*} 1&=\frac{x^2}{9}+\frac{y^2}{4}\\ &=\frac{\left(3\cos(\theta)\right)^2}{9}+\frac{\left(2\sin(\theta)\right)^2}{4}\\ &=\frac{9}{9}\cos^2(\theta)+\frac{4}{4}\sin^2(\theta)\\ &=\cos^2(\theta)+\sin^2(\theta)\\ &=1 \end{align*} $$

And if you vary $r$ (let $0\leq r<1$), you'll be inside that ellipse.

  • $\begingroup$ I don't know if this truly answered your question. Please tell me if (and how) it didn't. $\endgroup$ – Arturo don Juan Nov 11 '14 at 0:14
  • $\begingroup$ I thought about it before posting the question, but weren't very sure about it. Now I see that it is actually a valid argument. Thanks. $\endgroup$ – Vladimir Vargas Nov 11 '14 at 0:18

To solve the problem, one could try defining: \begin{align} \frac{x}{3}=u, && \frac{y}{2}=v. \end{align} Therefore $$\iiint\limits_D (4x^2+9y^2+36z^2)\,dV=\iiint\limits_{D^*} (36u^2+36v^2+36z^2)6\,dV^*.$$ Now taking the spherical transformation: $$216\iiint\limits_{D^{**}}\rho^4\sin\theta \,dV^{**} = \frac{864}{5}\pi.$$


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