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I am trying to find the volume of the ellipsoid $\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 + \left(\frac{z}{c}\right)^2=1$ by making the substitution $u=x/a$, $v=y/b$ and $w=z/c$.

With this substitution, the equation becomes $u^2+v^2+w^2=1$. Projecting this on the $uv$-plane, we get a circle of radius $1$. Hence, the triple integral is

$I=\int_{u=-1}^{1} \int_{v=-\sqrt{1-u^2}}^{\sqrt{1-u^2}} \int_{w=-\sqrt{1-u^2-w^2}}^{\sqrt{1-u^2-w^2}} abc \, \mathrm{dw} \, \mathrm{dv} \, \mathrm{du}$.

The $abc$ comes from the Jacobian. Then, to evaluate this, I decided to switch to cylindrical coordinates. So, $u=r\cos\theta$, $v=r\sin\theta$, and $w=w$. This gave me

$I=\int_{\theta=0}^{2\pi}\int_{r=0}^{1}\int_{w=-\sqrt{1-r^2}}^{\sqrt{1-r^2}} abc r \, \mathrm{dw} \, \mathrm{dr} \, \mathrm{d\theta}$, where the $r$ comes from the Jacobian of the second transformation.

I then evaluated this:

\begin{align*} I &= \int_{\theta=0}^{2\pi}\int_{r=0}^{1}\left[abcrw\right]_{w=-\sqrt{1-r^2}}^{\sqrt{1-r^2}} \, \mathrm{dr} \, \mathrm{d\theta} \\ &= \int_{\theta=0}^{2\pi}\int_{r=0}^{1} 2abcr\sqrt{1-r^2} \, \mathrm{dr} \, \mathrm{d\theta} \\ &= \int_{\theta=0}^{2\pi} \frac{2abc}{3} \, \mathrm{d\theta} \\ &= \frac{4\pi}{3}abc \end{align*}

My answer is correct, which I am happy about, but is my solution correct? Is it clear to understand? Also, how else could I calculate the volume? I think spherical coordinates could also be useful, but I don't know if it would be easier than using cylindrical coordinates.

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    $\begingroup$ Why would you think that cylindrical coordinates are any easier than spherical coordinates? Once you rescale from xyz to uvw, you get the unit sphere $\endgroup$
    – Moko19
    Jun 23 '20 at 14:03
  • $\begingroup$ I don't think cylindrical coordinates are easier than spherical coordinates. I just feel more comfortable using them, which is why I went with it (though this may be a bad practice for a math undergrad). However, I don't think using spherical coordinates would have an inherent advantage in this situation. Maybe I'm wrong. $\endgroup$
    – rmdnusr
    Jun 23 '20 at 14:06
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Scaling an axis just scales all volumes in proportion - that's pretty elementary. It also follows easily when you think about the Jacobian (which you know about). So once you've scaled to create the unit ball, the volume of the ellipsoid is just that volume times $abc$, the scaling factor. No need to integrate at all (assuming you know the volume of a sphere).

Your solution does demonstrate your ability to manipulate those coordinate changes. That's a skill that might come in handy when there are no shortcuts.

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