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I am preparing for an exam and one of the practice questions (not homework, just for my benefit) asks to simply calculate $\iiint z^2dV$ over the ellipsoid $x^2+4y^2+z^2 \leq 1$. I'm not entirely sure how to take the limits of the integration. Could someone guide me through the process?

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4 Answers 4

This question uses a small trick using two changes of variables.

First do the change of variables $\tilde{y} = 2 y$. The Jacobian determinant is then $\frac12$. In other words

$$ \iiint_{x^2 + 4y^2 + z^2 \leq 1} z^2 \mathrm{d}x\mathrm{d}y\mathrm{d}z = \frac12 \iiint_{x^2 + \tilde{y}^2 + z^2 \leq 1} z^2 \mathrm{d}x \mathrm{d}\tilde{y} \mathrm{d}z $$

Next change to spherical coordinates, since you are now integrating in the sphere of radius one (with $x = r\sin\theta\cos\phi$, $y = r\sin\theta\sin\phi$ and $z = r\cos\theta$)

$$ \iiint_{x^2 + \tilde{y}^2 + z^2 \leq 1} z^2 \mathrm{d}x \mathrm{d}\tilde{y}\mathrm{d}z = \int_0^{2\pi}\int_0^\pi\int_0^1 r^2\cos^2\theta \cdot r^2\sin\theta \mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi = \underbrace{\frac15}_{r} \cdot \underbrace{2\pi}_{\phi} \int_0^\pi \cos^2\theta \sin\theta \mathrm{d}\theta $$

Now

$$ \int_0^\pi \cos^2\theta \sin\theta \mathrm{d}\theta = -\left.\frac13 \cos^3\theta \right]^\pi_0 = \frac23 $$

So we get that the integral evaluates to $\frac12 \frac15 2\pi \frac23 = \frac{2\pi}{15}$.

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When you change variables in a triple integral then your volume element is replaced as follows: $$dxdydz\to Jdrdudv$$ where $J$ is the Jacobian determinant. For spherical coordinates $$X\left(u,v\right)=\left(\begin{array}{c} r\sin u\cos v\\ r\sin u\sin v\\ r\cos u \end{array}\right)$$ Jacobian is given by the following formula $J=r^2\sin^2 u$. $r$ ranges from 0 to $$\rho\left(u,v\right)=\frac{1}{\left(\sin^{2}u\cos^{2}v+4\sin^{2}u\sin^{2}v+\cos^{2}u\right)}$$

Hence your integral shall be expressed as follows: $$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\rho\left(u,v\right)}r^{3}\sin u\cos^{2}udrdudv=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\sin u\cos^{2}u}{\left(\sin^{2}u\cos^{2}v+4\sin^{2}u\sin^{2}v+\cos^{2}u\right)^{2}}dudv $$

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First, I think you are missing something small from the volume differential (try drawing a picture...I think you need $\sin(\theta)$). From the spherical coordinates I think you end up with a restriction on only one of the three parameters. This means that you don't really have restrictions for the other two (ex. just have $0$ to $\pi$ for $\phi$). Also, when you integrate over one variable, you can keep the others constant. Hope this helps. =)

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You want to use spherical polar coordinates and you'll see it becomes quite doable.

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2  
I don't know how to apply this here... I'm not sure how the triple integral $(rcos(\phi))^2*rdrd(\phi)d(\theta)$ helps? –  Donnie Jun 3 '12 at 22:17

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