Mass of ellipsoid 
We are given the elipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1$$ with density function $$p(x,y,z)=x^2+y^2+z^2$$ Find the mass of the elipsoid.


I used the transformation $x=ar\sin\theta \cos\phi$, $y=br\sin \theta \sin \phi$ and $z=cr\cos\theta$. Spherical coordinates but I multiplied $x,y,z$ by $a,b,c$ respectively.
I calculate the jacobian of this transformation, it is $abcr^2\sin\theta$.
So the mass is $$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta drd\theta d\phi$$
But I'm having trouble calculating this integral. I thought maybe we can use that $$\sin^2x+\cos^2x=1$$ but because the coefficients are different it's difficult to use that here. How would I calculate this integral?
 A: $$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta drd\theta d\phi=$$
$$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}a^3bcr^4\sin^3\theta \cos^2\phi drd\theta d\phi+\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}ab^3cr^4\sin^3 \theta \sin ^2\phi  dr d\phi+\\\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}abc^3r^4\cos^2 \theta \sin\theta drd\theta d\phi=$$
$$a^3bc\int_{0}^{2\pi} sin^3\theta d\theta\int_{0}^{\pi}cos^2\phi d\phi\int_{0}^{1}r^4dr +ab^3c\int_{0}^{2\pi} sin^3 \theta d\theta \int_{0}^{\pi}\sin ^2\phi d\phi\int_{0}^{1}r^4dr +\\abc^3\int_{0}^{2\pi}cos^2 \theta \sin\theta d\theta\int_{0}^{\pi}d\phi\int_{0}^{1}r^4dr $$
A: Hint: There's absolutely no reason to fear this integral. It'll start to look less terrifying if you break it down into more manageable bitesize bits. Start by bringing constant factors outside the integral, distributing terms over parentheses, and breaking the integral of the sum down into a sum of integrals:
$$\begin{align}
M&=\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\
&=abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^4\sin^3\theta \cos^2\phi+b^2r^4\sin^3 \theta \sin ^2\phi+c^2r^4\sin\theta\cos^2 \theta)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\
&=abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^4\sin^3\theta \cos^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\
&~~~~+ abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(b^2r^4\sin^3 \theta \sin ^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\
&~~~~+ abc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(c^2r^4\sin\theta\cos^2 \theta)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi
\\
&=a^3bc\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin^3\theta \cos^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\
&~~~~+ ab^3c\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin^3 \theta \sin ^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi\\
&~~~~+ abc^3\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin\theta\cos^2 \theta)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi
\end{align}$$
Is that starting to look more manageable to you?

Hint #2: the sum of integrals arrived at above is not nearly as tedious as it looks. Notice that each of the multiple integrals you need to calculate is separable. For example, the first integral reduces to a product of one-dimensional integrals:
$$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(r^4\sin^3\theta \cos^2\phi)\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi = \left(\int_{0}^{2\pi}\cos^2\phi\mathrm{d}\phi\right) \left(\int_{0}^{\pi}\sin^3\theta\mathrm{d}\theta\right) \left(\int_{0}^{1}r^4\mathrm{d}r\right)\\
=I_1\,I_2\,I_3.$$
If you think carefully about the other two multiple integrals, you'll realize that you're with integration for this problem, since they just involve the factors $I_{1,2,3}$ again.
CORRECTION: oops, my last comment isn't exactly true. I misread the variables in the last integral. My apologies for any confusion.
A: You may integrate $\theta$ first, then $\phi$, and finally $r$.
$$I(r,\cos\theta,\phi):=(a^2r^2(1-\cos^2\theta) \cos^2\phi+b^2r^2(1-\cos^2 \theta) \sin ^2\phi+c^2r^2\cos^2 \theta)$$
$$J(r,\phi)=\int_{0}^{\pi}I(r,\cos\theta,\phi)\sin\theta d\theta=
\int_{-1}^{1}I(r,x,\phi)dx=\frac{2}{3}r^4(a^2 + b^2 + c^2 + (a^2 - b^2)\cos(2\phi))$$
$$K(r)=\int_0^{2\pi}J(r,\phi)d\phi=\frac{4}{3}(a^2 + b^2 + c^2)\pi r^4$$
$$L=\int_0^1 K(r)r^2 dr=\frac{4}{15}(a^2 + b^2 + c^2)\pi$$
Final result is 
$$\frac{4}{15}abc(a^2 + b^2 + c^2)\pi$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#66f}{\large\int_{{x^{2} \over a^{2}} + {y^{2} \over b^{2}} + {z^{2} \over c^{2}}\ \leq\ 1}
\pars{x^{2} + y^{2} + z^{2}}\,\dd x\,\dd y\,\dd z}
\\[3mm]&=\verts{abc}\int_{x^{2} + y^{2} + z^{2}\ \leq\ 1}
\pars{a^{2}x^{2} + b^{2}y^{2} + c^{2}z^{2}}\,\dd x\,\dd y\,\dd z
\\[3mm]&=\verts{abc}\int_{r\ \leq\ 1}\pars{%
a^{2}\,{1 \over 3}\,r^{2} + b^{2}{1 \over 3}\,r^{2} + c^{2}{1 \over 3}\,r^{2}}
\,\dd x\,\dd y\,\dd z
\\[3mm]&={1 \over 3}\verts{abc}\pars{a^{2} + b^{2} + c^{2}}\int_{r\ \leq\ 1}r^{2}
\,\dd x\,\dd y\,\dd z
\\[3mm]&={1 \over 3}\verts{abc}\pars{a^{2} + b^{2} + c^{2}}
\int_{0}^{1}r^{2}\pars{4\pi r^{2}}\,\dd r
=\color{#66f}{\large{4\pi \over 15}\verts{abc}\pars{a^{2} + b^{2} + c^{2}}}
\end{align}
