Evaluate the triple integral $$\iiint_E (x^2 +y^2 +z^2) dxdydz $$
Where $E$ is the solid ellipsoid defined by $$(\frac xa)^2+(\frac yb)^2+(\frac zc)^2 \le 1$$
with a, b and c being positive-valued constants.
I changed the variables to $u=x/a, v=y/b, w=z/c$ , to give me $$\iiint_S \left(a^2u^2+b^2v^2+c^2w^2\right)abc\:dudvdw$$
with $S$ being defined by $u^2+v^2+w^2\le 1$
and converting to spherical coordinates, I get the following integral:$$abc\int _0^{\pi }\int _0^{2\pi }\int _0^1\left(a^2\rho ^4sin^3\phi \:cos^2\theta \:+b^2\rho ^4sin^3\phi \:sin^2\theta +c^2\rho ^4sin\phi \:cos^2\phi \right)\:d\rho \:d\theta \:d\phi$$which I evaluated to get $$\frac{4abc\pi}{15}(a^2+b^2+c^2)$$
Are my steps correct? I'm given I hint where I can use symmetry arguements to simplify integrations, however I'm not sure where there even is one.
Best Answer
Your answer is correct, and there is another solution used Gauss-Green theorem, also divergence theorem.
Firstly, changing variables to $ u=\frac{x}{a}, v=\frac{y}{b}, w=\frac{z}{c}$, we have \begin{align} &\quad\iiint_{B(0,1)}\left(a^2u^2+b^2v^2+c^2w^2\right)abc\:dudvdw\\&=abc\int_0^1\int_{\partial B(0,r)} r\,(a^2u,b^2v,c^2w)\cdot\frac{(u,v,w)}{r} dS\,dr\\&=abc\int_0^1\int_{B(0,r)} r(a^2+b^2+c^2 )\,dudvdw\, dr\\&=abc(a^2+b^2+c^2)\int_0^1\frac{4}{3}\pi r^4\,dr\\&=\frac{4abc}{15}(a^2+b^2+c^2)\pi. \end{align}