Transformation of a sphere and computing an integral by using sphere coordinates Let $V \subset\mathbb R^3$be the ellipsoid $$9x^2+4y^2+z^2≤36.$$ How can I express $V$ as a transformation of a sphere and how can I compute the sphere $$\int_v x^2\,d\lambda^3(x,y,z)$$ with sphere coordinates?
A little help would be much appreciated.
 A: let
$$x=2\,\rho\,\cos\theta\sin\phi$$
$$y=3\,\rho\,\sin\theta\sin\phi$$
$$z=6\,\rho\,\cos\phi$$
we have
$$\frac{\partial(x,y,z)}{\partial(\rho,\theta,\phi)}=36\rho^2\sin\phi$$
and
$$V=144\int_{0}^{2\pi }{\int_{0}^{\pi }{\int_{0}^{1}{{{\rho }^{4}}}}}{{\cos }^{2}}\theta \,{{\sin }^{3}}\varphi \,d\rho \,d\phi \,d\theta $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#f00}{\iiint_{9x^{2} + 4y^{2} + z^{2} < 36}x^{2}\,\dd x\,\dd y\,\dd z} =
\iiint_{x^{2} + y^{2} + z^{2} < 36}\
{x^{2} \over 3^{2}}\,{\dd x \over 3}\,{\dd y \over 2}\,\dd z =
{1 \over 54}\pars{\iiint_{r < 6}{r^{2} \over 3}\,\dd x\,\dd y\,\dd z}
\\[3mm] = &\
{1 \over 162}\int_{0}^{6}r^{2}\pars{4\pi r^{2}}\,\dd r
= \color{#f00}{{192 \over 5}\,\pi} \approx 120.6372
\end{align}
