Transformation of a sphere and computing an integral by using sphere coordinates

Question

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.

Answer 1

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 $$V=144\int_{0}^{2\pi }{\int_{0}^{\pi }{\int_{0}^{1}{{{\rho }^{4}}}}}{{\cos }^{2}}\theta \,{{\sin }^{3}}\varphi \,d\rho \,d\phi \,d\theta $$

Answer 2

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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}