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Use Spherical coordinates to evaluate the triple integral $$\iiint_{\mathrm{x^2+y^2+z^2<z}}\sqrt{x^{2}+y^{2}+z^{2}}\, dV,$$

What I tried

Converting $x^2+y^2+z^2<z$ to Spherical coordinates gives $\rho^2<\rho \cos\phi$, hence $\rho<cos\phi$. And the iterated integral are $$0<\rho<\cos\phi$$. Then let $\rho=0$ to give $0<cos\phi$. And solving for $\phi$, it becomes $$0<\phi<\pi/2 $$. While $$0<\theta<2\pi$$

My integral becomes$$ \int_{\ 0}^{2\pi} \int_{0}^{\pi/2}\int_{0}^{\cos\phi} \rho^{3} \sin{\phi} dpd\phi d\theta$$

and solving my final answer becomes $\pi/6$.

I'm unsure whether I am right though. Could anyone help. Thanks

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  • $\begingroup$ Did you remember the Jacobian? $\endgroup$ Commented Oct 22, 2014 at 11:53
  • $\begingroup$ Im unsure about the Jacobian. Could u explain. Thanks $\endgroup$ Commented Oct 22, 2014 at 11:56
  • $\begingroup$ $\int\int\int \quad f(r,\theta,\phi) \quad p^2 \sin\phi dr d\theta d\phi$ where $p^2 \sin\phi$ is your jacobian of transformation $\endgroup$ Commented Oct 22, 2014 at 11:58
  • $\begingroup$ Oh yes, i got that but i dont seem to be getting the right answer. $\endgroup$ Commented Oct 22, 2014 at 12:10
  • $\begingroup$ Can you edit in your description of the region? E.g: $\{x,y,z\}:$ blah blah, $\{r,\theta,\phi\}:$ blah blah $\endgroup$ Commented Oct 22, 2014 at 12:11

1 Answer 1

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Your bounds are correct. Your integral in spherical coordinates becomes...

$$ = \int_0^{2\pi} \int_0^{\pi/2} \int_0^{\cos(\phi)} \rho \cdot \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta $$

Since nothing involves $\theta$, that integral can be factored out.

$$ = \int_0^{2\pi}\,d\theta \int_0^{\pi/2} \int_0^{\cos(\phi)} \rho^3 \sin(\phi) \,d\rho \,d\phi = 2\pi \int_0^{\pi/2} \int_0^{\cos(\phi)} \rho^3 \sin(\phi) \,d\rho \,d\phi $$

$$= 2\pi \int_0^{\pi/2} \frac{1}{4}\cos^4(\phi) \sin(\phi) \,d\phi = 2\pi \cdot \frac{1}{4} \cdot \frac{-1}{5}\left(\cos^5(\pi/2) - \cos^5(0)\right) = \frac{\pi}{10}$$

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