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Find the volume of the solid that lies in the first octant above the cone $z=\sqrt{3(x^2+y^2)}$ and inside the sphere $$x^{2}+y^{2}+z^{2}=4z $$ using spherical coordinates:

So here is what I have done, I would honestly just like to make sure if I am doing this properly because normally the "cone" is written just as $z=\sqrt{x^2+y^2}$.

$$\int_{0}^{\frac{\pi}{6}} \int_{0}^{2\pi} \int_{0}^{4{\cos(\phi)}} {\rho}^2\sin(\phi) \,{\rm d}{\rho}\,{\rm d}{\theta}\, {\rm d}{\phi}$$

$$= 2\pi \int_{0}^{\frac{\pi}{6}} \sin(\phi) \, {\rm d}\phi \, \left[\frac{\rho^3}{3}\right]_0^{4\cos(\phi)} = 2\pi \cdot 64 \int_0^{\frac{\pi}{6}} \sin(\phi)\left(\frac{\cos^3(\phi)}{3}\right) \, {\rm d}\phi \\ = - 2\pi \cdot 64 \cdot \frac{1}{3} \left[\frac{\cos^4(\phi)}{4}\right]_0^{\frac{\pi}{6}} = - 2\pi \cdot 64 \cdot \frac{1}{3} \cdot \frac{1}{4}\left[\cos^4\left(\frac{\pi}{6}\right) - \cos^4(0)\right] = \left(\frac{14\pi}{3}\right) = \frac{14\pi}{3}$$

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  • $\begingroup$ You mean to find the volume enclosed by the finite region of the intersection of the cone $z = \sqrt{3(x^2 + y^2)}$ and the ellipsoid-ish body $x^2 + y^2 + z^2 = 4z$. Right? $\endgroup$ – amcalde Dec 7 '15 at 15:12
  • $\begingroup$ yes except its a sphere $\endgroup$ – juliodesa Dec 7 '15 at 15:15
  • $\begingroup$ Corrected the RHS of sphere equation. $\endgroup$ – Narasimham Dec 7 '15 at 15:54
  • $\begingroup$ it was supposed to say 4z @Narasimham $\endgroup$ – juliodesa Dec 7 '15 at 16:01
  • $\begingroup$ At two places OP states it is a sphere. $\endgroup$ – Narasimham Dec 7 '15 at 16:50
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Your problem has azimuthal symmetry so I can immediately write: $$I = \int\!\!\!\int\!\!\!\int_\Omega dxdydz =2\pi \int_{\theta=0}^{\theta_1}\int_0^{r(\theta)}r^2 \sin \theta d\theta= \frac{2\pi}{3} \int_{\theta=0}^{\theta_1}(r(\theta))^3 \sin \theta d\theta$$ The trick is to calculate what $\theta_1$ is (where do the surfaces intersect?) And also the function $r(\theta)$ which we can do via direct substitution.

$$x^2+y^2+z^2 = 4z$$ becomes $$r^2=4r\cos \theta$$ So we have $$r(\theta) = 4\cos \theta$$ The angle is calculated: $$\theta_1=\arccos\frac{z}{r} = \arccos\frac{\sqrt{3(x^2+y^2)}}{\sqrt{x^2+y^2+z^2}}=\arccos\frac{\sqrt{3}}{2}$$ So $$I = 2\pi \int_{\theta=0}^{\pi/6} \frac{(4 \cos \theta )^3}{3} \sin \theta d\theta = -\frac{128\pi}{3} \int_{\theta=0}^{\pi/6} (\cos \theta )^3 d(\cos \theta)$$

$$I=-\frac{32\pi}{3} [(\cos(\pi/6))^4 - (\cos 0 )^4] = \frac{32\pi}{3} \frac{7}{16} = 14\pi/3$$

So I get the same thing. Hopefully this is what you were looking for. Let me know if you have any other questions about my answer.

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  • $\begingroup$ yes!!!! Thank you very much. @amcalde $\endgroup$ – juliodesa Dec 7 '15 at 18:43

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