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I want to compute the volume enclosed by the sphere $x^2+y^2+z^2\le4$ with $0\le z\le 1.$

If I use the rings method I get:

$$\pi \int_0^1 \left( \sqrt{4-z^2}\right)^2dz=\frac{11}{3}\pi $$

If I set up a triple integral in cylindrical coordinates I get:

$$\int_0^{2\pi} \int_0^1 \int_0^{\sqrt{4-z^2}} r\,dr\,dz\,d\theta=\frac{11}{3} \pi$$

So far so good, but I gan't get it done with a triple integral in spherical coordinates. I assume the second integral is wrong but I don't know why:

$$\int_0^{2\pi} \int_0^{\pi/6} \int_0^{2} \rho^2 \sin(\phi)\,d\rho\,d\phi\,d\theta=\left( \int_0^{2\pi}d\theta \right)\left( \int_0^{\pi/6}\sin(\phi) \,d\phi\right)\left( \int_0^{2} \rho^2 \,d\rho \right)\neq \frac{11}{3} \pi$$

I can't see where my calculations are incorrect. Please help!

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You should split the triple integral into two parts. $$V = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^{\frac{1}{cos \phi}} \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta+ \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_0^2 \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta $$ Note that the first one is the volume of a cone.

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  • $\begingroup$ Hey thanks a lot. Your comment about the first one being the volume of a cone made me realize why I was getting it wrong! $\endgroup$ – Francisco Apr 12 '14 at 21:16

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