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I have to find the volume between the two spheres. Their equations are

$x^2+y^2+z^2=2$ and $x^2+y^2+(z-\frac1 2)^2=\frac 1 4$ for $z>0$

The first one has center $(0,0,0)$ with $r=\sqrt 2$ and the second one $(0,0,\frac 1 2)$ with $r=\frac 1 2$.

After some fancy calculations I found that $0<\theta<2\pi, 0<\phi<\frac \pi 2, \cos\phi<r<\sqrt 2$

I had this exercise as homework and I tried to find the integration limits. I am not very sure about them cause it is the first time I try to apply spherical coordinates. I could just subtract the outer sphere from the inner one but I wanted to do it with one integral.

I forgot the result I found. It is $\frac{11\pi}{12}$

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Since the second sphere is entirely inside the first sphere, one should definitely use the difference of the volumes of the two enclosed balls. The result is $\frac43\pi(R^3-r^3)$ with $R^2=2$ and $r^2=\frac14$ hence your $\frac{11\pi}{12}$ is doubtful. –  Did May 14 '12 at 14:56
    
Integrals are very nice tools, but using integrals here is just too hairy. The difference method is much simpler, and perfectly correct to use in this case. –  Cameron Buie May 14 '12 at 15:56
    
@CameronBuie Hairy integrals? That's funny. –  jeanleauveux22 May 14 '12 at 16:27
    
By the way, the numerical result mentioned at the end of the post is incorrect. –  Did May 16 '12 at 17:58
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