# Setting up a triple integral using cylindrical & spherical coordinates

1. Jan 28, 2010

### calcuseless

1. The problem statement, all variables and given/known data

Inside the sphere x2 + y2 + z2 = R2 and between the planes z = $$\frac{R}{2}$$ and z = R. Show in cylindrical and spherical coordinates.

2. Relevant equations

$$\iiint\limits_Gr\,dz\,dr\,d\theta$$

$$\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta$$

3. The attempt at a solution

$$2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_?^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta$$
Is my upper limit for r correct? How do I find the lower limit for z?

$$\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_?^R\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta$$
How do I find the lower limit for rho?

2. Jan 28, 2010

### LCKurtz

Yes the upper limit is OK. Isn't the equation of the plane z = R/2? That would be z on the lower surface.

Again, the equation of that plane is z = R/2. Express that in spherical coordinates and solve it for $\rho$. You will find that $\rho$ depends on $\phi$, but that's OK.

3. Jan 31, 2010

### calcuseless

Ah, I didn't understand that concept. Thank you!

$$2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_\frac{R}{2}^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta$$
Is there supposed to be a 2 out in front? I don't know why I put it there and I don't think it is needed for this problem.

4. Jan 31, 2010

### LCKurtz

I don't know why you put it there either. Go with your instincts. If you don't think it should be there, don't put it there.
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