# requesting help for converting from Cylindrical to Spherical

#### zollen

$$\displaystyle \int_{0}^{2pi} \int_{0}^{sqrt(a^2-b^2)} \int_{b}^{sqrt(a^2-r^2)} r dz dr d \theta$$

I find it difficult to imagine the spherical coordinates with the above scenario.. Ant tips would be much appreciated. Thanks.

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#### zollen

Here is the original question:
Find the volume of the region B bounded above by the sphere x^2 + y^2 + z^2 = a^2 and below by the plane z = b, where a > b > 0

I have solved the problem with cylindrical coordinates, now I want to convert it to spherical coordinates.

#### 123qwerty

I think it's $$\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{b\sec \phi}^{a} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta$$, though I might be wrong - I haven't done these for a while.

#### zollen

$$\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{b\sec \phi}^{a} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta$$

I am not sure I understand your inner most limit
$$\displaystyle \int_{b\sec \phi}^{a}$$

This is what I come up, let me know if I got it wrong...

$$\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{0}^{a * cos{(\phi)} - b} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta$$

#### 123qwerty

$$\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{b\sec \phi}^{a} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta$$

I am not sure I understand your inner most limit
$$\displaystyle \int_{b\sec \phi}^{a}$$

This is what I come up, let me know if I got it wrong...

$$\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{0}^{a * cos{(\phi)} - b} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta$$
Could you mention how you got your limits? Cos I'm not sure about that (no pun intended) In particular, I'm not sure which triangle you got your cosine from.

In any case, your shape is bounded below by $$\displaystyle z = b$$, so I don't see how your lower limit can be 0.

Here's how I got mine:

#### zollen

I thought we are trying to calculate the volume above z=b, not under.

#### 123qwerty

I thought we are trying to calculate the volume above z=b, not under.
Yes, that's why I started integrating from the point $$\displaystyle b \sec \phi$$ and ended at $$\displaystyle a$$... That's the portion above z = b.

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