requesting help for converting from Cylindrical to Spherical

Jan 2017
209
3
Toronto
\(\displaystyle
\int_{0}^{2pi} \int_{0}^{sqrt(a^2-b^2)} \int_{b}^{sqrt(a^2-r^2)} r dz dr d \theta
\)

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

Attachments

Jan 2017
209
3
Toronto
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.
 
Dec 2012
850
310
Hong Kong
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.
 
Jan 2017
209
3
Toronto
\(\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{b\sec \phi}^{a} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta\)

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...

\(\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{0}^{a * cos{(\phi)} - b} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta\)
 
Dec 2012
850
310
Hong Kong
\(\displaystyle \int_0^{2\pi} \int_0^{\cos^{-1}(b/a)} \int_{b\sec \phi}^{a} \rho^2 \sin \phi \mathop{}\mathrm{d}\rho \mathop{}\mathrm{d}\phi \mathop{}\mathrm{d}\theta\)

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) :p 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:

 
Jan 2017
209
3
Toronto
I thought we are trying to calculate the volume above z=b, not under.
 
Dec 2012
850
310
Hong Kong
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.