# Homework Help: Finding the volume of a sphere sliced by a plane

1. Apr 20, 2012

### tasveerk

1. The problem statement, all variables and given/known data
What is the volume of the region above the plane z = 2 and inside the sphere x2 +y2 +z2 = 9.

2. Relevant equations

3. The attempt at a solution
I am trying to use a spherical triple integral. I am mainly having trouble finding ∅.
This is what I have been trying ∫arccos(2/3)02∏32 ρ2sin∅dpdθp∅
The answer is 8∏/3. I am pretty sure there are errors in my setup

2. Apr 20, 2012

### sharks

You should plot the graph to help you visualize the volume and its boundaries.
Using spherical coordinates:
$$\iiint {\rho}^2 \sin \phi \,.d\rho d\phi d\theta$$

Then describe the region (the crucial step in determining the limits):
For $\phi$ and $\theta$ fixed, $\rho$ varies from $\frac{2}{\cos \phi}$ to 3
For $\theta$ fixed, $\phi$ varies from 0 to $\cos^{-1} \frac{2}{3}$
$\theta$ varies from 0 to 2∏

In order to get the exact answer, you need to work with the inverse angles accordingly, without using their decimal counterparts, which will obviously give you a decimal answer.

Last edited: Apr 20, 2012
3. Apr 20, 2012

### tasveerk

Would you mind explaining where you got 2/cos∅ from? Also, can you recommend a good utility to graph these functions?
Thanks!

4. Apr 20, 2012

### sharks

What is the definition of z in terms of spherical coordinates?
Example, x in terms of spherical coordinates is given by: $\rho \sin \phi \cos \theta$

Personally, i just draw them on paper, and i recommend you do the same, unless you're allowed to bring special graphing equipments with you in the exams.
Maybe you could try this: http://www.wolframalpha.com/
Although, i think someone else might be better able to advise you on finding the best software for graphing in 3D.

5. Apr 21, 2012

### ehild

The integration would be much easier in cylindrical coordinates. The volume element can be the thin slice of thickness dz at height z with radius ρ=√(9-z2).

ehild

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6. Apr 21, 2012

### sharks

Using ehild's suggestion, you would get:

$$\iiint rdrd\theta dz$$

Description of region:
For r and θ fixed, z varies from z=$\sqrt{(9-r^2)}$ to z=3.
For θ fixed, r varies from r=0 to r=√5
θ varies from 0 to 2∏

You should get the same answer.

7. Apr 21, 2012

### ehild

I meant the integral
$$V=\int_2^3{\int_0^{\sqrt{9-z^2}}{r \int_0^{2\pi}{d\theta}dr}dz}$$
which is equivalent to integrating the volume element dV=ρ2πdz with respect to z from z=2 to z=3 where ρ=√(9-z^2)

$$V=\pi\int_2^3{(9-z^2)dz}$$

ehild

Last edited: Apr 21, 2012