# Setting up a triple integral with spherical coordinates

1. Jan 6, 2012

### krackedude

1. The problem statement, all variables and given/known data
http://img28.imageshack.us/img28/7118/capturenbc.jpg [Broken]

2. Relevant equations

x2 + y2 + z2 = p2

http://img684.imageshack.us/img684/3370/eq0006m.gif [Broken]

3. The attempt at a solution

Using the relevant equations I converted the given equation to:

∫∫∫e(p3/2) * p2 * sin(∅) dp dθ d∅

Then the bounds of the first integral, dp, would be from 1 to 2, because of the radius of the spheres.

I cannot figure out what the bounds for the second and third integrals would be though.

Last edited by a moderator: May 5, 2017
2. Jan 6, 2012

### Joffan

What shape is defined by z2 = 2(x2 + y2) ?

3. Jan 6, 2012

### krackedude

Last edited by a moderator: May 5, 2017
4. Jan 6, 2012

### SammyS

Staff Emeritus
You have $x^2+y^2+z^2\le1\,.$

I think you mean $x^2+y^2+z^2\ge1\,.$

To find the limits of integration for θ and φ , you need to know how x, y, and z are related to ρ, θ, and φ.

Last edited by a moderator: May 5, 2017
5. Jan 6, 2012

### Joffan

OK , it's a cone, so which integration does that imply a limit for, in spherical coordinates?

6. Jan 7, 2012

### krackedude

So, normally for a cone would have a θ from 0 <= θ <= 2∏, but the question says first octant, so would that mean 0 <= θ <= ∏/4?

And ∅ would be: 0 <= ∅ <= ∏/4?

EDIT: wait, wouldn't θ be limited by 3x <= y <= 4x? So, then how could i find those limits?

Last edited: Jan 7, 2012
7. Jan 7, 2012

### SammyS

Staff Emeritus
"... but the question says first octant, so would that mean 0 <= θ <= ∏/4?"

No. This is three dimensions. The coordinate planes cut R3 into 8 Octants. In the first Octant x, y, and z are all positive.
Therefore, in the first Octant, 0 <= θ <= π/2 .​

The cone determines the limits for φ . z2 tan2(φ) = x2 + y2.

The more difficult task will be to express y=3x and y = 4x in spherical coordinates. But it's not all that difficult .

8. Jan 7, 2012

### Joffan

$\phi$ is the one I was looking for limited by the cone, yes. The limit angle is not $\pi/4$, though.

And the other two planes do indeed limit $\theta$. Just need a little trig in each case to work out the angles (or express them as inverse trig functions, I guess).