Setting up a triple integral with spherical coordinates

[krackedude]

Homework Statement

http://img28.imageshack.us/img28/7118/capturenbc.jpg

Homework Equations

x² + y² + z² = p²

http://img684.imageshack.us/img684/3370/eq0006m.gif

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.

Thanks for your help!

 

[Joffan]

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

 

[krackedude]

Well, it's a cone. Here's a picture:

http://img829.imageshack.us/img829/7083/graphi.png

 

[SammyS]

Homework Statement

http://img28.imageshack.us/img28/7118/capturenbc.jpg

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.

 

[Joffan]

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

 

[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?

 

[SammyS]

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?

"... 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 .

 

[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).