Triple Integral in Spherical Coordinates

[BrownianMan]

Evaluate the integral by changing to spherical coordinates.

Not sure how to go about figuring out the limits of integration when changing to spherical coordinates.

 

[psholtz]

Do you know how to express x, y and z as functions of spherical coordinates?

Start there...

 

[BrownianMan]

I get,

0 <= theta <= pi
0 <= phi <= pi/4
-2 <= rho <= 2

and for the integrand, I get rho^5 * cos(phi) * sin(phi).

Is this right?

 

[psholtz]

Your answer is close to what I get, but I would get different limits of integration.

For the transformation, I would use:

$$x = r \sin \theta \cos \phi$$

$$y = r \sin \theta \sin \phi$$

$$z = r\cos \theta$$

If you work out the "integrand", I get:

$$r^3 \cos \theta$$

but then we're integrating the volume element, which in spherical coordinates works out to: $$r^2\sin \theta dr d\theta d\phi$$, so the full quantity under the integral sign would be:

$$r^5 \sin \theta \cos \theta dr d\theta d\phi$$

which I think is what you got.

I would just say:

(a) I don't think you can have "negative" radii;
(b) I'm not sure where your phi = pi/4 comes from.

So check the limits of integration...

 

[BrownianMan]

Would it be,

0 <= theta <= 2pi
0 <= phi <= 2pi
0 <= rho <= 2

 

[psholtz]

The radius limits are correct.

We start at 0, and move out to a radius of 2.

For the angle limits, consider the following:

Start at the origin (0,0,0) and move up the z-axis to 2: (0,0,2).

Now, move the radius vector "down" (i.e, rotate it), until instead of pointing "up", it's now pointing along the "down" axis, i.e., it's at (0,0,-2)... Two questions:

(a) which angle did we just rotate through? (i.e., phi or theta?)
(b) by how many radians did we rotate through?

Now we've created a "half disk", but in order to get the full sphere, we still need to rotate it through the other angle. Soo... let's do that, and by how many radians do we rotate it this time?

You have the right idea, but I think using the limits that you're using, you'll get an answer that 2x too big.

Of course, I'm not sure it matters in this particular case since I think the integral works out to 0, and 2x0 is still 0.

But it's good to understand the fundamentals of why things are the way they are.