# Homework Help: Spherical coordinates find volume

1. Jul 18, 2008

### fk378

1. The problem statement, all variables and given/known data
Use spherical coordinates to find the volume of the solid that lies above the cone z= sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = z.

3. The attempt at a solution

I'm having trouble solving for rho (p). I know it starts from 0, and it reaches to the sphere and to the cone. But I don't know where to go from there.

2. Jul 18, 2008

### Knissp

Assuming you meant $$\rho_o$$, the question becomes one of finding $$\phi$$.
In this case, you have to look at your cone. Essentially, it says $$z = r$$, where r is in cylindrical coordinates. Now use the conversions to spherical, $$z=\rho cos(\phi)$$ and $$r = \rho sin (\phi)$$. Equating those gives you $$cos(\phi) = sin(\phi)$$, or $$\phi = \pi/4$$. So $$0<\phi<\pi/4$$.

Last edited: Jul 18, 2008
3. Jul 18, 2008

### fk378

It is a sphere, though. If you move the z over to the left side of the equation, complete the square, you get a sphere with center (0,0,1/2). Also, I need to find rho, not phi.

4. Jul 18, 2008

### Knissp

Sorry.
$$x^2 + y^2 +z^2 = \rho^2$$

$$z = \rho cos(\phi)$$

$$x^2 + y^2 +z^2 = z$$ implies $$\rho^2 = \rho cos(\phi)$$

or $$\rho = cos(\phi)$$

so $$0 < \rho < cos(\phi)$$

5. Jul 18, 2008

### fk378

Why are we just looking at the sphere though and not the cone boundary?

6. Jul 18, 2008

### Knissp

Because you said "below the sphere x^2 + y^2 + z^2 = z" which implies the sphere is the upper bound, so its value for $$\rho$$ is the maximum value that $$\rho$$ can take on, thus establishing the upper limit of integration. Hope that made sense :)

7. Jul 18, 2008

### fk378

That makes sense...but we are also looking at the cone too. And the value of p for the cone doesn't depend on theta...it is constant. How do we know if we need to divide the triple integral expression into 2 separate ones, one for each different p?

8. Jul 18, 2008

### Knissp

Okay, this took me a while to find, but here's an example of a time when you need to split the expression for different values of $$\rho$$.

There are several evident differences between your problem and this one. Most importantly, notice that your solid is "above the cone z= sqrt(x^2 + y^2)" and the example I show is enclosed by the planes. There is a uniquely different value of $$\rho$$ that the function takes on for different values of $$\phi$$ namely because the solid that is being integrated must satisfy either being inside a cone or inside a sphere.

Essentially, in the example I give, there are uniquely different values of $$\rho$$ for different values of $$\phi$$, whereas in your case, no such thing occurs. If you try to find a different $$\rho$$, you get what I did in an earlier post:
Notice that the $$\rho$$ cancels, thus showing that there is no other unique value to make different bounds of integration.

http://img528.imageshack.us/my.php?image=11fm5.jpg

Last edited: Jul 18, 2008