# Finding Area Between Cone and Sphere

## Homework Statement

Find the volume bounded between the sphere of radius a centered at (0,0,0) and the cone z=sqrt(x2 +y2).

## The Attempt at a Solution

So, subbing our definition for z into the the equation for a sphere of radius a centered at (0,0,0):

2x2 + 2y2 = a2. Converting to cylindrical coordinates,

2r2 = a2

r = a/sqrt(2). I use the bounds,

0≤z≤a
0≤θ≤2*pi
0≤r≤a/sqrt(2)

I use 1*r as my integrand to find volume and I get a3*pi/2. Am I doing this right?

I like Serena
Homework Helper
Hi TranscendArcu!

You have the upper bound of r as a constant. It isn't in cylindrical coordinates.
Furthermore, you have the upper bound of z as "a", but in that neighbourhood you should have an r that follows the sphere instead of the cone.

I recommend you do your calculations in spherical coordinates.
That's way easier.

Okay. Would my bounds be,

0≤rho≤a
0≤theta≤2*pi
pi/4≤phi≤3*pi/4

I'm not at all certain about any of these bounds since this is the first time I've done spherical coordinates. I'm most concerned about my bounds on phi. How does one actually find these bounds (I was just doing it intuitively, which invariably leads to incorrectness)?

Anyway, I would then integrate with my bounds of (rho)^2 * sin(phi) d(rho)d(theta)d(phi)

I like Serena
Homework Helper
To find the bound for phi, pick a point on the cone and calculate the phi that comes with it.

You should note that the definition of your cone implies that z>0.

Btw, your problem does not state which part you should have.
The part within the cone (and the sphere), or the part outside of the cone (and inside the sphere)?

Last edited:
So, Looking at the cone at letting (x,y) be (1,1). I find z=sqrt(2). I also see that (1,1,sqrt(2)) has distance sqrt(2) to the z-axis. If tan(phi) = sqrt(2)/sqrt(2), then phi = pi/4. This means that 0≤phi≤pi/4, right?

If I integrate I find that the volume within both the cone and the sphere is

2*pi*(a^3)/3 * (-sqrt(2)/2 + 1)

Still wrong, or better?

Presumably, if I wanted to find the volume outside the cone, but still within the sphere, I would merely change my bounds on phi. Would I have,

-pi/2≤phi≤pi/4

?

I like Serena
Homework Helper
Just checked.
I get the same!

I like Serena
Homework Helper
Presumably, if I wanted to find the volume outside the cone, but still within the sphere, I would merely change my bounds on phi. Would I have,

-pi/2≤phi≤pi/4

?

Not quite.
What is the range of phi?
And to what axis is it relative?

So I think phi is relative to the z-axis. I suppose that any point on negative z-axis has phi = pi and any point on the positive z-axis has phi = 0. So, I guess that means my new bounds for the other integral would have bounds pi/4≤phi≤pi. Is that right?

I like Serena
Homework Helper
Yep!