Volume between a sphere and cone using triple integral

[Smusko]

Homework Statement

Evaluate the volume inside the sphere a^2 = x^2+y^2+z^2 and the cone z=sqrt(x^2+y^2) using triple integrals.

Homework Equations

a^2 = x^2+y^2+z^2

z=sqrt(x^2+y^2)

The solution is (2/3)*pi*a^3(1-1/sqrt(2))

The Attempt at a Solution

I first got the radius of the circle of intersection between the cone and the sphere and equated it to a/sqrt(2). I also got the angle between the x-y plane to the intersection circle to pi/4

Then I tried a myriad of ways to set up the triple integral to get the volume, none that has been successful.

Basically I am having trouble figuring out what volumes I need to calculate and maybe add/subtract from each other to get the desired final volume.

For example, In single variable calculus when you want the area between two curves you just subtract the area under the lower curve from the area under the higher curve. But how do I get the volume between two 3-dim objects? Is there a similar straightforward way or does it vary from problem to problem?

*EDIT*
I changed the picture since the last one was of the surface z= x^2+y^2.
Hopefully I got the right one this time.

*Edit Nr 2*
Yep, I did away. The updated image turned out to be z = sqrt(x+y) so I just removed it to avoid further embarrassment, also I solved the problem with help form spamiam, much love.

Turned out I was just being stupid as usual.

 

[spamiam]

Have you learned spherical coordinates yet? It makes things a lot easier.

 

[Smusko]

Yes I have, also cylindrical coordinates and the general way to swap between domains.

 

[spamiam]

Great! Can you write out inequalities describing the region of integration in spherical coordinates?

 

[tiny-tim]

Welcome to PF!

Hi Smusko! Welcome to PF!

(have a square-root: √ and a pi: π and try using the X² icon just above the Reply box )

Evaluate the volume inside the sphere a^2 = x^2+y^2+z^2 and the cone z=sqrt(x^2+y^2) using triple integrals.
…
The bold lines are the two surfaces …

Nooo …

you've drawn the paraboloid z = x²+y² instead! ​

 

[Smusko]

@Spamiam:

The cirle in spherical:
0 < p < a
pi/4 < Phi < 3pi/4
0 < theta < 2pi

The cone in Cartesian:
sqrt(x² + y²) < Z < 0

Then switching to circular:
0 < r < a/sqrt(2)
0 < theta < 2pi
---------------------------------

@ tiny - tuim:
Thank you. I'm currently slightly euphoric from discovering such an awesome resource for education and discussion.

But now I'm in shock at my own idiocy. Need to figure out how this affects the problem, since I have used the right equations but imaged the wrong figure.

 

[spamiam]

I think everything should still work out, despite the faulty picture. But you have a mistake for the \phi inequality: remember that \phi is measured from the positive z-axis downward.

 

[Smusko]

So from 0< Phi <pi/2 should work, or from pi/4 < phi < pi/2 then take that iteration twice?

Still I don't know what to calculate. How do I get the the union of the two volumes?

 

[spamiam]

No...what was the angle between the cone and the z axis? Maybe this will help a bit:

 

[Smusko]

Thanks, that was a great video, though I knew how the coordinates worked. Although the music gave me seizures.

DANG I got it, solved the problem. Thanks a allot! :D . Phi went from 0 to pi/4.
AAARRRGHHH, I'm so stupid, this was so easy.
And now I realize the new picture also is wrong. I'm going to go shoot myself. Brb.

 

[spamiam]

Apologies for the music. Never hire Stephen Wolfram as a DJ. Did it help you figure out the limits for \phi at least?

EDIT: Great, you got it!