Converting triple integral to spherical

[CINA]

Homework Statement

Evaluate this integral using spherical coordinates: http://img262.imageshack.us/img262/9361/xyzm.th.jpg

Homework Equations

http://img40.imageshack.us/img40/9508/conss.th.jpg

The Attempt at a Solution

http://img264.imageshack.us/img264/9457/attempt.th.jpg

Can anyone check and see if this is right? Specifically the limits of integration, that's what I'm having trouble with.

 

[fantispug]

Well the best way of checking the limits in a case like this is to draw the regions you are integrating over (ignoring the integrand).

What is the integration region of the first integral?

I'll help you with this one: x goes from 0 to 1, z goes from -sqrt(1-x^2) to sqrt(1+x^2) so these two integrals trace out a semicircle in the x-z plane (in the x>0 region). Since y goes from -sqrt(1-x^2-z^2) to sqrt(1-x^2-z^2) this rotates the semicircle along the y-axis to give a...


What is the second (transformed) integration region?

Are these the same? If so good, if not change the limits so they are.

 

[CINA]

I don't really get what you are saying.

I know that sqrt(1-x^2-z^2) and -sqrt(1-x^2-z^2) makes a sphere with radius 1, which can be turned into rho=1.

I just can't figure out what phi and theta run to. Since I think it's a sphere (right?) can I assume phi and theta run complete cycles? Is that right?

Can someone please tell me these two limits (and if rho does indeed =1) so that I can see if I'm doing this right?

 

[CINA]

Wait, if it's a sphere of rho=1, it then projects a shadow on the x-z plane of a circle with radius one, but the only area of interest is from x=0 to x=1.

So a semi-sphere that exists only in the positive-x?

Is that the shape and the correspondingly correct limits?

 

[lanedance]

hemi-sphere in positive x sounds good to me, now what do your spherical limits become...

 

[CINA]

rho from 0 to 1

phi from 0 to pi

and theta from 0 to pi

Right?

 

[CINA]

Right!?

 

[lanedance]

depends on your convention if phi is teh angle around teh horizontal & the x-axis aligns with phi = pi/2, which i think it does, then yes