Spherical coordinates: volume bound by z=r andz^2+y^2+x^2=4

[Unemployed]

Homework Statement

Using spherical coordinares, find the smaller volume bounded by the cone z=r and the sphere z^2+y^2+x^2=4

Homework Equations

x^2+y^2+z^2=4 ; rho=2, z=rhocosphi

The Attempt at a Solution

Shot in the dark:

Tried function integrating (rho squared - rhocosphi) d-rho d-phi d-theta with these limits

rho=-2 to 2, phi=0 to p/4, theta=0 to pi/2

Got 2/3 pi squared

Have trouble visualizing, the function is probably wrong and possibly the formula..

help.

 

[LCKurtz]

Some things wrong. For one thing, by convention you don't want to consider negative values of ρ. Another, what is the spherical element dV?

Have you drawn a picture? That will help with the limits.

 

[chrisbaird]

I think you have a typo because z=r does not define a cone. In general, to do a regular three-dimensional volume calculation, the integrand is just one, and you integrate over all volume elements in the limits. The volume element in spherical coordinates is:

r^2 sin(theta) dr d-theta d-phi

 

[Unemployed]

Thanks. So the general formula is (rho^2 sin phi drho d-phi d-theta)

z=rho-cos-phi
r=rho-sin-phi
rho cos phi=rho-sin-phi

tan phi=1

So now with the limits rho from 0-2
phi from 0-1 or is it 0-pi/4?
and theta from 0-pi/2

I got 8/3 *(cos 0-cos 1)*pi/2

1.296

 

[LCKurtz]

I think you have a typo because z=r does not define a cone.

You might want to reconsider that.

 

[LCKurtz]

Thanks. So the general formula is (rho^2 sin phi drho d-phi d-theta)

z=rho-cos-phi
r=rho-sin-phi
rho cos phi=rho-sin-phi

tan phi=1

Yes, good.

So now with the limits rho from 0-2
phi from 0-1 or is it 0-pi/4?

But you just showed tan(φ) = 1 so what is φ?

and theta from 0-pi/2

No. Have you drawn a picture yet?

 

[Unemployed]

So phi is 0-pi/4
and theta is also pi/4?

 

[LCKurtz]

So phi is 0-pi/4
and theta is also pi/4?

Yes for phi. You are just guessing for theta until you draw a picture.

 

[Unemployed]

0 to pi?

 

[LCKurtz]

0 to pi?

Sorry. I won't play the guessing game. Draw the dang picture.

 

[Unemployed]

i meant 2pi

 

[Unemployed]

Sorry. I won't play the guessing game. Draw the dang picture.

I did draw a picture. It is still very hard to visualize. I am guessing 2pi because the cone is centered on the origin and it is the positive x-z plane. Please help me.

 

[LCKurtz]

If you have java running on your computer, take a look at this link. It will help you visualize what varying θ and φ do. The only thing is, since it apparently a physics application, they reverse θ and φ from the standard mathematics notation. So click on the slider to vary φ to see what the results would be varying θ in your problem.

http://www.pha.jhu.edu/~javalab/spherical/spherical.html