Moment of inertia of a spherical segment

[bobred]

Homework Statement

Am I going about this the right way?
There is a sphere of radius 3 and a region that lies between the planes z=1 and z=2 and has a density of cz. We are asked to work in cylindrical coordinates.

Homework Equations

Let \rho=\sqrt{9-z^2}, is the following the right formula?

I=\displaystyle{\int_R}cz\rho^2\,dV where dV=\rho\,d\theta\,d\rho\,dz

The Attempt at a Solution

Would this be the integral?

I=2\pi c z\displaystyle{\int^2_1}\displaystyle{\int^{\sqrt{9-z^2}}_0}\rho^3\,d\rho\,dz

Thanks

 

[HallsofIvy]

Homework Statement

Am I going about this the right way?
There is a sphere of radius 3 and a region that lies between the planes z=1 and z=2 and has a density of cz. We are asked to work in cylindrical coordinates.

Are we to assume that the center of the sphere is at (0, 0, 0)?[/quote]

Homework Equations

Let \rho=\sqrt{9-z^2}, is the following the right formula?[/quote]
Normally, there is no "\rho" in cylindrical coordinates. Do you mean "r= \sqrt{9- z^2}". Assuming the sphere has center at (0, 0, 0), yes, that is correct.

I=\displaystyle{\int_R}cz\rho^2\,dV where dV=\rho\,d\theta\,d\rho\,dz

The Attempt at a Solution

Would this be the integral?

I=2\pi c z\displaystyle{\int^2_1}\displaystyle{\int^{\sqrt{9-z^2}}_0}\rho^3\,d\rho\,dz

Thanks[/QUOTE]
Well, you don't have the "z" outside the integral: you should have
2\pi c\int_{z= 1}^2 z\left(\int_{r= 0}^{9- z^2} r^3 dr\right) dz

 

[HallsofIvy]

Homework Statement

Am I going about this the right way?
There is a sphere of radius 3 and a region that lies between the planes z=1 and z=2 and has a density of cz. We are asked to work in cylindrical coordinates.

Are we to assume that the center of the sphere is at (0, 0, 0)?[/quote]

Homework Equations

Let \rho=\sqrt{9-z^2}, is the following the right formula?

Normally, there is no "\rho" in cylindrical coordinates. Do you mean "r= \sqrt{9- z^2}". Assuming the sphere has center at (0, 0, 0), yes, that is correct.

I=\displaystyle{\int_R}cz\rho^2\,dV where dV=\rho\,d\theta\,d\rho\,dz

The Attempt at a Solution

Would this be the integral?

I=2\pi c z\displaystyle{\int^2_1}\displaystyle{\int^{\sqrt{9-z^2}}_0}\rho^3\,d\rho\,dz

Thanks

Well, you can't have the "z" outside the integral: you should have
2\pi c\int_{z= 1}^2 z\left(\int_{r= 0}^{9- z^2} r^3 dr\right) dz

 

[bobred]

Thanks, the z should have been iinside and it is centered at the origin

James