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