Moment of inertia about z-axis in spherical coordinates

[clairez93]

Homework Statement

Use spherical coordinates to find the moment of inertia about the z-axis of a solid of uniform density bounded by the hemisphere \rho=cos\varphi, \pi/4\leq\varphi\leq\pi/2, and the cone \varphi=4.

Homework Equations

I_{z} = \int\int\int(x^{2}+y^{2})\rho(x, y, z) dV

The Attempt at a Solution

I tried to convert that equation to cylindrical coordinates and got this (k representing density because it's uniform)

I_{z} = k \int^{2\pi}_{0}\int^{\pi/2}_{\pi/4}\int^{1}_{0}\rho^2 sin^{2}\varphi*d\rho*d\varphi*d\theta

Plugged that into my calculator and got:

\frac{k\pi(\pi+2)}{12}

The book answer is:
\frac{k\pi}{192}

What am I doing wrong?

 

[LCKurtz]

Homework Statement

Use spherical coordinates to find the moment of inertia about the z-axis of a solid of uniform density bounded by the hemisphere \rho=cos\varphi, \pi/4\leq\varphi\leq\pi/2, and the cone \varphi=4.

You mean \varphi = \pi / 4 for the cone.

Homework Equations

I_{z} = \int\int\int(x^{2}+y^{2})\rho(x, y, z) dV

The Attempt at a Solution

I tried to convert that equation to cylindrical coordinates and got this (k representing density because it's uniform)

I_{z} = k \int^{2\pi}_{0}\int^{\pi/2}_{\pi/4}\int^{1}_{0}\rho^2 sin^{2}\varphi*d\rho*d\varphi*d\theta

That doesn't look like cylindrical coordinates to me. But then, why would you want cylindrical coordinates anyway? Assuming that the \rho^2\sin^2(\phi)
is the moment arm, check your spherical coordinate dV.

Plugged that into my calculator and got:

\frac{k\pi(\pi+2)}{12}

The book answer is:
\frac{k\pi}{192}

What am I doing wrong?

Aside from getting the dV wrong, using a calculator?

[Edit] Looking closer your limits for \rho are also wrong.

 

[clairez93]

You mean \varphi = \pi / 4 for the cone.

Yes, sorry, typo.

That doesn't look like cylindrical coordinates to me. But then, why would you want cylindrical coordinates anyway? Assuming that the \rho^2\sin^2(\phi)
is the moment arm, check your spherical coordinate dV.

Sorry, typo, I meant spherical coordinates.

I checked and dV should be

\rho^2\sin^2(\phi)d\rho*d\varphi*d\theta

So that should change the integral to:

I_{z} = k \int^{2\pi}_{0}\int^{\pi/2}_{\pi/4}\int^{1}_{0}(\rho^2 sin^{2}\varphi)^{2}*d\rho*d\varphi*d\theta

Aside from getting the dV wrong, using a calculator?

I usually use my calculator to check my setup, then once I know that is right, I go back and evaluate it by hand.

[Edit] Looking closer your limits for \rho are also wrong.

I'm not sure what to do for \rho, I thought since the radius of the hemisphere was 1, then \rho would go from 0 to 1.

 

[LCKurtz]

I checked and dV should be

\rho^2\sin^2(\phi)d\rho*d\varphi*d\theta

The sine should not be squared.

So that should change the integral to:

I_{z} = k \int^{2\pi}_{0}\int^{\pi/2}_{\pi/4}\int^{1}_{0}(\rho^2 sin^{2}\varphi)^{2}*d\rho*d\varphi*d\theta

I'm not sure what to do for \rho, I thought since the radius of the hemisphere was 1, then \rho would go from 0 to 1.

Have you drawn a picture of the desired volume? Your sphere is not centered at the origin and its equation isn't \rho = 1.