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.