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Moment of inertia about z-axis in spherical coordinates

  • Thread starter clairez93
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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 [tex]\rho=cos\varphi[/tex], [tex]\pi/4\leq\varphi\leq\pi/2[/tex], and the cone [tex]\varphi=4[/tex].

Homework Equations



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

The Attempt at a Solution



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

[tex]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[/tex]

Plugged that into my calculator and got:

[tex]\frac{k\pi(\pi+2)}{12}[/tex]

The book answer is:
[tex]\frac{k\pi}{192}[/tex]

What am I doing wrong?
 

Answers and Replies

  • #2
LCKurtz
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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 [tex]\rho=cos\varphi[/tex], [tex]\pi/4\leq\varphi\leq\pi/2[/tex], and the cone [tex]\varphi=4[/tex].
You mean [itex]\varphi = \pi / 4[/itex] for the cone.

Homework Equations



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

The Attempt at a Solution



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

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

Plugged that into my calculator and got:

[tex]\frac{k\pi(\pi+2)}{12}[/tex]

The book answer is:
[tex]\frac{k\pi}{192}[/tex]

What am I doing wrong?
Aside from getting the dV wrong, using a calculator?

[Edit] Looking closer your limits for [itex]\rho[/itex] are also wrong.
 
Last edited:
  • #3
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You mean [itex]\varphi = \pi / 4[/itex] 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 [itex]\rho^2\sin^2(\phi)[/itex]
is the moment arm, check your spherical coordinate dV.
Sorry, typo, I meant spherical coordinates.

I checked and dV should be

[itex]\rho^2\sin^2(\phi)d\rho*d\varphi*d\theta[/itex]

So that should change the integral to:

[tex]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[/tex]

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 [itex]\rho[/itex] are also wrong.
I'm not sure what to do for [itex]\rho[/itex], I thought since the radius of the hemisphere was 1, then [itex]\rho[/itex] would go from 0 to 1.
 
  • #4
LCKurtz
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I checked and dV should be

[itex]\rho^2\sin^2(\phi)d\rho*d\varphi*d\theta[/itex]
The sine should not be squared.

So that should change the integral to:

[tex]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[/tex]

I'm not sure what to do for [itex]\rho[/itex], I thought since the radius of the hemisphere was 1, then [itex]\rho[/itex] 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 [itex]\rho[/itex] = 1.
 

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