Moment of inertia of a spherical segment

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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 [tex]z=1[/tex] and [tex]z=2[/tex] and has a density of [tex]cz[/tex]. We are asked to work in cylindrical coordinates.


Homework Equations


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

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



The Attempt at a Solution



Would this be the integral?

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

Thanks
 

Answers and Replies

  • #2
HallsofIvy
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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 [tex]z=1[/tex] and [tex]z=2[/tex] and has a density of [tex]cz[/tex]. 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 [tex]\rho=\sqrt{9-z^2}[/tex], is the following the right formula?[/quote]
Normally, there is no "[itex]\rho[/itex]" in cylindrical coordinates. Do you mean "[itex]r= \sqrt{9- z^2}[/itex]". Assuming the sphere has center at (0, 0, 0), yes, that is correct.

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



The Attempt at a Solution



Would this be the integral?

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

Thanks[/QUOTE]
Well, you don't have the "z" outside the integral: you should have
[tex]2\pi c\int_{z= 1}^2 z\left(\int_{r= 0}^{9- z^2} r^3 dr\right) dz[/tex]
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
964

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 [tex]z=1[/tex] and [tex]z=2[/tex] and has a density of [tex]cz[/tex]. 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 [tex]\rho=\sqrt{9-z^2}[/tex], is the following the right formula?
Normally, there is no "[itex]\rho[/itex]" in cylindrical coordinates. Do you mean "[itex]r= \sqrt{9- z^2}[/itex]". Assuming the sphere has center at (0, 0, 0), yes, that is correct.

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



The Attempt at a Solution



Would this be the integral?

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

Thanks
Well, you can't have the "z" outside the integral: you should have
[tex]2\pi c\int_{z= 1}^2 z\left(\int_{r= 0}^{9- z^2} r^3 dr\right) dz[/tex]
 
  • #4
173
0
Thanks, the z should have been iinside and it is centered at the origin

James
 

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