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I'm trying to find how I should consider the triple integration of a solid volume under a sphere, which is not centered at the origin. Are the limits of integration going to be same as if it would be centered or is there some transformation you need take into accounts before proceeding? I don't know if I did this right but I tried to evaluate with the unit sphere (Considering the basic equation of 4/3 pi r^3 = 4.18879) and I found the same result using a triple integrals but somehow my teacher says I'm wrong and I can't proceed that way. I can't figure out why? I need to know where I'm going on to go on with my assignement as I'm trying to evaluate the moment of inertia of a solid bounded between the unit sphere centered at (0,1,0) and the sphere of radius 2 centered at (0,2,0).

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  • $\begingroup$ It's easier if you let the $y$ and $z$ axes switch roles, so that the sphere is centered at (0,0,2) instead. Then you can proceed as in this question: math.stackexchange.com/questions/1978512/… $\endgroup$ – Hans Lundmark Jul 26 '17 at 7:51
  • $\begingroup$ If the sphere isn't centered at the origin, you can just move the origin to the ccenter of the sphere (make up new coordinates, like $abc$, instead of $xyz$). If you don't, spherical coordinates aren't going to help you much. $\endgroup$ – Arthur Jul 26 '17 at 8:05
  • $\begingroup$ What is your axis of rotation? $\endgroup$ – Christian Blatter Jul 26 '17 at 10:11
  • $\begingroup$ Well I'm thinking about the z-axis here since the solid I'm looking to integrate is centered at (0,2,0) and as a radius of 2 so if you look at the 0zy plane you are looking at one circle which goes from 0 to 4 (y) with a little circle inside that range from 0 to 1 (y). What I'm not sure to understand is that triple integral I did (in the picture). How come the output is still 4/3 pi r^3 if this isn't a good approach? $\endgroup$ – Anthony Charest Jul 26 '17 at 15:22
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It's unclear what your thoughts were, and the linked "figure" is not of much help. At any rate we have to compute the moment of inertia $\Theta$ of a solid ball of density $1$ wit respect to an axis through the center of the ball. For a ball of radius $R$ one obtains $$\Theta_R={8\pi\over15}R^5\ .\tag{1}$$ Now your solid is the difference of a ball of radius $4$ and a ball of radius $1$ whereby the large ball is offset by the distance $d=2$ from the axis of the rotation. This offset can be taken care of by the parallel axis theorem: The value $\Theta_4$ has to be augmented by $d^2$ times the volume of the ball. It follows that the moment of inertia $\Theta_{\rm solid}$ is given by $$\Theta_{\rm solid}=\bigl(\Theta_4+4\cdot{4\pi\over3}4^3\bigr)-\Theta_1=\pi\left({8\over15}\>4^5+{16\over3}4^3-{8\over15}\right)={13\,304\>\pi\over15}\ .$$ Proofs of the formula $(1)$ can be found at many places on the internet.

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  • $\begingroup$ Oh I am so sorry I think I had the wrong translation for my question.! :| I'm trying to find the center of mass. Not the moment of inertia. I have found the center [0,1.888,0].But my thought are about the triple integral I don't understand why it is wrong the way I worked my limit and still get the to same volume (About the unit sphere). Since in spherical coordinate, phi. angle swipes from north to south and theta from east to west I really thought that setting those boundary for the element of solid with their respectful rho would work. $\endgroup$ – Anthony Charest Jul 26 '17 at 16:16
  • $\begingroup$ Please change your question. I let my answer stand as long as the question asks for the moment of inertia. If it's just about the centroid you don't have to integrate anything. The problem can be solved by mere thinking. $\endgroup$ – Christian Blatter Jul 26 '17 at 16:34
  • $\begingroup$ Ok! I made a post. More clear (I hope so!)math.stackexchange.com/questions/2372521/… $\endgroup$ – Anthony Charest Jul 26 '17 at 16:55

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