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We are given a cone $z^2 = x^2 +y^2$ and between the spheres $x^2 + y^2 + z^2 = 1$ and $x^2 + y^2 + z^2 = 4$

Also the mass density is equal to $z$ and we are asked z coordinate of centroid of volume inside the cone and above the xy plane.

Now for z coordinate of centroid $$I = \iiint \rho z\,dx\,dy\,dz$$ divided by mass $M$.

So now translating to spherical coordinates we get $$I=\iiint \rho zr^2\sin\theta\, dr\, d\theta\, d\phi$$

As volume inside the cone is asked then z lies between $\sqrt{x^2 + y^2}$ and the plane $z = 2$. Evaluating that after putting the value of x and y in spherical coordinate and $r$ between 2 and 1, $\rho$ = z I am getting the wrong answer. Have I made a mistake somewhere?

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  • $\begingroup$ Where is the plane $z=2$ coming from ? $\endgroup$ – Yves Daoust Jul 8 '17 at 16:08
  • $\begingroup$ @YvesDaoust if we consider the sphere r = 2 then for $x^2 + y^2 = 4$ z =2. So the cone is capped by the plane z =2. $\endgroup$ – user127 Jul 8 '17 at 16:13
  • $\begingroup$ First, the intersection of the cone and the sphere is given by $x^2+y^2+z^2=z^2+z^2=4$ hence $z=\sqrt2$. Second, it is nowhere said that the cone has planar bases. $\endgroup$ – Yves Daoust Jul 8 '17 at 17:27
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Considering the following image, projection on the $yz$ plane

enter image description here

enter image description here

we have \begin{align} &\int_1^2dr\int_0^{\pi/4}d\theta\int_0^{2\pi}d\phi\,(r\cos\theta)r^2\sin\theta=\\ &2\pi\int_1^2dr\int_0^{\pi/4}d\theta\,r^3\sin\theta\cos\theta=\\ &2\pi\int_1^2dr\,r^3\left.\left(\frac{1}{2}\sin^2\theta\right)\right|_0^{\pi/4}=\\ &\frac{\pi}{2}\int_1^2dr\,r^3=\\ &\frac{\pi}{2}\left.\left(\frac{1}{4}r^4\right)\right|_1^2=\frac{15}{8}\pi\\ \end{align}

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  • $\begingroup$ Can you please explain why the limit is from 0 to $\pi / 4$ ? Also shouldn't it be $rcos \theta$ squared as $\rho = z$? $\endgroup$ – user127 Jul 8 '17 at 16:52
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    $\begingroup$ In spherical coordinates $z$ is exactly $r\cos\theta$, and the cone has its generating lines $\pi/4$ away from the vertical axis. $\endgroup$ – enzotib Jul 8 '17 at 16:59
  • $\begingroup$ Okay I get it but why isn't there a $z^2$ ? The mass density is also z . $\endgroup$ – user127 Jul 8 '17 at 17:27
  • $\begingroup$ @user127: this is indeed the calculation for the mass, as you noticed, you need an extra $z$ for the $z$ coordinate of the centroid. $\endgroup$ – enzotib Jul 8 '17 at 17:43

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