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Let there be hollow hemisphere (there is no solid bottom) with radius $r$. Find the center of mass of the hollow hemisphere, assuming the mass distribution is uniform.

My attempt:

Let there be a hollow hemisphere centered at $(0,0,0)$ with the bottom plane of the hemisphere resting on the $x-z$ plane. Let it's total mass be M. By symmetry the center of mass of the hemisphere must lie on the $y$ axis.

Now let us divide the hemisphere into thin rings, starting from the bottom. Let $dA$ be the area of a thin ring, where is the center of mass is the ring's center. Also, let $dM$ be the mass of the ring.

From the diagram it is clear that:

$$x^2+r^2=R^2$$

$$dA=2\pi r\cdot dx \implies dA = 2\pi\cdot \sqrt{R^2-x^2}\cdot dx$$

Hence, $dM=\frac{M}{2\pi R^2} \cdot dA=\frac{M}{R^2} \cdot \sqrt{R^2-x^2} \cdot dx$

$$y_{com}=\frac{1}{M} \cdot \int dm \cdot x=\frac{1}{R^2}\int_0^Rx \cdot \sqrt{R^2-x^2} \cdot dx $$

Evaluating this gives $y_{com} = R/3$, which is proven wrong by a quick google search.

The method shown here uses an angle subtended by a small surface area on a ring, which after computation does give the correct answer of $R/2$. My question is, which step am I doing wrong? Why is the answer coming one way but not the other?

I did the same thing with a cone, and the same thing happened, if I approximated a hollow cone as a series of thin hollow cylinders on top of each other, I get the incorrect answer, but with the angle approach, it gives the correct answer. When is an approximation considered "good enough" mathematically? I would really appreciate an explanation.

Popular Power
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1 Answers1

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I think you want to say $y^2+r^2=R^2$ as you have assumed the base in xz-plane.

Now you write $dA = 2\pi r \ dy \ $ but that is surface area of a ring of width $dy$ on a cylinder of radius $r$, not on a spherical surface. This is the step where you have a mistake. The correct surface area of a ring of infinitesimal width $dy$ on a spherical surface is rather $dS = 2 \pi r\sqrt{(dr)^2+(dy)^2} = 2\pi r \left(\sqrt{\left(\cfrac{dr}{dy}\right)^2+1}\right) dy$

$ = 2 \pi r \left(\sqrt{\left(\cfrac{y}{r}\right)^2+1}\right) dy = 2 \pi R \ dy$

So the center of mass is $(0, \overline y, 0)$ where,

$ \overline y = \displaystyle \cfrac{1}{2 \pi R^2} \int_0^R 2 \pi R \ y \ dy$

Math Lover
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  • Can't we approximate a sphere as a series of stacked up cylinders? btw, thank you very much for the solution, it makes a lot of sense when I see that the hypotenuse (in this case $dS$ will always be greater than both $dr$ and $dy$, even in the case that they are infinitesimal. The only other thing I want to know is which "approximations" are valid. – Popular Power Aug 11 '21 at 17:54
  • @PopularPower this is a hollow hemisphere and so we are only considering the surface. Hence it cannot be considered as stacked up cylinders. The approximation that we are using is a triangle of height $dr$ and width $dy$ as you said and hypotenuse being the surface. – Math Lover Aug 11 '21 at 18:05
  • Thank you very much for your answer! I now understand why I was wrong, but is there a rigorous reason why? Like are there certain rules in calculus for approximations? – Popular Power Aug 11 '21 at 18:46
  • The straight line hypotenuse is approximation of arc length. This should help: https://tutorial.math.lamar.edu/classes/calcii/arclength.aspx – Math Lover Aug 11 '21 at 18:52
  • Thank you very much for the resource. I only have one doubt left. Approximating as cylinders does not work in the case of finding the surface area, why does it work when calculating the volume? It seems really counterintuitive to me. If it does not work here, why would it give the correct answer when calculating the volume? – Popular Power Aug 12 '21 at 04:16
  • When calculating volume you are stacking up discs. The part you are worried about is much smaller than the area of each disc. That is not the case when you are finding surface area. Area on one face of the disc is $ \pi r^2$ and on the other it is $\pi (r - dr)^2$. The width is $dy$ and the slant width is $\sqrt{(dr)^2 + (dy)^2}$. Please see if this helps - https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Apex)/07%3A_Applications_of_Integration/7.02%3A_Volume_by_Cross-Sectional_Area-_Disk_and_Washer_Methods – Math Lover Aug 12 '21 at 05:37
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    Thank you very much! All my doubts are cleared. So basically in case of approximation using cylinder, the error is first order, which is why the answer is not correct, while in case of volume, the error is 2nd order so it does not matter. – Popular Power Aug 12 '21 at 11:57