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A hole of radius r is bored through the center of a sphere of radius $R > r$. Find the volume of the remaining portion of the sphere.


I'm not sure on how to approach this problem. What is the most difficult part is knowing how to set up the bounds of integration. Mainly, how does one set up the bounds either when regarding cylindrical shells or washers? How do we exactly arrive at this (is it the points in which we do the sum of the tiny elements of area $dA$)? Is there a "fool-proof" way of knowing the bounds?

I was trying to solve this problem using the washer method in terms of y, but I'm not sure on how to set up my bounds of integration.

Thank you very much for your help in advance.

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    $\begingroup$ If you used cylindrical coordinates, you can obtain the volume by just changing the limits of integration of the $r$ term. $\endgroup$ – Tpofofn Jan 31 '14 at 3:09
  • $\begingroup$ You might consider solving the following classical gem instead: Find the volume of the remaining portion of a ball after a hole of length 1 is bored through its center (neither $r$ nor $R$ are given). $\endgroup$ – Gil Bor Jan 31 '14 at 3:20
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First calculate the volume of the solid of revolution for the contour function $\sqrt{R^2-x^2}$ between $\pm\sqrt{R^2-r^2}$ and then subtract the volume of the cylinder with radius $r$ and height $\sqrt{R^2-r^2}$. You'll find that the volume of that ring is $$V=\frac{4}{3}\pi\sqrt{R^2-r^2}^3.$$ Now call $h=2\sqrt{R^2-r^2}$ the height of the ring, then surprisingly $$V=\frac{\pi}{6}h^3.$$

Well that means that the volume of the ring solely depends on the height of the ring and not on $R$. Keeping $h$ constant and increasing $R$, the volumes stays unchanged, only the thickness $R-r$ decreases. So any two rings of the same material sharing the same height share also the same mass.

Edit: Note that $V$ is also the volume of a sphere with radius $h/2$.

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The part removed will be a cylinder plus two spherical caps.

If the radius of the hole is $r$ and the radius of the sphere is $R$, then the height of the cap at the center will be $R - \sqrt{R^2 - r^2}.$ (The height of the cylinder, consequently, is $2 \sqrt{R^2 - r^2}$.)

So using the washer method for this cap volume:

$$V_C = \pi \int_{R - \sqrt{R^2 - r^2}}^{R} dx (R^2 - x^2) = \frac{\pi}{3}(r^2 - R^2)(\sqrt{R^2 - r^2} - 3R).$$

So the remaining volume would be

$$V_{left} = \frac{4 \pi R^3}{3} - 2V_C - 2 \pi r^2 \sqrt{R^2 - r^2}.$$

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