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I recently watched this video on deriving the surface area of a sphere by summing up the surface area of thin cylinders and wanted to see if a similar method worked for deriving the volume of a sphere by summing up the volume of thin cylinders.

However, I just can't get it to work. The radii of the cylinders is the same as when finding the surface area in the video ($r\cos\theta$), the height of the thin cylinders is $r\mathrm d\theta$ (a thin arc of the sphere) and then I integrate between $\dfrac{\pi}{2}$ and $0$.

However, my answer comes out as $\dfrac{\pi^2r^3}{2}$ and not the standard formula.

I can derive the formula with volumes of revolution and other ways but was wondering why this method seems not to work when it works for the surface area. Thanks for your help in advance.

Edit: my working.

Radius of sphere is $r$ and so radius of circular cross sections is $r\cos\theta$.

A very thin arc will be $r\mathrm d\theta$.

Area of any circular cross section will be $\pi(r\cos\theta)^2$.

Volume of thin cylinders will be $$\pi(r\cos\theta)^2(r\mathrm d\theta)=\pi r^3\cos^2\theta \mathrm d\theta$$

Integrating all the thin volumes of the thin cylinders from $0$ to $\frac{\pi}{2}$ gives:

$$\pi r^3\int_{0}^{\frac{\pi}{2}}\cos^2\theta \mathrm d\theta$$

I use the cosine double angle identity to solve it and my answer comes out as $\frac{\pi^2r^3}{2}$ (after doubling to get the whole volume of the sphere not just the hemisphere).

Edit 2:

I've solved the problem. I found the heights of the cylinders would approximately be $rd\theta cos\theta$ (the tiny arcs are approximately the hypotenuse of a tiny right angle triangle). This gives the integral as $$\pi r^3\int_{0}^{\frac{\pi}{2}}cos^3\theta d\theta=\frac{2\pi r^3}{3}$$ giving the volume of the whole sphere as $$\frac{4\pi r^3}{3}$$

Robby
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  • Please show us your work – Math Lover Dec 04 '21 at 10:51
  • I will edit my post above but really it's the same method as in the video but instead of using the circumference of a circle to find the surface area, I use the area of a circle to find the volume. – Robby Dec 04 '21 at 11:07
  • You cannot use the same method that you used for surface area to find volume. Once you update your work, will see what you are exactly doing. – Math Lover Dec 04 '21 at 11:12
  • Thanks, I've updated it but why can't I use the same method? If, for finding surface area, I can just add up the surface areas of thin cylinders with height $rd\theta$, then why can't I do the same for volume by summing up the volumes of the same thin cylinders? – Robby Dec 04 '21 at 11:28
  • That's because if you take area parallel to xy-plane and integrate over $dz$, note that $dz \ne r d\theta$. $r d\theta$ is the slant length on the surface of the sphere. For volume, you are interested in the height. – Math Lover Dec 04 '21 at 11:41
  • Thanks for that. However, I'm confused as to why the method works for the surface area since I use $rd\theta$ for the height of the cylinders to add up the surface areas of the thin cylinders. – Robby Dec 04 '21 at 11:45
  • because when you are finding surface area, you are interested in slant height on the surface which is $r d\theta$. There are many questions on the site that explain this. I will search and share a link in sometime. – Math Lover Dec 04 '21 at 11:48
  • https://math.stackexchange.com/questions/1691147/why-is-surface-area-not-simply-2-pi-int-ab-y-dx-instead-of-2-pi-in – Math Lover Dec 04 '21 at 15:21
  • https://math.stackexchange.com/questions/750406/volume-vs-surface-area-integrals – Math Lover Dec 04 '21 at 15:21
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    Thanks for that. I've solved the problem using your logic. I found the heights of the cylinders would approximately be $rd\theta cos\theta$. This gives the integral as $$\pi r^3\int_{0}^{\frac{\pi}{2}}cos^3\theta d\theta=\frac{2\pi r^3}{3}$$ giving the volume of the whole sphere as $$\frac{4\pi r^3}{3}$$ – Robby Dec 04 '21 at 17:16

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