Wikipedia:Reference desk/Archives/Mathematics/2023 February 2

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February 2

Setting up integrals for sphere volume and surface area

Dear Wikipedians:

As a little exercise I'm trying to set up integrals for calculating the volume and surface area of a classic sphere of radius R. My thought is to approximate the sphere as a series of flat circular discs along the h axis each of thickness dh. With this style of approximation I set up the following two integrals for the volume and surface area respectively:

$V=\int _{-R}^{R}\pi {\sqrt {R^{2}-h^{2}}}^{2}\,dh$

$S=\int _{-R}^{R}2\pi {\sqrt {R^{2}-h^{2}}}\,dh$

The first integral worked out perfectly for the volume. The second integral failed miserably for the surface area (had a π² term in the result). Now I'm wracking my brain over what went wrong with my second integral without peeking at the correct answer.

Any pointers from you would be much appreciated.

66.241.128.130 (talk) 14:44, 2 February 2023 (UTC)

• Go back one step. You derived those equations from the (more general and correct)

$V=\int _{h{=}-R}^{R}V(h,dh)$

$S=\int _{h{=}-R}^{R}S(h,dh)$

where $V(h,dh),S(h,dh)$ are the volume and surface of the elementary slice of sphere whose height is between h and h+dh (taken with the limit dh → 0).

Are you sure you used the correct expression for $S(h,dh)$? Try to draw what the surface of that elementary slice looks like. Tigraan^{Click here for my talk page ("private" contact)} 16:22, 2 February 2023 (UTC)

Thanks Tigraan. The surface of that elementary slice looks like a circular cylinder/ring of height dh and circumference 2πr where r is the radius at that elementary slice which is calculated from Pythagorean theorem from R and h. The surface area of the circular cylinder/ring is therefore 2πrdh and stacks of these cylindrical/ring surface areas added together in limit will then approach the surface area of the sphere. Or is that not the correct elementary slice for calculating the surface area? 66.241.128.130 (talk) 16:31, 2 February 2023 (UTC)

If you try that reasoning on finding the lengths of the side of a triangle by integrating up the height you'd find they are always equal to the height! NadVolum (talk) 16:36, 2 February 2023 (UTC)

Ohhhh, there is a slope! Okay let me try to account for the slope this time! Thanks for the pointer! 66.241.128.130 (talk) 16:47, 2 February 2023 (UTC)

(edit conflict) Or, more simply, consider the surface area(s) of the sides of a parallelepiped. Their slants should play a role.  --Lambiam 16:50, 2 February 2023 (UTC)

See also Lambert cylindrical equal-area projection.  --Lambiam 17:16, 2 February 2023 (UTC)

Thank you Lambiam for the parallelpiped and Lambert projection articles, I will have a read at them. 66.241.128.130 (talk) 18:27, 2 February 2023 (UTC)

Got it! The slanted height of the cylinder/ring is ${\frac {dh}{\cos \theta }}$. Therefore $S=\int _{-R}^{R}2\pi {\sqrt {R^{2}-h^{2}}}{\frac {\,dh}{\cos \theta }}=2\pi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\sqrt {R^{2}-(R\sin \theta )^{2}}}{\frac {\,d(R\sin \theta )}{\cos \theta }}=2\pi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}R^{2}\cos \theta \,d\theta =4\pi R^{2}$ 66.241.128.130 (talk) 17:57, 2 February 2023 (UTC)

It is so very mathematically interesting that the volume slice works without taking into account the slant/slope, whereas the surface area slice does NOT work without taking into account the slant/slope. 66.241.128.130 (talk) 18:28, 2 February 2023 (UTC)

It's because the error for omitting the slope is proportional to the slice thickness in the surface area calculation, but it's a lower order term in the volume calculation. So in the volume calculation, the error goes to zero.2600:4040:7B33:6E00:B965:2E7F:BEFB:DE77 (talk) 01:10, 4 February 2023 (UTC)

Thanks for this insight. It reminds me of the big-O and small-o notations that I've learned for calculus and computer science algorithms runtime analyses, vanishes in higher order stuff. 66.241.128.130 (talk) 21:18, 6 February 2023 (UTC)

By the way, a completely different way to find the surface area is to use $S={\frac {d}{dR}}V,$ which works in any number of dimensions (for hypersurface and hypervolume). For example, $2\pi R={\frac {d}{dR}}\pi R^{2}.$ This also works for regular polytopes, taking $R$ to be (for example) the inradius.  --Lambiam 17:13, 2 February 2023 (UTC)

Hahaha yes, I've noticed that the surface area of a sphere is the derivative of its volume for quite some time now. Thanks for pointing it out. 66.241.128.130 (talk) 18:19, 2 February 2023 (UTC)

Resolved