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Suppose a sphere with radius R is centered at the origin, whose cross section is as follows (R is the constant radius while r is variable):enter image description here then its volume can be easily calculated:

$V=\int_{-R}^{R}\pi r^2 dy = \int_{-R}^{R}\pi (R^2-y^2) dy = \cfrac{4 \pi R^3} {3}$

This can be intuitively understood as summing up thin cylinders with base radius $r$ and height $dy$. However, if I try to do find the surface area by doing something similar: sum up rings with radius $r$ and thickness $dy$, I end up with

$A=\int_{-R}^{R} 2 \pi r dy=\int_{-R}^{R} 2 \pi \sqrt{R^2-y^2} dy = \pi^2 R^2$

which is completely wrong. Why is that the same approach works for deriving the volume formula while results in a wrong answer for surface area? Can this approach be modified to make it work for surface area? Thanks!

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  • $\begingroup$ dy is wrong. Should be $ds=\sqrt{dx^2+dy^2}$. $\endgroup$ – herb steinberg Nov 2 '18 at 1:11
  • $\begingroup$ TL;DR: the little rings you’re adding up don’t approximate the surface area well enough. It’s the same sort of error that’s made in the “proof” that $\pi=4$. $\endgroup$ – amd Nov 2 '18 at 1:19
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    $\begingroup$ Try thinking about arclength first. Can you get the length of a slanted (but non-vertical) line segment by adding up only the $\Delta y$'s? $\endgroup$ – Ted Shifrin Nov 2 '18 at 2:08
  • $\begingroup$ See this answer math.stackexchange.com/a/1692595/72031 $\endgroup$ – Paramanand Singh Nov 2 '18 at 2:51
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For your surface area the thickness $dy$ or $dx$ does not count the slope of the surface.

You need to use $ds=\sqrt{1+(\frac {dy}{dx})^2}dx$

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This is famous high school integral's fail example. Although you sum up $\int 2\pi r dy$, cone is expressed like same $\int2\pi r dy$, why so.

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