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!

  • $\begingroup$ dy is wrong. Should be $ds=\sqrt{dx^2+dy^2}$. $\endgroup$ Nov 2, 2018 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, 2018 at 1:19
  • 1
    $\begingroup$ 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? $\endgroup$ Nov 2, 2018 at 2:08
  • $\begingroup$ See this answer math.stackexchange.com/a/1692595/72031 $\endgroup$
    – Paramanand Singh
    Nov 2, 2018 at 2:51

2 Answers 2


For your surface area the thickness $dy$ or $dx$ does not count the slope of the surface.

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.