# Why is this way of deriving surface area of sphere wrong when a similar method can be used to derive volume?

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): 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!

• dy is wrong. Should be $ds=\sqrt{dx^2+dy^2}$. – herb steinberg Nov 2 '18 at 1:11
• 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 $\pi=4$. – amd Nov 2 '18 at 1:19
• 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? – Ted Shifrin Nov 2 '18 at 2:08
• See this answer math.stackexchange.com/a/1692595/72031 – Paramanand Singh Nov 2 '18 at 2:51

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.