# Talk:Surface area

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Field: Geometry

## Two dimensions

Is it really correct that two dimensional structures such as triangles have "surface area" ? I do not think so, "surface area" is a three dimensional concept. Ar

I've moved the table of areas of plane figures to the talk page of "Area". Arcfrk (talk) 08:48, 11 March 2008 (UTC)

## Surface Area To Volume Ratios

There is a problem with the last section. It states that if you increase the radius the ratio decreases. However, if you change the units of measure, the ratio can increase with a larger radius. A radius of 100 meters has a SA:V ratio of .03, but a radius of 1 kilometer has a ratio of 3. Also, it should be clear that this is assuming cells have a spherical shape. —Preceding unsigned comment added by 70.188.231.137 (talk) 04:05, 30 March 2008 (UTC)

SA:V is measured in inverse distance units. It is not dimensionless. A sphere with a radius of 100 meters has a ratio of 0.03/meter while the sphere with a radius of 1 kilometer has a ratio of 3/kilometer = 3/(1000 meters) = 0.003/meter. Measuring in the same units, the sphere ten times larger has a ten times smaller ratio, as it should. This similarity law holds for any shape, not just spheres. In the case of cells the only assumption is that a big cell is the same shape as a little one. This is more or less true of cells. It is definitely not true of multicellular structures, which is why one can easily distinguish a mouse bone from an elephant bone even when the mouse bone is magnified to elephantine size. -Dmh (talk) 05:32, 23

And I am SMART —Preceding unsigned comment added by 66.112.37.98 (talk) 22:33, 2 March 2011 (UTC)

## What. The. Hell.

I came here to verify a formula, but I ended up stumbling upon a page a 4th grader could have written. What in the world happened to this article?

S lijin (talk) 01:54, 19 May 2009 (UTC)

I came here to verify a formula, but I ended up stumbling upon a page a professor could have written. What in the world happened to this article? i can not understand any of this, perhaps someone could submit something eaiser to understand Summer911 (talk) 05:32, 10 March 2010 (UTC)

## Moved from the article

Shape Area formula derivation
Sphere The surface area of a sphere is the integral of infinitesimal circular rings of width $dx$

The radius of the circular ring is $f(x) = \sqrt{r^2-x^2}$. The length of the circular ring is equal to $2\pi\cdot f(x)$
The width of the ring can be determined by using Pythagoras' formula for a rectangular triangle with side lengths $dx$ and $f'(x) \cdot dx$, which leads to $\sqrt{1+f'(x)^2}\,dx$
The infinitesimal surface area of the circular ring thus is equal to $2\pi f(x)\cdot \sqrt{1+f'(x)^2}\,dx$
The derivative of $f(x)$ is equal to $f'(x) = \frac{-x}{\sqrt{r^2-x^2}}$
The surface area of the sphere can be calculated as

$\int_{-r}^r 2\pi f(x)\cdot \sqrt{1+f'(x)^2}\,dx$ = $\int_{-r}^r 2\pi \sqrt{r^2-x^2} \cdot \sqrt(1+\frac{x^2}{r^2-x^2})\,dx = \int_{-r}^r 2\pi \sqrt {r^2}\,dx = 2\pi r \int_{-r}^r 1\,dx$

The antiderivative needed is the simple linear function $x$
Thus, the sphere surface area amounts to

Asphere = $2\pi r[r-(-r)] = 4\pi r^2$

## References recovered partially

http://web.archive.org/web/20120427201949/http://www.math.usma.edu/people/rickey/hm/CalcNotes/schwarz-paradox.pdf — Preceding unsigned comment added by 94.197.120.122 (talk) 22:10, 9 October 2013 (UTC)

http://web.archive.org/web/20111215152255/http://mathdl.maa.org/images/upload_library/22/Polya/00494925.di020678.02p0385w.pdf — Preceding unsigned comment added by 94.197.120.122 (talk) 22:19, 9 October 2013 (UTC)