Talk:Surface area

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[edit] 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)

[edit] 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

== What. The. Hell. ==hi

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)

[edit] Moved from the article

Shape

Area formula derivation

Sphere

The surface area of a sphere is the integral of infinitesimal circular rings of width

d x

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 $d x$ 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

A_sphere = $2π r [r - ( - r)] = 4π r 2$

Talk:Surface area

From Wikipedia, the free encyclopedia

[edit] Two dimensions

[edit] Surface Area To Volume Ratios

[edit] Moved from the article

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