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Surface area is the measure of how much exposed area an object has. It is expressed in square units. If an object has flat faces, its surface area can be calculated by adding together the areas of its faces. Even objects with smooth surfaces, such as spheres, have surface area.
## Formula

#### 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 $2picdot\; f(x)$

The width of the ring can be determined by using Pythagoras' formula for a rectangular triangle with side lengths $dx$ and $f\text{'}(x)\; cdot\; dx$, which leads to $sqrt\{1+f\text{'}(x)^2\},dx$

The infinitesimal surface area of the circular ring thus is equal to $2pi\; f(x)cdot\; sqrt\{1+f\text{'}(x)^2\},dx$

The derivative of $f(x)$ is equal to $f\text{'}(x)\; =\; frac\{-x\}\{sqrt\{r^2-x^2\}\}$

The surface area of the sphere can be calculated as

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

The antiderivative needed is the simple linear function $x$

Thus, the sphere surface area amounts to

A_{sphere} = $2pi\; r[r-(-r)]\; =\; 4pi\; r^2$
#### Cylinder

The surface area of a (circular) cylinder of radius r and height h is
#### Cube

Sa = Total Surface Area, L = Side Length, V = Volume, Sf = Surface Area Of a Single Face

$Sa\; =\; L^26$

$Sa\; =\; Sf6$

$Sa\; =\; (V^\{-3\})^26$## Surfaces whose area cannot be defined

## In chemistry

Surface area is important in chemical kinetics. Increasing the surface area of a substance generally increases the rate of a chemical reaction. For example, iron in a fine powder will combust, while in solid blocks it is stable enough to use in structures. For different applications a minimal or maximal surface area may be desired.
## In biology

## External Links

## See also

The radius of the circular ring is $f(x)\; =\; sqrt\{r^2-x^2\}$. The length of the circular ring is equal to $2picdot\; f(x)$

The width of the ring can be determined by using Pythagoras' formula for a rectangular triangle with side lengths $dx$ and $f\text{'}(x)\; cdot\; dx$, which leads to $sqrt\{1+f\text{'}(x)^2\},dx$

The infinitesimal surface area of the circular ring thus is equal to $2pi\; f(x)cdot\; sqrt\{1+f\text{'}(x)^2\},dx$

The derivative of $f(x)$ is equal to $f\text{'}(x)\; =\; frac\{-x\}\{sqrt\{r^2-x^2\}\}$

The surface area of the sphere can be calculated as

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

The antiderivative needed is the simple linear function $x$

Thus, the sphere surface area amounts to

A

- $2\; pi\; r\; h\; +\; 2\; pi\; r^2$

$Sa\; =\; L^26$

$Sa\; =\; Sf6$

$Sa\; =\; (V^\{-3\})^26$

While areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a lot of care. Various approaches to defining the surface area were developed in the late nineteenth and the early twentieth century by Henri Lebesgue and Hermann Minkowski. For a very wide class of geometric surfaces called piecewise-smooth all these approaches result in the same notion of area. However, if a surface is very irregular or rough, then it may not be possible to assign any area at all to it. A typical example is given by a surface with spikes spread throughout in a dense fashion. Many surfaces of this type occur in the theory of fractals. Extensions of the notion of area which partially fulfill its function and may be defined even for very badly irregular surfaces are studied in the geometric measure theory. A specific example of such an extension is the Minkowski content of a surface.

The surface area-to-volume ratio (SA:V) of a cell imposes upper limits on size, as the volume increases much faster than does the surface area, thus limiting the rate at which substances diffuse from the interior across the cell membrane to interstitial spaces or to other cells. If you consider the math, you'll see the relation between SA and V much more intuitively: V = 4/3 π r^{3}; SA = 4 π r^{2}, where r is the radius of the cell. Do the math and the resulting ratio becomes 3/r. If a cell has a radius of 1 μm, the SA:V ratio is 3. Increase the cell's radius to 10 μm and the SA:V ratio becomes 0.3. With a cell radius of 100, SA:V ratio is 0.03. Using the previous simple example, we can see how the surface area falls off steeply with increasing volume.

What limitations does this place on a living cell? For small cells, SA:V ratio allows for relatively good exchange of nutrients and wastes. For larger cells and organisms, SA:V ratio forces the cell or organism to find more efficient ways to exchange nutrients and waste products, e.g. specific conduits that carry blood, hormones, lymph, etc. from deep regions to the surface of an organism.

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Last updated on Wednesday October 08, 2008 at 04:03:54 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Wednesday October 08, 2008 at 04:03:54 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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