Surface area

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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, can have well-defined surface area.

1 Area formulas
2 Ill-defined areas
3 In chemistry
4 In biology
5 See also

[edit] Area formulas

Note: For 2D figures, the surface area and the area are the same.

Shape

Common equations for surface area (2-Dimensional Objects):
Equation	Variables
A rectangle:	<math>l \cdot w </math>	l = length, w = width
A circle:	<math>\pi \cdot r^2 </math>	r = radius
Any regular polygon:	<math>P \cdot a/2</math>	P = length of the perimeter, a = length of the apothem of the polygon (the distance from the center of the polygon to the center of one side)
A parallelogram:	<math>B \cdot h</math>	B (base) = any side, h (height) = the distance between the lines that the sides of length B lie on
A trapezoid:	<math>(B+b) \cdot h/2</math>	B and b = lengths of the parallel sides, h = distance between the lines on which the parallel sides lie
A triangle (1):	<math>B \cdot h/2</math>	B = any side, h = distance from the line on which B lies to the other point of the triangle
A triangle (2) (Heron's formula):	<math> \sqrt{[p \cdot (p-a) \cdot (p-b) \cdot (p-c)]}</math>	a, b and c = sides of triangle, p = half of the perimeter, or (a+b+c)/2

Common equations for surface area (3-Dimensional Objects):
Shape	Equation	Variables
A cube:	<math>6 \cdot s^2 </math>	s = length of any side
A rectangular prism:	<math>2 \cdot (l \cdot w + l \cdot h + w \cdot h)</math>	l = length, w = width, h = height
A sphere:	<math>4 \cdot \pi \cdot r^2</math>	r = radius of sphere, which is the first derivative of the volume of a sphere
A cylinder:	<math>2 \cdot \pi \cdot r \cdot (h+r)</math>	r' = radius of circular base, h = height
A cone (lateral surface area):	<math>\pi \cdot r \cdot [(r + \sqrt{(r^2+h^2)}]</math>	r = radius of circular base, "h" = height
A cone:	<math>\pi \cdot r^2 \ + \pi \cdot r\cdot s </math>	r = radius of circular base, s = slant height of the cone

Shape	Area formula derivation
Sphere	The surface area of a sphere is the integral of infinitesimal circular rings of width <math>dx</math> The radius of the circular ring is <math> f(x) = \sqrt(r^2-x^2)</math>. The length of the circular ring is equal to <math>2\pi\cdot f(x)</math> The width of the ring can be determined by using Pythagoras' formula for a rectangular triangle with side lengths <math>dx</math> and <math>f'(x) \cdot dx</math>, which leads to <math>\sqrt(1+f'(x)^2)dx</math> The infinitesimal surface area of the circular ring thus is equal to <math>2\pi f(x)\cdot \sqrt(1+f'(x)^2)dx</math> The derivative of <math>f(x) </math> is equal to <math>f'(x) = \frac{-x}{\sqrt(r^2-x^2)}</math> The surface area of the sphere can be calculated as <math> \int_{-r}^r 2\pi f(x)\cdot \sqrt(1+f'(x)^2)\,dx</math> = <math> \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</math> The antiderivative needed is the simple linear function <math>x</math> Thus, the sphere surface area amounts to A_sphere = <math>2\pi r[r-(-r)] = 4\pi r^2</math>

Shape

Area formula derivation

Sphere

The surface area of a sphere is the integral of infinitesimal circular rings of width <math>dx</math>

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

<math> \int_{-r}^r 2\pi f(x)\cdot \sqrt(1+f'(x)^2)\,dx</math> = <math> \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</math>

The antiderivative needed is the simple linear function <math>x</math>
Thus, the sphere surface area amounts to

A_sphere = <math>2\pi r[r-(-r)] = 4\pi r^2</math>

[edit] Ill-defined areas

If one adopts the axiom of choice, then it is possible to prove that there are some shapes whose area cannot be meaningfully defined; see Lebesgue measure for more details.

[edit] 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.

[edit] In biology

Image:Mitochondrion 186.jpg

The inner membrane of the mitochondrion has a large surface area due to infoldings, allowing higher rates of cellular respiration (electron micrograph).

The surface area of an organism is important in several considerations, such as regulation of body temperature, and digestion. Animals use their teeth to grind food down into smaller particles, increasing the surface area available for digestion. The epithelial tissue lining the digestive tract contains microvilli, greatly increasing the area available for absorption. Elephants have large ears, allowing them to regulate their own body temperature. In other instances animals will need to minimize surface area, for example people will fold their arms over their chest when cold to minimize heat loss.

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 r³; SA = 4 π r², 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.