Hello :P
I suppose you mean areal surface area of the penis.
Imagining develop a penis on a plane, with "area of the penis" there refers to the area occupied by the development of the penis on the floor.
Suppose we develop an erect penis so as to prevent the variety that represents the surface is twisted around itself or other problems occur topological [in this case is really the development of an area].
From the mathematical point of life is a penis? We can assume that a subset Ω ⊂ IR ³ or a solid.
How can we describe this solid Ω which we calculate the surface area? Given the particularly bizarre form which unfortunately [or fortunately] is not attributable to any known primary function must use one or more functions that approximate the shape.
At the application layer after performing the steps trying to determine "the equation of the penis, or recourse to methods provided by the theory of functions tries to find a function or equation that best approximates the values obtained from the measures.
We hypothesize that the penis has a uniform cylindrical shape, at least for the trunk. Under this hypothesis, intersects the cylinder forming the trunk with a plan will get a circle of radius R. A better approximation can be obtained by considering an elliptical rather than circular cylinder with a director [eccentricity may be more or less pronounced].
We now give some definitions that will be useful later
DEFINITION 1
We define the total length ℓ of the penis the following size
ℓ: = ℓ ₁ + ℓ ₂
where
ℓ ₁ penis length is measured from the bottom to the base of the glans
ℓ ₂ is the length of the glans
DEFINITION 2
We define the penis in this way
Ω: = Ω ₁ U Ω ₂
where
Ω ₁: = ((x, y, z) ∈ IR ³ x ² + y ² = R ², 0 ≤ z ≤ ℓ ₁) is the trunk of the penis [R is the radius of the circle described previously]
Ω ₂ = ((x, y, z) ∈ IR ³ ...) is the glans
I left because of ellipses describe the glans from the mathematical point of view is a real problem.
As I mentioned previously, it is necessary to find the equation of a surface or a function that approximates the shape.
The first function that comes to mind is the Gauss bell function described by equation
ƒ (x, y) = exp (- y ² - x ²)
http://img147.imageshack.us/i/oeb2d8tmp.jpg/
or even the paraboloid of equation
ƒ (x, y) = ℓ ₂ - y ² - x ²
http://img147.imageshack.us/i/immaginext.jpg/
If we use this second function to approximate the glans will
Ω ₂ = ((x, y, z) ∈ IR ³: ℓ ₁ ≤ z ≤ ℓ ₂ - y ² - x ²)
Now we have to do some 'healthy crafts.
Regarding the surface area of the trunk will
A_Ω ₁ = 2πR ℓ ₁
We now calculate the surface area of the glans by calculating the surface area underlying the paraboloid
ƒ (x, y) = ℓ ₂ - y ² - x ²
To do this we will use a formula that represents the same two variables calcoalo of the length of a curve subtended a function of a variable.
In a variable length of a curve subtended a function ƒ (x) is given by the following equation:
L: = INTEGRAL between α & β √ (1 + ƒ '(x)) dx
where α & β are the endpoints of the curve
In two variables underlying the surface area is a function ƒ (x, y) given by the following equation:
A double integral over T = (√ (1 + [∂ / ∂ x [ƒ (x, y)]] ² + [∂ / ∂ y [ƒ (x, y) dxdy )]]²)
where T is the set bounded by the edge of the area.
In our case we
T: = ((x, y) ∈ IR ²: x ² + y ² ≤ ℓ ₂)
ƒ (x, y) = ℓ ₂ - y ² - x ²
∂ / ∂ x [ƒ (x, y)] = ∂ / ∂ x [ℓ ₂ - y ² - x ²] = - 2x
∂ / ∂ y [ƒ (x, y)] = ∂ / ∂ y [ℓ ₂ - y ² - x ²] = - 2y
and therefore
A_Ω ₂ = double integral over T (√ (1 + [∂ / ∂ x [ƒ (x, y)]] ² + [∂ / ∂ y [ƒ (x, y) dxdy = )]]²)
Double integral over T = (√ (1 + (-2x) ² + (-2y) ²)) dxdy =
Double integral over T = (√ (1 + 4x ² + 4y ²)) dxdy =
At this point we switch to polar coordinates to simplify the calculations
(X: = ρcos (θ)
(Y = ρsen (θ)
T: = ((ρ, θ) ∈ IR ²: 0 ≤ ρ ≤ √ (ℓ ₂), 0 ≤ θ ≤ 2π)
Recalling that by applying the generic change of coordinates
(X: = φ (u, v)
(Y = ψ (u, v)
is called "Jacobian of transformation" for the following
J: = |. . . ∂ / ∂ u [φ (u, v)]. . ∂ / ∂ v [φ (u, v)]. . . |
. . . . |. . . ∂ / ∂ u [ψ (u, v)]. . ∂ / ∂ v [ψ (u, v)]. . . |
and recalling that by applying the above change of coordinates is the following relation
Double integral on T ƒ (x, y) dxdy =
Double integral over T = ƒ (φ (ρ, θ), ψ (ρ, θ)) * det | J | dρdθ
in our case we
(X = φ (ρ, θ) = ρcos (θ)
(Y = ψ (ρ, θ) = ρsen (θ)
∂ / ∂ θ [ρcos (θ)] =-ρsen (θ)
∂ / ∂ ρ [ρcos (θ)] = cos (θ)
∂ / ∂ θ [ρsen (θ)] = ρcos (θ)
∂ / ∂ ρ [ρsen (θ)] = sin (θ)
J: = |. . . ∂ / ∂ u [φ (ρ, θ)]. . ∂ / ∂ v [φ (ρ, θ)]. . . |
. . . . |. . . ∂ / ∂ u [ψ (ρ, θ)]. . ∂ / ∂ v [ψ (ρ, θ)]. . . |
J = |. . . ∂ / ∂ ρ [ρcos (θ)]. . ∂ / ∂ θ [ρcos (θ)]. . |
. . . |. . . ∂ / ∂ ρ [ρsen (θ)]. . ∂ / ∂ θ [ρsen (θ)]. . |
J = |. . cos (θ). . -Ρsen (θ). . |
. . . |. . sin (θ). . . Ρcos (θ). . |
det | J | = ρcos ² (θ) + ρsen ² (θ) = ρ [cos ² (θ) + sen ² (θ)] = ρ
and therefore
Double integral over T = (√ (1 + 4x ² + 4y ²)) dxdy =
Double integral over T = (√ (1 + 4 (ρcos (θ)) ² + 4 (ρsen (θ)) ²)) = ρdρdθ
Double integral over T = (√ (1 + 4ρ ² cos ² (θ) + 4ρ sen ² ² (θ))) = ρdρdθ
Double integral over T = (√ (1 + 4ρ ² (cos ² (θ) + sen ² (θ)))) = ρdρdθ
Double integral over T = (√ (1 + 4ρ ²)) = ρdρdθ
= (INTEGRAL between 0 & 2π dθ) * (INTEGRAL between 0 & √ (ℓ ₂) (ρ √ (1 + 4ρ ²))) = dρ
= ([Θ] _calcolato between 0 & 2π) * ((1 / 12) √ ((1 + 4ρ ²)³)]_ calculated 0 & √ (ℓ ₂)) =
= (2π) * (1 / 12) [√ ((1 + 4 (√ (ℓ ₂ ))²)³) - √ ((1 + 0) ³)] =
= (Π / 6) [√ ((1 + 4 ℓ ₂) ³) - 1]
We then discovered how good the surface area of the glans:
A_Ω ₂ = (π / 6) [√ ((1 + 4 ℓ ₂) ³) - 1]
The total area of the surface of the penis will therefore be the sum of the surface area of the trunk and the surface area of the glans
A: = + A_Ω A_Ω ₁ ₂ = 2πR ℓ ₁ + (π / 6) [√ ((1 + 4 ℓ ₂) ³) - 1]
This final formula which we have just obtained or
A = 2πR ℓ ₁ + (π / 6) [√ ((1 + 4 ℓ ₂) ³) - 1]
is the formula to calculate the area of a penis
____________________________
Now I do have a question: when do we go from theory to practice? : P