The simplest way to define a surface integral is to go through a derivation of the following question, how can one find the mass of a surface?

Let **z = f(x,y) **define a surface in the **xyz plane** above a region **R** in the xy plane. The density per unit area of the surface is given by the function **P(x,y,z) **. Think of a patch of the surface above an infinitesimal region in the xy plane around a point ** (x,y) with area dA** (best way to visualize this is to draw a grid of small rectangles on f(x,y), the patch is one of those rectangles). The surface area of this patch is:

**Sqrt(1+ (pDx)**^{2} + (pDy)^{2})dA

**pDx - partial derivative of z ( or f(x,y)) with respect to x.**

pDy - partial derivative of z with respect to y.

To find the mass of the patch I multiply density times area:

**P(x,y,f(x,y)) * Sqrt(1+ (pDx)**^{2} + (pDy)^{2})dA

If that is the mass of one of those rectangles, to get the total mass of the surface I sum up the patches of surface above all infinitesimal regions in **R**.

∫∫_{R}** P(x,y,f(x,y)) * Sqrt(1+ (pDx)**^{2} + (pDy)^{2})dA (1)

Surface integrals are

double integrals
Surface integrals usually have the following notation:

∫∫_{S}**P(x,y,z)**d**S** (2)

**(1) & (2)** are equivalent expressions. (

**Side note: interestingly enough, if P(x,y,z) is 1, then (2) becomes just the surface area of S**).

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The parametric version of the surface integral form:

If a surface **S** is defined in the parametric form

** r(u,v) = < x(u,v), y(u,v), z(u,v) > **

where (u,v) lies in a region **R** in the uv plane, the Surface integral is given by:

∫∫_{S}**P(x,y,z)**d**S** = ∫∫_{R}**P(x(u,v), y(u,v), z(u,v)) * |r**_{u} X r_{v}| dA

where:

** r**_{u} = < xDu, yDu, zDu >

r_{v} = < xDv, yDv, zDv > ** **

xDu – partial derivative of x function with respect to u, same for y and z.

xDv – partial derivative of x function with respect to v, same for y and z .

X – means the cross product of those two vectors r_{u} and r_{v}