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:
(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
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:
∫∫SP(x,y,z)dS = ∫∫RP(x(u,v), y(u,v), z(u,v)) * |ru X rv| dA
where: ru = < xDu, yDu, zDu >
rv = < 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 ru and rv