Surface Integrals (continued)
Also note the partial derivatives ru = ∂ r / ∂ u and rv = ∂ r / ∂ v
Now the element of surface area is dS = | ru du x rv dv | = | ru x rv | du dv
but ru x rv = N = normal to the surface, S and du dv = dA
So the relation between the element of surface area, dS, and the element of area, dA
dS = | N | dA (See the figure below)
Evaluation of the surface integral Is = ∫ f(x, y, z) dS is a two-step process.
i.e. If the surface is given by z = g(x,y), then f(x,y,z) = f(x, y, g(x,y))
dS = | N(u,v)| du dv = | N(u,v)| dA where N(u,v) = ∂r/∂u x ∂r/∂v
take u = x and v = y so N(x,y) = ∂r/∂x x ∂r/∂y
Here N(x,y) represents the normal vector to the surface, S, at each arbitrary
point (x,y,z) and r = xi + yj + g(x,y)k is the position vector from the origin of the
coordinate system to any arbitrary point (x,y,z) on the surface, S. See the figure below.
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