Math Help

Surface Integrals (continued)

Note: Domain D shown in the previous figure is the “area” projected on to the

relevant plane. In this case it is projected on to the x-y plane since the surface, S,

was described by z = g(x,y).

If, on the other hand, the surface was described by y = g(x,z), then D would be

the “area” projected on to the x-z plane, etc.

You have at least two options in evaluating surface integrals.

1. You could use a transformation dS = | r_u x r_v | dA

I_S = ∫ ∫ f(x,y,z) dS = ∫ ∫ f(x,y,z) | r_u x r_v | dA

S D

2. You could attempt direct evaluation of the surface integral

I_S = ∫ ∫ f(x,y,z) dS

When the surface, S, is described as follows:

z = g(x,y) (recall let u = x and v = y)

the position vector to an arbitrary point on S is r = xi + yj + g(x,y)k

Then one can use x and y rather than u and v as the parameters describing the surface, S.

So the normal vector to S is N(u,v) = N(x,y) = r_x x r_y which yields the

following 3 x 3 determinant. Note: N is not a unit vector.

i j k i j k

N(x,y) = det ∂x/∂x ∂y/∂x ∂g/∂x = det 1 0 ∂g/∂x

∂x/∂y ∂y/∂y ∂g/∂y 0 1 ∂g/∂y

N(x,y) = - ∂g/∂x i - ∂g/∂y j + 1 k and | N(x,y) | = √[1 + (∂g/∂x)² + (∂g/∂x)²]

Thus dS = √[1 + (∂g/∂x)² + (∂g/∂x)²] dx dy and

I_S = ∫ ∫ f(x, y, z) dS = ∫ ∫ f(x, y, g(x,y)) √[1 + (∂g/∂x)² + (∂g/∂x)²] dx dy

S D

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