Math 314 Lecture #34 §16.7: Surface Integrals
Math 314 Lecture #34 §16.7: Surface Integrals Outcome A: Compute the surface integral of a function over a parametric surface. Let S be a smooth surface described parametrically by ~r(u, v), (u, v) ∈ D, and suppose the domain of f(x, y, z) includes S. The surface integral of f over S with respect to surface area is ZZ ZZ f(x, y, z) dS = f(~r(u, v))k~ru × ~rvk dA. S D When S is a piecewise smooth surface, i.e., a finite union of smooth surfaces S1,S2,...,Sk, that intersect only along their boundaries, the surface integral of f over S with respect to surface area is ZZ ZZ ZZ ZZ f dS = f dS + f dS + ··· + f dS, S S1 S2 Sk where the dS in each double integral is evaluated according the the parametric description of Si. Example. A parametric description of a surface S is ~r(u, v) = hu2, u sin v, u cos vi, 0 ≤ u ≤ 1, 0 ≤ v ≤ π/2. Here is a rendering of this surface. Here ~ru = h2u, sin v, cos vi and ~rv = h0, u cos v, −u sin vi, so that ~i ~j ~k 2 2 ~ru × ~rv = 2u sin v cos v = h−u, 2u sin v, 2u cos vi, 0 u cos v −u sin v and (with u ≥ 0), √ √ 2 4 2 k~ru × ~rvk = u + 4u = u 1 + 4u . The surface integral of f(x, y, z) = yz over S with respect to surface area is ZZ Z 1 Z π/2 √ yz dS = (u2 sin v cos v)(u 1 + 4u2) dvdu S 0 0 Z 1 √ sin2 v π/2 = u3 1 + 4u2 du 0 2 0 1 Z 1 √ = u3 1 + 4u2 du.
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