Math 314 Lecture #34 §16.7: Surface Integrals

Home , Parametric surface, Surface integral

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. 2 0 The substitution t = 1 + 4u2 gives dt = 8udu, t = 1 when u = 0, t = 5 when u = 1, and u2 = (t − 1)/4. Continuing the integration gives ZZ 1 Z 5 1 √ yz dS = (t − 1) t dt S 2 1 32 1 Z 5 = t3/2 − t1/2 dt 64 1 1 2t5/2 2t3/2 5 = − 64 5 3 1 1 2(55/2 − 1) 2(53/2 − 1) = + . 64 5 3 Outcome B: Find the orientation for an oriented surface, in particular, the positive orientation for a closed oriented surface. Before we can discuss surface integrals of vector ﬁelds, we need to deal with the notion of “orientation” of surfaces. A parametric smooth surface S : ~r(u, v), (u, v) ∈ D, comes equipped with a unit normal vector, namely ~r × ~r ~n = u v . k~ru × ~rvk Here is a rendering of this unit normal vector over a grid of points on a surface. The other unit normal vector for S is given by opposite direction of −~n. A surface S is oriented if there is a choice of a unit normal vector ~n at every point (x, y, z) on S for which ~n various continuously over S. An orientation for an oriented surface S is a choice of a continuously varying unit normal vector ~n over S. Notice geometrically that an oriented surface has two sides: one side corresponds to ~n and the other side corresponds to the opposite orientation −~n. Every surface S that is the graph of z = f(x, y), (x, y) ∈ D, is oriented with the upward orientation −f ~i − f ~j + ~k ~n = x y . p 2 2 (fx) + (fy) + 1 Not every surface is oriented: the M¨obius strip, rendered below, has only one side.

A surface S is closed if it is the boundary of a bounded solid E. The positive orientation of a closed surface is the unit normal vector that points outward from the bounded solid E.

Outcome C: Compute the flux of a vector field over a parametric surface. Recall for a parametric curve C : ~r(t) we wrote d~r = ~r 0(t)dt where ~r 0(t) is tangent to C. Analogously, for a parametric oriented surface S : ~r(u, v), (u, v) ∈ D, with unit normal vector ~n, we write dS~ = ~ndS. Suppose a vector field F~ (x, y, z) is continuous on a domain that contains S. The flux of F~ across S (in the direction specified by the choice of orientation) is the scalar ZZ ZZ F~ · dS~ = F~ · ~ndS. S S Recall that dS = k~ru × ~rvk dA; if we choose ~n = ~rv × ~rv/k~ru × ~rvk, then ~ dS = ~ndS = ~n k~ru × ~rvkdA = ~ru × ~rv dA,

and the ﬂux of F~ across S in the direction ~n becomes ZZ ZZ ~ ~ ~ F · dS = F (~r(u, v)) · (~ru × ~rv) dA. S D

Example. Let F~ = hx, −z, yi, and let S be the part of the sphere x2 + y2 + z2 = 4 that lies in the ﬁrst octant, with orientation toward the origin. With D = {(x, y): x2 + y2 ≤ 4, x, y ≥ 0}, a standard parameterization of S is p ~r(x, y) = hx, y, 4 − x2 − y2i, (x, y) ∈ D.

With z = f(x, y) = p4 − x2 − y2, a normal vector for S is

x y ~rx × ~ry = , , 1 . p4 − x2 − y2 p4 − x2 − y2

~ The ﬂux of F across S in the direction ~n = ~rx × ~ry/k~rx × ~ryk is

ZZ 1 ZZ p x y F~ · dS~ = hx, − 4 − x2 − y2, yi · , , 1 dA p 2 2 p 2 2 S 2 D 4 − x − y 4 − x − y 1 ZZ x2 = dA [switch to polar coordinates] p 2 2 2 D 4 − x − y 1 Z π/2 Z 2 r3 cos2 θ = √ drdθ 2 2 0 0 4 − r 4π = . 3 Did we account for the speciﬁed orientation correctly? No, because ~n points away from the origin, and so the requested ﬂux is the negative of what we got, namely, −4π/3.