18.02 Multivariable Calculus (Spring 2009): Lecture 31Surface Area and flux across cylinders and spheres.April 24Reading Material: From Simmons: 21.4. From Lecture Notes: V8, V9.Last time: Surface Area. Vector fields in 3D. Surface Integrals and flux.Today: Surface Area and flux across cylinders and spheres.In the last class we learned that the flux across an oriented surface S isflux of~F across S =ZZS(~F · ˆn) dSShorthand: We often use the notation ˆn dS = d~S, so you could also writeflux of~F across S =ZZS~F · d~S.This is a typical surface integral. We already know how to compute one of then, namely the areaof a surface:Area (S) =ZZS1dSso it will not be difficult to generalize this formula.2 Evaluating flux IntegralsAssume S is a surface given by the graph of a function f (x, y). Assume S is oriented by using theupward normal.To convert the flux integralZZS~F · d~Sinto a double iterated integral we use (3.2) above and we observe thatd~S = ˆn dS =~N|~N||~N| dA = (−fxˆi − fyˆj +ˆk) dA1and as a consequence we simply haveZZS~F · d~S =ZZR~F (x, y, f(x, y)) · (−fxˆi − fyˆj +ˆk) dA,where R is the shadow cast by S on the xy-plane.Exercise 1. Consider the vector field~F = zˆk and the surface S given by the graph of z = x2+ y2above the region R = {(x, y)/ − 1 ≤ x, y ≤ 1} oriented upward. Compute the flux of~F across S.Solution:Exercise 2. Assume that now~F = xˆj and S is given by the graph of z = 1 − x2− y2for x, y, z ≥ 0and S is oriented outward. Again compute the flux of~F across S.Solulition:2Last time we gave formulas to compute the area of surfaces and flux across a surface given as graphsof functions of type z = f(x, y). In this section we consider instead a piece of a cylinder and a pieceof a sphere.3 Surface area and flux: cylindersWe first recall that the cylindrical coordinates are obtained by using polar coordinates in the xyplane and the z coordinate along the z-axis, to summarize:rectangular cylindrical = polar + z(x, y, z) (r, θ, z)Conversion: Like polar:(r, θ, z) −→ (x, y, z)x = r cos θy = r sin θz = zand the infinitesimal element of volume isdV = r dθ dr dz. (3.1)It is clear that if we want to describe the piece of cylinder C of radius a and hight h between theangles θ1and θ2we just need to use the coordinates above where we FIX r = a. To measure its area we have tocomputeArea =ZZC1dSand from (3.1) it is not hard to deduce thatdS = a dθ dz, (3.2)since r = a and there is no variation in r. As a consequenceArea =ZZC1dS =hZ0θ2Zθ1a dθ dz = a(θ2− θ1)h,3as expected.Now assume the cylinder is oriented outwardthen it is clear that at the point P = (a cos θ, a sin θ, z) the outward normal vector ˆn = cos θˆi + sin θˆjsince at each hight we have the pictureGiven a vector filed~F = Mˆi + Mˆj + Pˆk. the flux through the oriented cylinder D of~F isFlux =ZZC~F · ˆn dS =hZ0θ2Zθ1[M(a cos θ , a sin θ, z) cos θ + N(a cos θ, a sin θ, z) sin θ]a dθ dz.Note: For simplicity here I am assuming that the cylinder is touching the xy-plane, in other words0 ≤ z ≤ h. But it is also possible to have a more general situation when h1≤ z ≤ h2, for somearbitrary h1and h2.Exercise 3. Consider the vector field~F = xˆi + zˆk and the cylinder Cwhere 0 ≤ θ, 0 ≤ z ≤ h and a = 1. We orient it outward. Compute the flux of~F across C.4Solution:4 Surface area and flux: sphereswhere(ρ, φ, θ) −→ (r, θ, z) −→ (x, y, z)r = ρ sin φ x = r cos θ = ρ sin φ cos θθ = θ y = r sin θ = ρ sin φ sin θz = ρ cos φ z = z = ρ cos φWe also recall thatdV = ρ2sin φ dρ dθ dφ. (4.1)Now consider for example the surface S given by the piece of sphere centered at the origin of radiusρ = a with longitude φ between φ1and φ2and latitude θ between θ1and θ25To compute the area of its surface we deduce from (4.1) thatdS = a2sin φ dθ dφ (4.2)since there is no variation in ρ and we writeArea =ZZS1 dS =φ2Zφ1θ2Zθ1a2sin φ dθ dφ.Supp ose now we orient S by outward normal ˆn. To find this ˆn we first reason in rectangularcoordinates: Since the sphere is described by x2+ y2+ z2− a2= 0 it is easy, for example by takingthe gradient, that a normal outward vector is~N = xˆi + yˆj + zˆkThe size of this vector is |~N| =px2+ y2+ z2= a on the sphere, henceˆn =1a(xˆi + yˆj + zˆk).We can then write ˆn by using spherical coordinates asˆn = sin φ cos θˆi + sin φ sin φθˆj + cos φˆk. (4.3)Now assume a vector field~F = Mˆi + Mˆj + Pˆk is also given and we want to compute its flux acrossS. We have from (4.2) and (4.3) thatFlux =ZZS~F · ˆn dS=φ2Zφ1θ2Zθ1[M(a sin φ cos θ, a sin φ sin φθ, a cos φ) sin φ cos θ + N(•) sin φ sin φθ + P (•) cos φ]a2sin φ dθ dφ,where we use • for short instead of rewriting two more times the same coordinates that appear inthe argument of M.Exercise 4. Consider the piece S of the sphere centered at the origin, radius 2 with 0 ≤ θ ≤ π/2and 0 ≤ φ ≤ π/2, oriented inward. Compute the flux of the vector field~F = xˆi across the surface S.6Solution:Handy Fact: In general given an oriented surface S with normal vector ˆn and a vector field~F , if~F · ˆn = C, for some constant C, thenFlux =ZZS~F · ˆn dS = C Area
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