18.02 Multivariable Calculus (Spring 2009): Lecture 30Surface Area. Vector fields in 3D. Surface Integrals and fluxApril 23Reading Material: From Simmons: 21.4. From Lecture Notes: V8, V9.Last time: Cylindrical and Spherical Coordinates.Today: Surface Area for a graph. Vector fields in 3D. Surface Integrals and flux.2 Surface AreaConsider the surface S given by the graph of z = f(x, y). This surface casts a shadow on a regionR of the xy-plane:Question: How do we compute the area of S?Simple Case: Assume that S is flat and that its shadow is a rectangle R:1Looking at the picture we have that1Area RArea S= cos θ = ˆn ·ˆkand from hereArea S =Area Rˆn ·ˆk.(2.1)To consider the general case we again tile the surface into smaller pieces ∆S that are almost flatrectangles, each one of which casting a rectangular shadow ∆R as in the pictureIf we set ∆A = Area ∆R then from (2.1) we have that for each pieceArea ∆S ∼∆Aˆn ·ˆk.(2.2)If the surface is the graph of z = f(x, y), then an upward normal is given by~N = −fx(x, y)ˆi − fy(x, y)ˆj +ˆkand as a consequenceˆn =~N|~N|.It follows thatˆn ·ˆk =1|~N|andArea ∆S = |~N|∆A =qf2x+ f2y+ 1 ∆A.Finally summing all over the tiles and letting the tiles be smaller and smaller, we obtain the followingformula for the surface area:Area S =ZZRqf2x+ f2y+ 1 dA.We can use the following “suggestive” notation for this areaArea S =ZZS1 dS,1To convince yourself of this cons ider the two extreme cases ˆn //ˆk and ˆn ⊥ˆk.2where, from the discussion above, we can think thatdS = |~N|dA =qf2x+ f2y+ 1 dA. (2.3)We summarize what we found in this theoremTheorem 1. If S is a surface given by the graph of z = f(x, y) that casts a shadow on a region Rin the xy-plane, thenArea S =ZZS1 dS =ZZRqf2x+ f2y+ 1 dA. (2.4)Exercise 1. Consider the quarter of paraboloid z = x2+ y2in the first quadrant and below z = 1.Compute its area.Solution:3 Work and Flux in 3DThe definition of work of a vector filed~F = Mˆi + Nˆj + Pˆk along a curve C in 3D is a simplegeneralization of the 2D definition. More precisely if we parametrize the curve C by~r(t) = x(t)ˆi + y(t)ˆj + z(t)ˆk t1≤ t ≤ t2thenWork =t2Zt1M(x(t), y(t), z(t))dxdt+ N(·)dydt+ P (·)dzdtdt.3The definition of flux in 3D is instead geometrically different than the one in 2D. In fact:2D flux = flow across C.3D flux = flow across surface S in the direction of a fixed normal on SRemark. It is very important to specify a so called orientation on the surface we are considering,that is we need to specify a normal vector ˆn on each point of the surface a nd this normal vect orneeds to be a continuous one.One should be aware of the fact that not all surfaces are orientable. For example the Klein bottle(http://www.daviddarling.info/encyclopedia/K/Klein-bottle.html) and the Mobius strip(http://www.daviddarling.info/encyclopedia/M/Mobius-band.html) are example of non orientable sur-faces.4 Computing the 3D fluxSimple case: We assume that~F is constant and S is an oriented piece of a plane, that is we fix anormal vector ˆn on it4If~F ⊥ S and in the same direction of ˆn, thenflow across S = |~F |( area (S))If not thenflow across S = (~F · ˆn)( area(S))that is only the component of~F ⊥ to S is effective in the computation of the flow:Exercise 2. Consider the constant vector field~F =ˆi +ˆj +ˆk and the surface S given by the orientedsquare of side 2 in the pictureCompute the flux of~F across S.Solution:General Case: In order to com pute the flux of a non constant field~F across any (reasonable)5oriented surface S we use the usual technique of breaking S into tiny pieces that are approximatelyflat and we use the simple case discussed above:We then sum everything up again. It is not surprising that the notation for the flux across anoriented surface S isflux of~F across S =ZZS(~F · ˆn) dS.Shorthand: 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.5 Evaluating flux Integrals on graphsAssume 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 (2.3) above and we observe thatd~S = ˆn dS =~N|~N||~N| dA = (−fxˆi − fyˆj +ˆk) dAand 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.6Exercise 3. 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 4. 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:7Study Guide 1. Consider the following questions:• at this point you are ready to compute any surface integral of the typeZZSg dSif the surface is a given as a graph of a function f above a region R of the xy−plane. Can youwrite a formula for it as an iterated double integral?• Can you write a formula w hen S is a piece of cylinder or a piece of
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