18 02 Multivariable Calculus Spring 2009 Lecture 30 Surface Area Vector fields in 3D Surface Integrals and flux April 23 Reading 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 Area Consider the surface S given by the graph of z f x y This surface casts a shadow on a region R 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 1 Looking at the picture we have that1 Area R cos n k Area S and from here Area R 2 1 n k To consider the general case we again tile the surface into smaller pieces S that are almost flat rectangles each one of which casting a rectangular shadow R as in the picture Area S If we set A Area R then from 2 1 we have that for each piece Area S A n k 2 2 If the surface is the graph of z f x y then an upward normal is given by fx x y i fy x y j k N and as a consequence n N N It follows that 1 N n k and A Area S N q fx2 fy2 1 A Finally summing all over the tiles and letting the tiles be smaller and smaller we obtain the following formula for the surface area ZZ q fx2 fy2 1 dA Area S R We can use the following suggestive notation for this area ZZ Area S 1 dS S 1 To convince yourself of this consider the two extreme cases n k and n k 2 where from the discussion above we can think that q dA fx2 fy2 1 dA dS N 2 3 We summarize what we found in this theorem Theorem 1 If S is a surface given by the graph of z f x y that casts a shadow on a region R in the xy plane then ZZ ZZ q Area S 1 dS fx2 fy2 1 dA 2 4 S R Exercise 1 Consider the quarter of paraboloid z x2 y 2 in the first quadrant and below z 1 Compute its area Solution 3 Work and Flux in 3D The definition of work of a vector filed F M i N j P k along a curve C in 3D is a simple generalization of the 2D definition More precisely if we parametrize the curve C by r t x t i y t j z t k then Zt2 Work M x t y t z t t1 3 t 1 t t2 dy dz dx N P dt dt dt dt The 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 S Remark 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 n on each point of the surface and this normal vector needs 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 surfaces 4 Computing the 3D flux Simple case We assume that F is constant and S is an oriented piece of a plane that is we fix a normal vector n on it 4 If F S and in the same direction of n then flow across S F area S If not then flow across S F n 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 i j k and the surface S given by the oriented square of side 2 in the picture Compute the flux of F across S Solution General Case In order to compute the flux of a non constant field F across any reasonable 5 oriented surface S we use the usual technique of breaking S into tiny pieces that are approximately flat 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 an oriented surface S is ZZ flux of F across S F n dS S so you could also write Shorthand We often use the notation n dS dS ZZ flux of F across S F dS S This is a typical surface integral We already know how to compute one of then namely the area of a surface ZZ Area S 1dS S so it will not be difficult to generalize this formula 5 Evaluating flux Integrals on graphs Assume S is a surface given by the graph of a function f x y Assume S is oriented by using the upward normal To convert the flux integral ZZ F dS S into a double iterated integral we use 2 3 above and we observe that n dS dS N dA f xi f y j k dA N N and as a consequence we simply have ZZ ZZ F dS F x y f x y f xi f y j k dA S R where R is the shadow cast by S on the xy plane 6 Exercise 3 Consider the vector field F z k and the surface S given by the graph of z x2 y 2 above 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 xj and S is given by the graph of z 1 x2 y 2 for x y z 0 and S is oriented outward Again compute the flux of F across S Solulition 7 Study Guide 1 Consider the following questions at this point you are ready to compute any surface integral of the type ZZ g dS S if the surface is a given as a graph of a function f above a region R of the xy plane Can you write a formula for it as an iterated double integral Can you write a formula when S is a piece of cylinder or a piece of sphere 8
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