lecture30-09 - 18.02 Multivariable Calculus(Spring 2009...

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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
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Looking at the picture we have that 1 Area R Area S = cos θ = ˆ n · ˆ k and from here Area S = Area R ˆ n · ˆ k. (2.1) To consider the general case we again tile the surface into smaller pieces Δ S that are almost flat rectangles, each one of which casting a rectangular shadow Δ R as in the picture If we set Δ A = Area Δ R then from (2.1) we have that for each piece Area Δ S Δ A ˆ n · ˆ k. (2.2) If the surface is the graph of z = f ( x, y ), then an upward normal is given by N = - f x ( x, y ) ˆ i - f y ( x, y ) ˆ j + ˆ k and as a consequence ˆ n = N | N | . It follows that ˆ n · ˆ k = 1 | N | and Area Δ S = | N | Δ A = f 2 x + f 2 y + 1 Δ A. Finally summing all over the tiles and letting the tiles be smaller and smaller, we obtain the following formula for the surface area: Area S = R f 2 x + f 2 y + 1 dA.
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