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Unformatted text preview: 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 that 1 Area R Area S = cos = n k and from here Area S = Area R n k. (2.1) 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 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 ~ N =- f x ( x, y ) i- f y ( x, y ) j + k and as a consequence n = ~ N | ~ N | . It follows that n k = 1 | ~ N | and Area S = | ~ N | A = q f 2 x + f 2 y + 1 A....
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