surfacearea - Math 21a Handout on Surface Area The text...

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Math 21a Handout on Surface Area The text argued that after parameterizing a surface (or part of one) by a function X (u, v), with u and v coordinates on a region R in R 2 , then the integral of a function f on the surface is the same as the integral fdS S ∫∫ fu v X X d u d v uv R ((,) ) | | X × ∫∫ (1) The purpose of this handout is to provide three examples of the preceding formula, but all with the same surface and same function f. Rather, each example will use a different parameterization of the surface, and give evidence for the (not obvious) fact that the integral in (1) is truly independent of your chosen way of parameterizing the surface. In particular, the surface in question is the top half of the unit sphere in R 3 , that is where x 2 + y 2 + z 2 = 1, z 0. (2) Meanwhile, the function f is just the z-coordinate in these examples. Example 1 : This example computes the integral in 1 using the parameterization where X(u, v) = (u, v, (1 - u 2 - v 2 ) 1/2
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surfacearea - Math 21a Handout on Surface Area The text...

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