# UCR MATH 10B - Flux integral computations (3 pages)

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**View the full content.**# Flux integral computations

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## Flux integral computations

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- Pages:
- 3
- School:
- University of California, Riverside
- Course:
- Math 10b - Calculus of Several Variables II

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Flux integral computations PROBLEM 1 Set up and evaluate the flux integral Z Z F dS S where F x y z x y z and S is the surface z 4 x2 y 2 defined for x2 y 2 4 with the upward normal orientation Solution This surface is the graph of the function f x y 4 x2 y 2 for x2 y 2 4 so we shall use the graph parametrization X x y x y 4 x2 y 2 We shall use the following very important general formula for the upward normal of a surface defined by an equation of the form z f x y N x y f1 f2 1 For the surface of interest here it follows that N x y 2x 2y 1 2x 2y 1 Since we also have F X x y x y 4 x2 y 2 it follows that the flux F N is given by 2x2 2y 2 4 x2 y 2 4 x2 y 2 Therefore the flux integral is equal to Z Z 4 x2 y 2 dA D where D is the disk defined by x2 y 2 4 One simple way to evaluate this integral is by means of polar coordinates and it follows that the flux integral must be equal to Z 2 0 Z 2 4 r 2 r dr d 0 Evaluating this iterated integral from the inside out we see that it is equal to Z 2 0 Z 2 3 4r r dr d 0 Z 2 r4 2r 4 2 2 0 0 d Z 2 8 4 d 2 12 24 0 PROBLEM 2 A modified version of Problem 15 on page 438 of the text Let S be the boundary of the cylinder defined by x2 y 2 9 and 0 z 4 and take the preferred orientation pointing outward from S Compute the flux integral of F x y z x y z 1 over S Solution This surface has three separate smooth pieces and we need to check that each of them is parametrized so that the normal is pointing outward i Top face of the cylinder This is given by x2 y 2 9 and z 4 so it can be parametrized as the graph of f x y 4 Over this piece of the surface the outward pointing normal direction is the same as the upward pointing direction so the given parametrization yields the outward pointing normal and if we apply the formula from the preceding section we see that N 0 0 1 on this piece of the surface ii Lateral face of the cylinder This is given by x2 y 2 9 and 0 z 4 so it can be parametrized by X z 3 cos 3 sin z for 0 2 and 0 z 4 If we compute the

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