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Unformatted text preview: MATH 32B  Lecture 4  Winter 2008 Final Exam Solutions  March 17, 2008 NAME: STUDENT ID #: DISCUSSION SECTION: This is a closedbook and closednote examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 9 problems for a total of 200 points. 1 2 1. (20 points) Compute the line integral Z C ( e x sin y 3 y ) dx + Z C ( e x cos y 3) dy where C is the upper half of the circle (i.e. y 0) x 2 + y 2 = ax and a is a positive constant. Solution. If we calculate Q x P y = 3 so the vector field is not quite conservative, however it looks like a good problem for Greens theorem. The path is not closed, but if we add a path from ( a, 0) to (0 , 0) it will become closed. The easiest path to choose is the straight line so our integral is equal to Z C 1 ( e x sin y 3 y ) dx +( e x cos y 3) dy + Z C 2 ( e x sin y 3 y ) dx +( e x cos y 3) dy where C 1 the path around the semicircle in the negative orientation. and C 2 is the line from (0 , 0) to ( a, 0). We can compute the first integral by Greens theorem, = ZZ D Q x P y = 3 Area ( D ) = 3( a 2 8 ) . We compute the second integral directly, we can parameterize the line by r ( t ) = h t, i t a, then r ( t ) = h 1 , i so our integral is Z C 2 ( e x sin y 3 y ) dx + ( e x cos y 3) dy = Z a (0)(1) + ( e t 3)(0) dt = 0 . thus our original integral is 3 a 2 8 . 3 2. (25 points) Determine whether F ( x, y, z ) = ( 2 x cos y 2 z 3 ) i + (3 + 2 ye z x 2 sin y ) j + ( y 2 e z 6 xz 2 ) k is conservative on R . If so, find a potential function f with F = f . Solution. Note that the domain of F is R 3 so to show its conservative we just need to show that curl F = 0. R y = 2 ye z = Q z , P z = 6 z 2 = R x and Q x = 2 x sin y = P y thus curl F = 0 and F is conservative. To find f note f z = y 2 e z 6 xz 2 so f = y 2 e z 2 xz 3 + g ( x, y ) . Then f x = 2 z 3 + g x ( x, y ) = 2 x cos y 2 z 3 . Thus g x = 2 x cos y so g = x 2 cos y + h ( y ), and f = y 2 e z 2 xz 3 + x 2 cos y + h ( y ) this implies f y = 2 ye z x 2 sin y + h ( y ) = 3 + 2 ye z x 2 sin y so h = 3 and h = 3 y + C so the potential function is f = y 2 e z 2 xz 3 + x 2 cos y + 3...
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This note was uploaded on 05/01/2008 for the course MATH 32B taught by Professor Rogawski during the Spring '08 term at UCLA.
 Spring '08
 Rogawski
 Math

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