notes001 (86)

# notes001 (86) - and j r ( t ) j = 2 and we have Z C ( x + y...

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November 5, 2010 NAME______________________________ correct notation) in order to get full credit. 1) Change the Cartesian integral Z 1 1 Z p 1 y 2 p 1 y 2 x 2 + y 2 ± dx dy into an equivalent polar integral and then evaluate the integral. Solution: By drawing a picture of the region of integration, we see that Z 1 1 Z p 1 y 2 p 1 y 2 x 2 + y 2 ± dx dy = Z 2 & 0 Z 1 0 r 3 = ± 2 . 2) Evaluate the line integral Z C ( x + y ) ds where C is the curve x 2 + y 2 = 4 starting at the point (2 ; 0) and ending at the point (0 ; 2) . Solution: The curve can be parameterized as r ( t ) = 2 cos ( t ) i + 2 sin ( t ) j 0 t ± 2 . Thus r 0 ( t ) = ± 2 sin ( t ) i + 2 cos ( t ) j

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Unformatted text preview: and j r ( t ) j = 2 and we have Z C ( x + y ) ds = Z &= 2 (2 cos ( t ) + 2 sin ( t )) (2) dt = 8 . 3) Let V ( x; y ) be the velocity eld V ( x; y ) = x i + y j . 1 Find the &ow along and the &ux across the curve r ( t ) = cos ( t ) i + sin ( t ) j & t & & . Solution: Here we have M ( x; y ) = x , N ( x; y ) = y , dx=dt = sin ( t ) and dy=dt = cos ( t ) . The &ow is thus Z & ((cos ( t )) ( sin ( t )) + (sin ( t )) (cos ( t ))) dt = 0 and the &ux is Z & ((cos ( t )) (cos ( t )) (sin ( t )) ( sin ( t ))) dt = & . 2...
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## This note was uploaded on 02/05/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.

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notes001 (86) - and j r ( t ) j = 2 and we have Z C ( x + y...

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