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

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MATH 2203 °Quiz 6 (Version 1) Solutions November 5, 2010 NAME______________________________ Instructions: All work must be shown in detail (and be well°written using 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 dr d° = ± 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 (

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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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