16.5 Ex 33 - 36

# 16.5 Ex 33 - 36 - 1008 C H A P T E R 16 M U LTI P L E I N T...

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1008 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) Therefore, D is deFned by the inequalities 1 y x 2 , 3 y + x 6 Since x = u v + 1 and y = u v v + 1 ,wehave y x = u v v + 1 u v + 1 = v and y + x = u v v + 1 + u v + 1 = u (v + 1 ) v + 1 = u Therefore, the corresponding domain D 0 in the u v -plane is the rectangle D 0 : 1 v 2 , 3 u 6 y Φ u x D 0 D 6 3 1 3 6 6 3 1 2 We express the function f ( x , y ) = x + y in terms of u and v : f ( x , y ) = u v v + 1 + u v + 1 = u v + u v + 1 = u (v + 1 ) v + 1 = u We compute the Jacobian of 8 .Since x = u v + 1 and y = u v v + 1 x u = 1 v + 1 , x ∂v =− u (v + 1 ) 2 , y u = v v + 1 , y = u (v + 1 ) u v · 1 (v + 1 ) 2 = u (v + 1 ) 2 Therefore, Jac (8) = ∂( x , y ) u ,v) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 v + 1 u (v + 1 ) 2 v v + 1 u (v + 1 ) 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1 v + 1 · u (v + 1 ) 2 + u (v + 1 ) 2 · v v + 1 = u + u v (v + 1 ) 3 = u ( 1 + v) (v + 1 ) 3 = u ( 1 + 2 Now we can apply the Change of Variables ±ormula to compute the integral: ZZ D ( x + y ) dx dy = D 0 u · u ( 1 + 2 dud v = Z 2 1 Z 6 3 u 2 · 1 ( 1 + 2 v = ± Z 2 1 1 ( 1 + 2 d v Z 6 3 u 2 du ! = ± 1 1 + v ¯ ¯ ¯ ¯ 2 1 u 3 3 ¯ ¯ ¯ ¯ 6 3 ! = µ 1 3 + 1 2 ¶µ 216 3 27 3 = 63 6 = 10 . 5 33. Show that T ( u = ( u 2 v 2 , 2 u maps the triangle D 0 ={ ( u : 0 v u 1 } to the domain D bounded by x = 0, y = 0, and y 2 = 4 4 x .Use T to evaluate D q x 2 + y 2 SOLUTION We show that the boundary of D 0 is mapped to the boundary of D .

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SECTION 16.5 Change of Variables (ET Section 15.5) 1009 y x D 2 1 v u D 0 1 T y 2 = 4 4 x v = u We have x = u 2 v 2 and y = 2 u v The line v = u is mapped to the following set: ( x , y ) = ( u 2 u 2 , 2 u 2 ) = ( 0 , 2 u 2 ) x = 0 , y 0
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## This homework help was uploaded on 04/13/2008 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.

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16.5 Ex 33 - 36 - 1008 C H A P T E R 16 M U LTI P L E I N T...

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