Ch16Rev Ex 42-46

Ch16Rev Ex 42-46 - 1050 C H A P T E R 16 M U LTI P L E I N...

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1050 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) = e 2 r cosh 2 ( s ) e 2 r sinh 2 ( s ) = e 2 r ( cosh 2 ( s ) sinh 2 ( s )) = e 2 r 42. Find a linear mapping 8( u ,v) that maps the unit square to the parallelogram in the xy -plane spanned by the vectors h 3 , 1 ± and h 1 , 4 ± . Then, use the Jacobian to fnd the area o± R ) ,where R =[ 0 , 4 ]×[ 0 , 3 ] . SOLUTION We denote the linear map by u = ( Au + C v, Bu + D v) (1) Φ (0, 0) (1, 0) (0, 1) u (1, 4) x y (3, 1) (0, 0) The image o± the unit square is the quadrangle whose vertices are the images o± the vertices o± the square. There±ore we ask that 0 , 0 ) = ( A · 0 + C · 0 , B · 0 + D · 0 ) = ( 0 , 0 ) 1 , 0 ) = ( A · 1 + C · 0 , B · 1 + D · 0 ) = ( 3 , 1 ) 0 , 1 ) = ( A · 0 + C · 1 , B · 0 + D · 1 ) = ( 1 , 4 ) ( 0 , 0 ) = ( 0 , 0 ) ( A , B ) = ( 3 , 1 ) ( C , D ) = ( 1 , 4 ) These equations imply that A = 3, B =− 1, C = 1, and D = 4. Substituting in (1) we obtain the ±ollowing map: u = ( 3 u + u + 4 The area o± the rectangle R 0 , 4 0 , 3 ] is 4 · 3 = 12, there±ore the area o± R ) is Area (8( R )) =| Jac (8) 12 The Jacobian o± the linear map 8 is Jac (8 ) = ¯ ¯ ¯ ¯ AC BD ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 31 14 ¯ ¯ ¯ ¯ = 12 ( 1 ) = 13 There±ore, Area (8( R )) = 13 · 12 = 156 . 43. Use the map u = µ u + v 2 , u v 2 to compute ZZ R ( ( x y ) sin ( x + y ) ) 2 dx dy R is the square with vertices ( π , 0 ) , ( 2 , ) , ( , 2 ) ,and ( 0 , ) . We express f ( x , y ) = (( x y ) sin ( x + y )) 2 in terms o± u and v .Since x = u + v 2 and y = u v 2 ,wehave x y = v and x + y = u . Hence, f ( x , y ) = v 2 sin 2 u . We fnd the Jacobian o± the linear trans±ormation: Jac (8) = ¯ ¯ ¯ ¯ ¯ 1 2 1 2 1 2 1 2 ¯ ¯ ¯ ¯ ¯ 1 4 1 4 1 2 R R 0 x y u v Φ (π, 0) (π, 2π) (2π, π) (π, π) (3π, π) (π, −π) (3π, −π) (0, π)

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Chapter Review Exercises 1051 To compute the vertices of the quadrangle R mapped by 8 onto R ,weFrstFndtheinverseof 8 by solving the following equations for u , v in terms of x and y : x = u + v 2 y = u v 2 u + v = 2 x u v = 2 y u = x + y ,v = x y Hence, 8 1 ( x , y ) = ( x + y , x y ) We now compute the vertices of P
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This note 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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Ch16Rev Ex 42-46 - 1050 C H A P T E R 16 M U LTI P L E I N...

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