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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 C H A P T E R 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 ( u , v) that maps the unit square to the parallelogram in the xy -plane spanned by the vectors 3 , 1 and 1 , 4 . Then, use the Jacobian to find the area of ( R ) , where R = [ 0 , 4 ] × [ 0 , 3 ] . SOLUTION We denote the linear map by ( u , v) = ( 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 of the unit square is the quadrangle whose vertices are the images of the vertices of the square. Therefore 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 following map: ( u , v) = ( 3 u + v, u + 4 v) The area of the rectangle R = [ 0 , 4 ] × [ 0 , 3 ] is 4 · 3 = 12, therefore the area of ( R ) is Area ( ( R )) = | Jac ( ) | · 12 The Jacobian of the linear map is Jac ( ) = A C B D = 3 1 1 4 = 12 ( 1 ) = 13 Therefore, Area ( ( R )) = 13 · 12 = 156 . 43. Use the map ( u , v) = u + v 2 , u v 2 to compute R ( ( x y ) sin ( x + y ) ) 2 dx dy , where R is the square with vertices ( π , 0 ) , ( 2 π , π ) , ( π , 2 π ) , and ( 0 , π ) . SOLUTION We express f ( x , y ) = (( x y ) sin ( x + y )) 2 in terms of u and v . Since x = u + v 2 and y = u v 2 , we have x y = v and x + y = u . Hence, f ( x , y ) = v 2 sin 2 u . We find the Jacobian of the linear transformation: Jac ( ) = 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 onto R , we first find the inverse of by solving the following equations for u , v in terms of x and y : x = u + v 2 y = u v 2
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