# A2 b 2 42 31 10 42 32 15 15 3 5 25 21

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 θdρ dφ dθ = 0 0 0 192 π 5 26. Let G1 be the region u2 + v 2 + w2 ≤ 1, let x = au, y = bv, z = cw, (y 2 + z 2 )dx dy dz = Ix = G (b2 v 2 + c2 w2 )du dv dw G1 2π π 1 abc(b2 sin2 φ sin2 θ + c2 cos2 φ)ρ4 sin φ dρ dφ dθ = 0 0 2π = 0 0 4 abc 2 2 (4b sin θ + 2c2 )dθ = πabc(b2 + c2 ) 15 15 27. Let u = y − 4x, v = y + 4x, then x = 1 u dAuv = v 8 1 8 S 5 2 2 0 1 2 uv dAuv = − 2 1 2 1 ∂ (x, y ) 1 1 (v − u), y = (v + u) so =− ; 8 2 ∂ (u, v ) 8 15 u du dv = ln v 42 28. Let u = y + x, v = y − x, then x = − 1 ∂ (x, y ) 1 1 (u − v ), y = (u + v ) so =; 2 2 ∂ (u, v ) 2 1 uv du dv = − 0 S 0 29. Let u = x − y, v = x + y , then x = 1 2 1 ∂ (x, y ) 1 1 (v + u), y = (v − u) so = ; the boundary curves of 2 2 ∂ (u, v ) 2 the region S in the uv -plane are u = 0, v = u, and v = π/4; thus 1 sin u dAuv = cos v 2 1 2 S ∂ (x, y, z ) = abc; ∂ (u, v, w) π /4 0 v 0 √ 1 sin u du dv = [ln( 2 + 1) − π/4] cos v 2 Exercise Set 16.8 602 1 ∂ (x, y ) 1 1 (v − u), y = (u + v ) so = − ; the boundary 2 2 ∂ (u, v ) 2 curves of the region S in the uv -plane are v = −u, v = u, v = 1, and v = 4; thus 30. Let u = y − x, v = y + x, then x = 1 2 v 4 1 2 eu/v dAuv = eu/v du dv = −v 1 S 15 (e − e−1 ) 4 31. Let u = y/x, v = x/y 2 , then x = 1/(u2 v ), y = 1/(uv ) so 4 1 dAuv = u4 v 3 2 1 S 1 1 ∂ (x, y ) = 4 3; ∂ (u, v ) uv 1 du dv = 35/256 u4 v 3 ∂ (x, y ) = 6; S is the region in the uv -plane enclosed by the circle u2 + v 2 = 1 ∂ (u, v ) 32. Let x = 3u, y = 2v , 2π (9 − x − y )dA = so R S 2 1 3 0 0 1 ∂ (x, y, z ) =− ; ∂ (u, v, w) u 33. x = u, y = w/u, z = v + w/u, 4 (9 − 3r cos θ − 2r sin θ)r dr dθ = 54π 0 S v2 w dVuvw = u 1 6(9 − 3u − 2v )dAuv = 6 1 v2 w du dv dw = 2 ln 3 u 34. u = xy, v = yz, w = xz, 1 ≤ u ≤ 2, 1 ≤ v ≤ 3, 1 ≤ w ≤ 4, x= uw/v, y = V= uv/w, z = dV = 2 G 1 3 1 4 1 1 ∂ (x, y, z ) =√ ∂ (u, v, w) 2 uvw √ √ dw dv du = 4( 2 − 1)( 3 − 1) v w/u, 1 2 uvw √ 35. (b) If x = x(u, v ), y = y (u, v ) where u = u(x, y ), v = v (x, y ), then by the chain rule ∂x ∂x ∂u ∂x ∂v ∂x ∂x ∂u ∂x ∂v + = = 1, + = =0 ∂u ∂x ∂v ∂x ∂x ∂u ∂y ∂v ∂y ∂y ∂y ∂y ∂u ∂y ∂v ∂y ∂y ∂u ∂y ∂v + = = 0, + = =1 ∂u ∂x ∂v ∂x ∂x ∂u ∂y ∂v ∂y ∂y ∂ (x, y ) = ∂ (u, v ) 1−v v ∂ (u, v ) = ∂ (x, y ) 36. (a) −u u 1 −y/(x + y )2 = u; u = x + y, v = 1 x/(x + y )2 = y , x+y 1 y 1 x =; + = (x + y )2 (x + y )2 x+y u ∂ (u, v ) ∂ (x, y ) =1 ∂ (x, y ) ∂ (u, v ) (b) ∂ (x, y ) = ∂ (u, v ) ∂ (u, v ) = ∂ (x, y ) √ √ u = 2v 2 ; u = x/ y, v = y, 2v √ 1 ∂ (u, v ) ∂ (x, y ) 1 1/ y −x/(2y −3/2 ) √ = 2; =1 = 0 1/(2 y ) 2y 2v ∂ (x, y ) ∂ (u, v ) v 0 603 Chapter 16 (c) ∂ (x, y ) = ∂ (u, v ) ∂ (u, v ) = ∂ (x, y ) 37. √ √ v = −2uv ; u = x + y, v = x − y, −v √ √ 1/(2 x + y ) 1/(2 x + y ) 1 ∂ (u, v ) ∂ (x, y ) 1 ; =1 =− =− √ √ 2 − y2 2uv ∂ (x, y ) ∂ (u, v ) 1/(2 x − y ) −1/(2 x − y ) 2x u u 11 ∂ (x, y ) ∂ (u, v ) = 3xy 4 = 3v so = ; ∂ (x, y ) ∂ (u, v ) 3v 3 1 S 38. ∂ (x, y ) 1 ∂ (x, y ) ∂ (u, v ) = 8xy so = ; xy = xy ∂ (x, y ) ∂ (u, v ) 8xy ∂ (u, v ) 1 8 dAuv = 1 8 S 39. 16 1 8xy 2π 2 1 sin u dAuv = v 3 = π 2 sin u du dv = − ln 2 v 3 1 so 8 4 du dv = 21/8 9 1 1 ∂ (x, y ) ∂ (u, v ) = −2(x2 + y 2 ) so =− ; ∂ (x, y ) ∂ (u, v ) 2(x2 + y 2 ) (x4 − y 4 )exy 1 2 x4 − y 4 xy 1 1 ∂ (x, y ) = e = (x2 − y 2 )exy = veu so ∂ (u, v ) 2(x2 + y 2 ) 2 2 veu dAuv = 4 1 2 3 veu du dv = 3 S 1 73 (e − e) 4 40. Set u = x + y + 2z, v = x − 2y + z, w = 4x + y + z , then 6 2 3 −6 −2 −3 dx dy dz = V= R ∂ (u, v, w) = ∂ (x, y, z ) 1 12 1 −2 1 4 11 = 18, and 1 ∂ (x, y, z ) du dv dw = 6(4)(12) = 16 ∂ (u, v, w) 18 41. (a) Let u = x + y, v = y , then the triangle R with vertices (0, 0), (1, 0) and (0, 1) becomes the triangle in the uv -plane with vertices (0, 0), (1, 0), (1, 1), and u 1 f (x + y )dA = f (u) 0 R 1 ueu du = (u − 1)eu (b) 0 42. (a) (b) ∂ (x, y ) = ∂ (r, θ) ∂ (x, y, z ) = ∂ (ρ, θ, φ) cos θ sin θ 0 ∂ (x, y ) dv du = ∂ (u, v ) 1 uf (u) du 0 1 =1 0 −r sin θ r cos θ sin φ cos θ sin φ sin θ cos φ = r, ∂ (x, y ) =r ∂ (r, θ) −ρ sin φ sin θ ρ sin φ cos θ 0 ρ cos φ cos θ ρ cos φ sin θ −ρ sin φ = −ρ2 sin φ; ∂ (x, y, z ) = ρ2 sin φ ∂ (ρ, θ, φ) Chapter 16 Supplementary Exercises 604 CHAPTER 16 SUPPLEMENTARY EXERCISES 3. (a) dA (b) dV R G 2 ∂z ∂x (c) + ∂z ∂y 2 dA R 4. (a) x = a sin φ cos θ, y = a sin φ sin θ, z = ρ cos φ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π (b) x = a cos θ, y = a sin θ, z = z, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h 1 1+ 7. 1− 0 √ √ 1−y 2 2x 2 f (x, y ) dx dy 8. 3 6−x f (x, y ) dy dx + 1−y 2 x 0 f (x, y ) dy dx 2 x 9. (a) (1, 2) = (b, d), (2, 1) = (a, c), so a = 2, b = 1, c = 1, d = 2 1 1 dA = (b) 0 R 10. 0 < sin √ 0 π 0= π 0 0 1 2x cos(πx2 ) dx = 11. 1/2 2 0 π 0 dy dx < 0 x2 y3 e 2 1 1 1 3du dv = 3 0 0 xy < 1 for 0 < x,y &...
View Full Document

Ask a homework question - tutors are online