Hw15-8 - Mathematics 1920 Section 15.8 6. The boundaries y...

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Unformatted text preview: Mathematics 1920 Section 15.8 6. The boundaries y = −2x + 4, y = −2x + 7 correspond to v = 4 and v = 7 respectively. The boundaries y = x − 2, y = x + 1 correspond to u = 2 and u = −1 respectively. Note that 1 (u + v ) 3 1 y= (−2u + v ) 3 x= Thus the Jacobian is J (u, v ) = = = ∂y ∂u ∂y ∂v 1 3 1 3 ∂x ∂u ∂x ∂v 1 3 −2 3 1 3 (2x2 − xy − y 2 )dxdy = R (x − y )(2x + y )dxdy R uv |J (u, v )|dudv = G = 1 3 1 = 3 11 = 2 2 7 uvdudv −1 2 4 u2 v 2 −1 7 dv 4 2 udu −1 2 11 u 22 33 = 4 2 = 15. x = u v and y = uv =⇒ y x = v 2 and xy = u2 . −1 ∂ (x,y ) ∂ (u,v ) = J (u, v ) = is J (u, v ) = = ∂x ∂u ∂x ∂v −1 ∂y ∂u ∂y ∂v v v −uv −2 u = v −1 u + v −1 u 2u = v y = x =⇒ uv = u =⇒ v = 1 and y = 4x =⇒ uv = v xy = 1 =⇒ u = 1 and xy = 4 =⇒ u = 2, thus 2 y 4 2 4/y 2 (x + y )dxdy + 1 1 y 2 2 2 2 (x + y )dxdy = 2 y/4 1 1 2 2 = 1 1 2 = 1 2 = 1 4u v =⇒ v = 2. u2 + u2 v 2 2 v 2u3 + 2u3 v v3 15v 2 15 = − 2+ 4v 4 255 = 16 18. (a)x = u cos v and y = u sin v =⇒ J (u, v ) = = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v cos v −u sin v sin v u cos v = u cos2 v + u sin2 v =u 2 u4 1 + u4 v dv 3 2v 2 1 15 15v + dv 2v 3 2 2 1 2u v dudv (b) x = u sin v and y = u cos v implies J (u, v ) = = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v sin v u cos v cos v −u sin v = −u sin2 v − u cos2 v = −u 19. The Jacobian is sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ cos φ −ρ sin θ 0 which evaluates to ρ2 sin φ. 23. a00 0b0 00c = abc J (u, v, w) = Note that the ellipsoid (R) in xyz-space transforms into unit sphere(G) in uvw-space. Thus we can transform the integral into uvw-space as a2 b2 c2 uvwdwdvdu |xyz |dxdydz = R G Now we can transform this unit sphere G in uvw space to spherical coordinates as π /2 222 π /2 1 222 a b c uvwdwdvdu = 8a b c G (ρ sin φ cos θ)(ρ sin φ sin θ) 0 0 0 (ρ cos φ)(ρ2 sin φ)dρdφdθ 4a2 b2 c2 3 a2 b 2 c 2 = 3 222 abc = 6 π /2 π /2 sin θ cos θ sin3 φ cos φdφdθ = 0 π /2 0 sin θ cos θdθ 0 ...
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Hw15-8 - Mathematics 1920 Section 15.8 6. The boundaries y...

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