X2 y 2 z 2 4 2 x2 y 2 4 22 sin2 2 2

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Unformatted text preview: e line integral by Green’s Theorem. 5. Since P (x, y ) = x2 Parametric Equations for Surfaces and Surface Areas 1. (a) Let x = √ 3 u, 3 y= √ 2 v 2 and z = w, then 3x2 + 2y 2 + z 2 = 1 =⇒ u2 + v 2 + w2 = 1 with w ≥ 0. Now, the ellipsoid is transformed into a unit sphere, so u (φ, θ) = sin φ cos θ, v (φ, θ) = sin φ sin θ, and w (φ, θ) = cos φ, with π 0 ≤ φ ≤ and 0 ≤ θ ≤ 2π . 2 Therefore, r (φ, θ) = √ √ π 3 2 sin φ cos θ, sin φ sin θ, cos φ , where 0 ≤ φ ≤ and 0 ≤ θ ≤ 2π . 3 2 2 6 z z= 1 − 3x2 − 2y 2 y x Figure 38 (b) The intersection on the xz -plane, y = 0, is the ellipse 3x2 + 2z 2 = 6, so r (x, z ) = x, 6 − 3x2 − 2z 2 , z , where 0 ≤ 3x2 + 2z 2 ≤ 6. (c) x2 + y 2 + z 2 = 4 is the sphere with radius of 2, so ρ = 2. π x2 + y 2 + z 2 = 4 =⇒ 2 x2 + y 2 = 4 =⇒ (2)2 sin2 φ = 2 =⇒ φ = 2 + y2 z= x 4 Therefore, r (φ, θ) = [2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ] , where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π . 4 (d) r (r, θ) = [r cos θ, r sin θ, r cos θ + 3] , where 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π 2. (a) Let z = f (x, y ) = 4 − x − 2y , then ∂z ∂z = −1, = −2 and ∂x ∂y ∂z ∂x 1+ 2 + ∂z ∂y 2 = √ 6. D = {(r, θ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π } √ A (S ) = 2π 2 6 dA = 0 √ √ 6 (r dr dθ) = 4 6π 0 D (b) Let z = f (x, y ) = x + y 2 , then ∂z ∂x 1+ 2 + D = {(x, y ) | 0 ≤ x ≤ y, 0 ≤ y ≤ 1} A (S ) = 0 2 = 2 + 4y 2 . √ √ 3 6− 2 2 + 4y 2 dx dy = 6 y 1 ∂z ∂y 0 (c) r (u, v ) = [uv, u + v, u − v ] and D = {(r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π } ru = [v, 1, 1] , rv = [u, 1, −1] and ru × rv = [−2, u + v, v − u] √ √ ru × rv = 4 + 2u2 + 2v 2 = 4 + 2r2 √ 1√ 2π √ π 6 6−8 4 + 2u2 + 2v 2 dA = 4 + 2r2 (r dr dθ) = A (S ) = 3 0 0 D 7 (d) r (u, v ) = [u cos v, u sin v, v ] and D = {(u, v ) | 0 ≤ u ≤ 1, 0 ≤ v ≤ π } ru = [cos v, sin v, 0] , rv = [−u sin v, u cos v, 1] and...
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