Week 1112 - Solution

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

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: 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...
View Full Document

Ask a homework question - tutors are online