Week 1112 - Solution

# f x y z yz f r cos r sin g r cos r sin

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Unformatted text preview: ru × rv = [sin v, − cos v, u] √ r u × r v = 1 + u2 √ √ 1√ π π 2 + ln 1 + 2 1 + u2 du dv = A (S ) = 2 0 0 Note: √ a2 + x2 dx = √ x√ 2 a2 a + x2 + ln x + a2 + x2 + C , where a ∈ R. 2 2 z S x y Figure 39 Surface Integrals 1. (a) 3x + 2y + z = 6 =⇒ z = g (x, y ) = 6 − 3x − 2y 3 D = (x, y ) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 − 2 x f (x, y, z ) = y =⇒ f (x, y, g (x, y )) = y dS = 2 ∂z ∂x ∂z ∂y + 3 3− 2 x 2 y dS = 0 2 √ + 1 dA = √ (y ) 14 dy dx √ 14 dy dx = 3 14 0 S (b) Triangle ABC lies on the plane x + y + z = 1. x + y + z = 1 =⇒ z = g (x, y ) = 1 − x − y D = {(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x} f (x, y, z ) = xz =⇒ f (x, y, g (x, y )) = x (1 − x − y ) 2 ∂z ∂x dS = + 1−x 1 xz dS = 0 0 S 8 2 ∂z ∂y + 1 dA = 2 x − x − xy √ √ 3 dy dx 3 dy dx = √ 3 24 (c) y = g (x, z ) = x2 + 4z D = {(x, z ) | 0 ≤ x ≤ 2, 0 ≤ z ≤ 2} f (x, y, z ) = x =⇒ f (x, g (x, z ) , z ) = x 2 ∂y ∂x dS = 2 2 x dS = (x) 0 ∂y ∂z + √ 0 2 + 1 dA = √ 4x2 + 17 dz dx √ √ 33 33 − 17 17 4x2 + 17 dz dx = 6 S (d) z = g (x, y ) = y + 3 D = {(r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π } f (x, y, z ) = yz =⇒ f (r cos θ, r sin θ, g (r cos θ, r sin θ)) = (r sin θ) (r sin θ + 3) 2 ∂z ∂x dS = 2π 2 ∂z ∂y + + 1 dA = √ √ 2 dy dx = 2 r dr dθ 1 (r sin θ) (r sin θ + 3) yz dS = √ 2 r dr dθ = √ 0 0 2π 4 S 2. (a) r (x, y ) = x, y, x2 + y 2 with D = {(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} F (x, y, z ) = ey , yex , x2 y =⇒ F (r (x, y )) = ey , yex , x2 y rx = [1, 0, 2x] , ry = [0, 1, 2y ] and rx × ry = [−2x, −2y, 1] 1 F • dS = 0 1 0 ey , yex , x2 y • [−2x, −2y, 1] dy dx = 11 − 10e 6 S (b) r (x, y ) = [x, y, 6 − 3x − 2y ] with D = 3 (x, y ) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 − x 2 F (x, y, z ) = [x, xy, xz ] =⇒ F (r (x, y )) = x, xy, 6x − 3x2 − 2xy...
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