By Temam R

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Hence O y2 dS z 2π = = a ⎩ 0 1 a ⎧ ⎨ 2π 0 0 2 sin2 ϕ a cosh · · cosh ⎫ ⎬ a a a sin2 ϕ dϕ · 3 d = 0 d ⎭ dϕ 1 1 πa3 · π · a4 = . 4 Calculate in each of the following cases the given surface integral over the surface given by a parametric description F = x ∈ R3 | x = r(u, v), (u, v) ∈ E . First ﬁnd the normal vector of the surface N(u, v). 1) The surface integral F xz 2 dS, where the surface F is given by for 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π. x = r(u, v) = (u cos v, u sin v, hv), 2) The surface integral F z 2 dS, where the surface F is given by √ √ x = r(u, v) = ( u cos v, u sin v, ev ) 3) The surface integral x = r(u, v) = √ 4) The surface integral F ln(2u) ln u ≤v≤ .

3. 2) We get from dS 10 + x2 dx dy = = F 1+ g = (x, 3) that E 1 = 1+ −1 x2 6 Then by the substitution x = dS = F = 1 3 20 3 = 20 3 = 5 3 = = = = = 5 6 0 0 √ 10 + x2 . The surface area is 1 1 −1 − x6 10 + x2 dy 2 2 6 (6 + 10 sinh2 t) · Arsinh( √110 ) Arsinh( √110 ) 0 = 1 (6 + x2 ) dx 10 + x2 dx. 0 √ x , 10 sinh t, t = Arsinh √ 10 Arsinh( √110 ) 0 2 10 + x2 dx = Arsinh( √110 ) 0 g Arsinh( √110 ) √ √ 10 cosh t · 10 cosh t dt (3 + 5 sinh2 t) cosh2 t dt 3 5 (1 + cosh 2t) + sinh2 2t dt 2 4 5 6 + 6 cosh 2t + (cosh 4t − 1) dt 2 {7 + 12 cosh 2t + 5 cosh 4t}dt 5 5 7t + 6 sinh 2t + sinh 4t 6 4 Arsinh( √110 ) 0 Arsinh( √110 ) 5 7t + 12 sinh t 1 + sinh2 t + 5 sinh t 1 + sinh2 t · (1 + 2 sinh2 t) 6 0 1 1 1 2 11 11 11 5 7 ln √ + + 12 · √ · +5· √ · · 1+ 6 10 10 10 10 10 10 10 √ √ 6 + 11 12 √ 6 √ 35 5 1 + 11 3 √ √ · 11 + · 11 = ln + 7 ln + · 11.

Or have you already graduated? P. Moller - Maersk. com 50 Calculus 2c-8 Surface integral Then the surface integral becomes |x|(z+a) x2 +y 2 O 2a dS 2π = a 0 π 2 = 2 −π 2 | cos ϕ|(z+a) (z) dϕ cos ϕ dϕ · 2a = 4 (z + a) a 2a 2a (z+a) a 2z 2 +4az+a2 dz z 2 +2az z 2 +2az · 2z 2 +4aza2 dz z 2 +2az 2z 2 + 4az + a2 dz 2z 2 +4az+a2 d 2z 2 +4az+a2 = z=a = = = 3 2a 2 (2z 2 +4az+a2 ) 2 3 z=a 3 3 2 2 2 2 2 8a + 8a + a − 2a2 +4a2 +a2 2 3 3 3 √ √ 2 2 17a2 2 − 7a2 2 = {17 17 − 7 7}a3 . 3 3 4) The curve M is a part of an hyperbola, cf.

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