Create an iterated double integral that gives the surface area of the part of th

Question

Create an iterated double integral that gives the surface area of the part of the hyperboloid x2 + y2 - z2 = 1 between z = 0 and Z = 3. Use the parametrization r(Theta,z) = cos Theta, sin Theta,z. Show your work, but do not solve the integral. Compute F middot dS where F = y, z, x and S is the surface parameterized by r(u, v) = u2 - u, u + v, v2 with 0 u 2 and - 1 v 1. Use the orientation ru x rv. Show your work.

Explanation / Answer

Given r(θ, z) = : r_θ = r_z = So, r_θ x r_z = ==> ||r_θ x r_z|| = √((z^2 + 1) + z^2) = √(2z^2 + 1). Hence, the surface area equals ∫(θ = 0 to 2Ï€) ∫(z = 0 to 3) √(2z^2 + 1) dz dθ ------------------ 2) Differentiating: r_u = r_v = So, r_u x r_v = . Hence, ∫∫s F · dS = ∫(u = 0 to 2) ∫(v = -1 to 1) · dv du = ∫(u = 0 to 2) ∫(v = -1 to 1) [(2uv + 2v^2) + -4uv^3 + (u^2 - v)(2u + 1)] dv du = ∫(u = 0 to 2) ∫(v = -1 to 1) [(0 + 2v^2) + 0 + (u^2 - 0)(2u + 1)] dv du, via odd terms in v = ∫(u = 0 to 2) ∫(v = -1 to 1) (2v^2 + 2u^3) dv du = ∫(u = 0 to 2) (4/3 + 4u^3) du = (4u/3 + u^4) {for u = 0 to 2} = 56/3.

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