# Find the mass of the wire formed by the intersection of the sphere x 2 + y 2 + z 2 = 1 and the plane x + z = 0 if the density of the wire is 4x 2 grams per unit length.

Answer: Answer: xÂ˛+(â’xâ’z)Â˛+zÂ˛ = 1 â†’ 2xÂ˛+2zx+(2zÂ˛â’1) = 0 Solve as a quadratic : x = { â’2z Â± âš(4zÂ˛â’8(2zÂ˛â’1) } / 4 = { â’z Â± âš(2â’3zÂ˛) } / 2 A more manageable form can be derived by setting z = âš(2/3).sint to remove âš This gives x = â’âš(1/6).sint + âš(1/2).cost (can omit Â± sign since â’ is covered by Ď€â’t) Using x+y+z = 0 gives y = â’xâ’z = â’âš(1/6)sint â’ âš(1/2).cost So a parameterisation in terms of t â [0,2Ď€] is x = â’âš(1/6).sint + âš(1/2).cost, y = â’âš(1/6).sint â’ âš(1/2).cost, z = âš(2/3).sint (ds/dt)Â˛ = (â’âš(1/6).costâ’âš(1/2).sint)Â˛ + (â’âš(1/6).cost+âš(1/2).sint)Â˛ + (âš(2/3).cost)Â˛ = 1 â´ â« Ďds = â« ( â’âš(1/6).sint + âš(1/2).cost )Â˛dt, [t=0,2Ď€] = 2Ď€/3

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