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.

x²+(â’xâ’z)²+z² = 1 → 2x²+2zx+(2z²â’1) = 0 Solve as a quadratic : x = { â’2z ± âš(4z²â’8(2z²â’1) } / 4 = { â’z ± âš(2â’3z²) } / 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 ± 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)² = (â’âš(1/6).costâ’âš(1/2).sint)² + (â’âš(1/6).cost+âš(1/2).sint)² + (âš(2/3).cost)² = 1 â´ â« Ďds = â« ( â’âš(1/6).sint + âš(1/2).cost )²dt, [t=0,2Ď€] = 2Ď€/3