Question

2) Potential energy U in a liquid film surface S can be calculated by the flux integral U-L ?.del joules, where ?(x),2) is the surface tension at the point (x.y,z). Use the Divergence Theorem to find the potential energy in a membrane surface shaped like a paraboloid with equation z- Zo -x -y' for z20 and oriented inward with a surface tension ?(x,y, z) Newtons Flux and Circulation Please handwrite and write legibly. Please I need solutions and answers as soon as possible. Many thanks and appreciation

Answer 1

Answer: Given Sigma (x, y, z) = yi^- - xj^- - zk^-/x^2 + y^2 + z^2 Newtons dw Sigma = partial differential/partial differential x(y/x^2 + y^2 + z^2) + partial differential/partial differential y (-x/x^2 + y^2 + z^2) + partial differential/partial differential z (-z/x^2 + y^2 + z^2) = -2xy/(x^2 + y^2 + z^2) + 2xy/(x^2 + y^2 + z^2) + 2xy/(x^2 + y^2 + z^2) + -1 (x^2 + y^2 + z^2) + z^2/(x^2 + y^2 + z^2) = z^2 - x^2 - y^2/x^2 + y^2 + z^2 Now, we have integral integral integral z^2 - x^2 - y^2/x^2 + y^2 + z^2 dx z = zo^z - x^z.y^z z > = 0 x^+ y^z = zo^z As in 0 < = r < = > 0 \r\n 0 < = theta < = 2 pi So in polar we have integral^x_0 integral k^z_0 integral^z^2_0 - r^2_0 z^2 - r^2/(r^2 + z^2)^2 r dz dr d theta