Question

If the mass per unit area of a surface is given by p-xy, find the mass if S is the part of the cylinder 2z2 9 which is in the first octant and contained within the cylinder y24. mass

If the mass per unit area of a surface is given by ρ=xyρ=xy, find the mass

∫∫SxydS∫∫SxydS

if S is the part of the cylinder x2+z2=9x2+z2=9 which is in the first octant and contained within the cylinder x2+y2=4x2+y2=4..

mass =

Answer 1

Given that The mass per area of a Surfaces is given by rho = xy S is the Surface x^2 + z^2 = 9 implies z = squareroot 9-x^2 Apply Partial derivation on both sides partial differential z/ partial differential x = partial differential/ partial differential x (squareroot 9-x^2) = -x/ squareroot 9-x^2 and partial differential z/ partial differential y = 0 The Surface area is ds = squareroot 1 + (partial differential z/ partial differential x)^2 + (partial differential z/ partial differential y)^2 dx dx = squareroot 1 + [x/ squareroot 9-x^2]^2 dx dy = squareroot 9-x62 + x^2/9-x^2 dx dy = squareroot 9/9-x^2 dx dy ds = 3/ squareroot 9-x^2 dx dy \r\n The part of S is in the first octant and Contained the Cylinder x^2 + y^2 = 4 So, D is bounded by the Circle x^2 + y62 = 4 and Y-axis D = {(x, y)/0 lessthanorequalto x lessthanorequalto squareroot 4-y^2, 0 lessthanorequalto y lessthanorequalto 2} therefore The mass is m = doubletintegral_s xy ds