Question

ohpolar0 124) The Silk Road...Villa Gabriela lugar... Math 392 Lecture 4.. ork MATHEMATICAL ASSOCIATION OF AMERICA webwork /19sp392 j/13.9/4 13.9: Problem 4 (1 point) The region W lies between the spheres z2 + y2 + z2 1 and z2 + y2 + z2 9 and within the cone z , z4y2 with z > 0; its boundary is the closed surface, S, oriented outward. Find the flux of Fiyj+kout of S ms flux= ((729(2-sqrt(2))pi))/5 You have attempted this problem 2 times Your overall r score is 0%. You have unlimited attempts remaining. Page generated at 05 18/2019 at 12:57am EDT WeBWork 1996-2016 | theme: math4 I ww version: 2.11 l pg_version: 2.11 The WeBWorK Project 12:57 AM ENG 5/n8/2019 A Type here to search

Answer 1

The given answer is not correct.

vec F = x^3 hat I + y^3 hat j + z^3 hat k S is the given surface Therefore div vec F = 3 x^2 + 3 y^2 + 3 z^2 By divergence theorem doubleintegral_s vec F hat n ds = tripleintegral_v div vec F dv = tripleintegral_v 3 (x^2 + y^2 + z^2) dv Now x = r sin Theta cos phi y = r sin Theta sin phi z = r cos Theta 1 lessthanorequalto r lessthanorequalto 3 S surface of the solid between x^2 + y^2 + z^2 = 1 & x^2 + y^2 + z^2 = 9 and with in z = squareroot x^2 + y z greaterthanorequalto 0 and r^2 sin^2 Theta = r^2 cos^2 Theta, {Because z^2 = x^2 + y^2] Rightarrow tan^2 Theta = 1 Rightarrow tan Theta = 1 Rightarrow Theta = pi/4 Therefore 0 lessthanorequalto Theta lessthanorequalto pi/4