Consider a cube C 0 2 times 1 3 times 0 2 that is C consist

Consider a cube C = [0, 2] times [1, 3] times [0, 2], that is, C consists of all points (x, y, z) R^3 such that 0 lessthanorequalto x lessthanorequalto 2, 1 lessthanorequalto y lessthanorequalto 3, and 0 lessthanorequalto z lessthanorequalto 2. Orient the boundary partial differential C with outward normals and evaluate a surface integral doubleintegral_partial differential C(2xze_1 + e^x+y e_2 + z^3 e_3) middot dS (e_1, e_2, e_3 is the standard basis for R^3).

Solution

given that cube C=[0,2] * [1,3]* [0,2]

and given 0 <= x <=2

1 <= y <=3

0 <= z <=2

Given that

||(2xze₁+e^x+ye₂+z³e₃) .dS

=|| (2xz+e^x+y+z³) .(e₁,e₂,e₃) .dS

by guass diverdenge therum we can write

|||divf dv =|| F.N dS here | is intigration symbal

||| div f dv = ||f(e_1,e₂,e₃) ds

we have F = 2xzi+e^x+yj+z³k

div F= d/dx (2xz) +d/dy(e^x+y) +d/dz(z³) here d/dx, d/dy, d/dz are partial derivatives

div F= 2z+e^x+y+3z²

||| div F dv =||| (2z+e^x+y+3z²) dv

   =||| (2z+e^x+y+3z²) dz dy dx for dv=dxdydz x in [0,2] y in [1,3] , z in [0,2]

   =||(2z²/2 +e^x+y z +3 z³/3)dy dx

   =||(z²+e^x+yz +z³) dydx

   =||(4+2e^x+y+8)dydx after substituting z limits [0,2]

   =| (4y +2e^x+y+8y)dx

   =|(4(3-1)+2e^x+(3-1)+8(3-1)) dx after sub stituting y limits [1,3]

   =| (8+2e^x+2+16)dx

=(8x+2e^x+2+16x)

   =(8(2-0)+2e^(2-0)+2+16(2-0)) after substituting x limits

   =16+2e⁴+32

=48+2e⁴