Question

Use the Divergence Theorem to compute the net outward flux of the field F=<-2x,y,-2z> across the surface S, where S is the boundary of the tetrahedron in the first octant formed by the planex+y+z=2.
The net outward flux across the boundary of the tetrahedron is:___________.

Answer 1

Answer:

The net outward flux across the boundary of the tetrahedron is: -4

Step-by-step explanation:

Given vector field F = ( -2x, y, - 2 z )

$div F = \nabla F = ( i \dfrac{\partial }{\partial x }+ j \dfrac{\partial}{\partial y} + k \dfrac{\partial}{\partial z}) \langle -2x, y, -2z \rangle$

$div F = \nabla F = ( \dfrac{\partial }{\partial x }(-2x)+ \dfrac{\partial}{\partial y}(y) + \dfrac{\partial}{\partial z}(-2z))$

= -2 + 1 -2

= -3

According to divergence theorem;

Flux = $\int \int \int div \ \ (F) \ dv$

x+y+z = 2; $1^{st}$  Octant

x from 0 to 2

y from 0 to 2 -x

z from 0 to 2-x-y

$= \int\limits^2_0 \int\limits^{2-x}_0 \int\limits^{2-x-y}_0 -3dzdydx$

$=-3 \int\limits^2_0 \int\limits^{2-x}_0 (2-x-y)dy dx$

$= -3 \int\limits^2_0[(2-x)y - \dfrac{y^2}{2}]^{2-x}__0 \ \ dx$

$= -3 \int\limits^2_0(2-x)^2 - \dfrac{(2-x)^2}{2} dx$

$= -3 \int\limits^2_0\dfrac{(2-x)^2}{2} dx = - \dfrac{3}{2} \int\limits^2_0(4-4x+x^2) dx$

$=- \dfrac{3}{2}(4x-x^2 + \dfrac{x^3}{3})^2_0$

$=- \dfrac{3}{2}(8-8+\dfrac{8}{3})$

$=- \dfrac{3}{2}(\dfrac{8}{3})$

= -4

Thus; The net outward flux across the boundary of the tetrahedron is: -4