Question

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = y i + (z − y) j + x k S is the surface of the tetrahedron with vertices (0, 0, 0), (8, 0, 0), (0, 8, 0), and (0, 0, 8)

Answer 1

Answer:

$\dfrac{-8^3}{6}$

Step-by-step explanation:

According to the divergence theorem;

The flux through the surface S is given by the formula:

$\iint _S F.dS = \iiint_E \ div (F) \ dV$

where the vector field is:

F = $\langle y,z-y,x \rangle$

Then the divergence of the vector field is:

$div (F) = \bigtriangledown.F = \Bigg [ \dfrac{\partial (y)}{\partial x} + \dfrac{\partial (z-y)}{\partial (y)}+ \dfrac{\partial (x)}{\partial (z)} \Bigg ]$

= 0 - 1 + 0

= -1

Thus, the flux through the surface of the tetrahedron is:

$\iint_S . FdS = \iiint _E(-1) \ dV \\ \\ = - \iiint_E \ dV$

To determine the  volume of the tetrahedron with vertices O(0,0,0), A(8,0,0), B (0,8,0) & C(0,0,6)

The equation of the plane P moving through the vertices A, B and C is:

$P = \dfrac{x}{8}+ \dfrac{y}{8}+ \dfrac{z}{8} = 1$

x + y + z = 8

Range:

For z: 0 ≤ z ≤ 8 - x - y

For y: 0 ≤ y ≤ 8 - x

For x; 0 ≤ x ≤ 8

Thus;

$\iiint_E \ dV = \int ^8_0 \int ^{8-x}_{0} \int ^{8-x-y}_{0}$

$\int ^8_0 \int ^{8-x}_{0} [z] ^{8-x-y}_{0} \ dydx = \int ^8_0 \int ^{8-x}_{0} \ (8 -x-y) \ dy dx$

$\int ^8_0 [ (8-x)^2 - \dfrac{(8-x)^2}{2} ] dx = \dfrac{1}{2} \int ^8_0 (8-x)^2 \ dx$

i.e.

$= \dfrac{1}{2} [ \dfrac{(8-x)^3}{(-1)^3}]^8_0$

$= \dfrac{-1}{6}[(8-8)^3-(0-8)^3]$

$= \dfrac{-8^3}{6}$