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:

[tex]\dfrac{-8^3}{6}[/tex]

Step-by-step explanation:

According to the divergence theorem;

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

[tex]\iint _S F.dS = \iiint_E \ div (F) \ dV[/tex]

where the vector field is:

F = [tex]\langle y,z-y,x \rangle[/tex]

Then the divergence of the vector field is:

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

= 0 - 1 + 0

= -1

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

[tex]\iint_S . FdS = \iiint _E(-1) \ dV \\ \\ = - \iiint_E \ dV[/tex]

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:

[tex]P = \dfrac{x}{8}+ \dfrac{y}{8}+ \dfrac{z}{8} = 1[/tex]

x + y + z = 8

Range:

For z: 0 ≤ z ≤ 8 - x - y

For y: 0 ≤ y ≤ 8 - x

For x; 0 ≤ x ≤ 8

Thus;

[tex]\iiint_E \ dV = \int ^8_0 \int ^{8-x}_{0} \int ^{8-x-y}_{0}[/tex]

[tex]\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[/tex]

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

i.e.

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

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

[tex]= \dfrac{-8^3}{6}[/tex]

This question is based on the Gauss Divergence theorem. Therefore, the surface integral  [tex]\int\limits {F.dS}[/tex] is  -85.33.

Given:

F(x, y, z) = y i + (z − y) j + x k S in outward orientation.

Tetrahedron with vertices (0, 0, 0), (8, 0, 0), (0, 8, 0), and (0, 0, 8).

We have to evaluate the surface integral  [tex]\int\limits {F.dS}[/tex] .

According to the Gauss divergence theorem ,

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

[tex]\int\int _s F.dS = \int \int \int_e div (F)\; dV[/tex]

Where the vector field is:

F  = ( y, z-y, x )

Therefore,  the divergence of the vector field is:

[tex]div(F) = \bigtriangledown .F = ( \dfrac{\partial( y)}{\partial (x)} + \dfrac{\partial(z-y)}{\partial(y)} + \dfrac{\partial(x)}{\partial(z)} )\\\\div(F) = \bigtriangledown .F = 0-1+0=-1[/tex]

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

[tex]\int\int _s F.dS = \int \int \int_e (-1)\; dV = -\int \int \int_e \; dV[/tex]

Now, 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:

[tex]P = \dfrac{x}{8} +\dfrac{y}{8} +\dfrac{z}{8} = 1[/tex]

x + y + z = 8

Range:

For z: 0 ≤ z ≤ 8 - x - y

For y: 0 ≤ y ≤ 8 - x

For x; 0 ≤ x ≤ 8

Thus,

[tex]\int\int\int_e dV = \int\limits^8_0\int\limits^{8-x} _ 0 \int\limits^{8-x-y}_0 \; dzdxdy\\= \int\limits^8_0\int\limits^{8-x} _ 0 [z]\limits^{8-x-y}_0 dx \\= \int\limits^8_0\int\limits^{8-x} _ 0 (8-x-y) dy dx\\= \int\limits^8_0 [ 8y-xy-\dfrac{y^{2} }{2} ]\limits^{8-x}_ 0 dx\\= \int\limits^8_0 ([ 8-x]^{2} - \dfrac{ [ 8-x]^{2}}{2} ) dx\\= \dfrac{1}{2} [\dfrac{(8-x)^{3} }{(-1)^{3} } ] \limits^8_0\\=\dfrac{-8^{3} }{6} \\\\= -85.33[/tex]

Therefore, the surface integral  [tex]\int\limits {F.dS}[/tex] is  -85.33.

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