Question

Use the Divergence Theorem to calculate the surface integral S F · dS; that is, calculate the flux of F across S. F(x, y, z) = x2yi + xy2j + 3xyzk, S is the surface of the tetrahedron bounded by the planes x = 0, y = 0, z = 0, and x + 2y + z = 2.

Answer 1

Answer:

-14 / 3

Step-by-step explanation:

- Divergence theorem, expresses an explicit way to determine the flux of a force field ( F ) through a surface ( S ) with the help of "del" operator ( D ) which is the sum of spatial partial derivatives of the force field ( F ).

- The given force field as such:

                      $F = (x^2y) i + (xy^2) j + (3xyz) k$

Where,

         i, j, k are unit vectors along the x, y and z coordinate axes, respectively.

- The surface ( S ) is described as a tetrahedron bounded by the planes:

                      $x = 0 \\y = 0\\x + 2y + z = 2$

                      $z = 0$

- The divergence theorem gives us the following formulation:

                      $_S\int\int {F} \,. dS = _V\int\int\int {D [F]} \,. dV$

- We will first apply the del operator on the force field as follows:

                      $D [ F ] = 2xy + 2xy + 3xy = 7xy$

- Now, we will define the boundaries of the solid surface ( Tetrahedron ).

- The surface ( S ) is bounded in the z - direction by plane z = 0 and the plane [ z = 2 - x - 2y ]. The limits of integration for " dz " are as follows:

                      dz: [ z = 0 - > 2 - x - 2y ]

- Now we will project the surface ( S ) onto the ( x-y ) plane. The projection is a triangle bounded by the axes x = y = 0 and the line: x = 2 - 2y. We will set up our limits in the x- direction bounded by x = 0 and x = 2 - 2y. The limits of integration for " dx " are as follows:

                     dx: [ x = 0 - > 2 - 2y ]

- The limits of "dy" are constants defined by the axis y = 0 and y = -2 / -2 = 1. Hence,

                    dy: [ y = 0 - > 1 ]

- Next we will evaluate the triple integral as follows:

                   $\int\int\int ({D [ F ] }) \, dz.dx.dy = \int\int\int (7xy) \, dz.dx.dy\\\\\int\int (7xyz) \, | \limits_0^2^-^x^-^2^ydx.dy\\\\\int\int (7xy[ 2 - x - 2y ] ) dx.dy = \int\int (14xy -7x^2y -14 xy^2 ) dx.dy\\\\\int (7x^2y -\frac{7}{3} x^3y -7 x^2y^2 )| \limits_0^2^-^2^y.dy \\\\\int (7(2-2y)^2y -\frac{7}{3} (2-2y)^3y -7 (2-2y)^2y^2 ).dy$

                 $7 (-\frac{(2-2y)^3}{6} + (2-2y)^2 ) -\frac{7}{3} ( -\frac{(2-2y)^4}{8} + (2-2y)^3) -7 ( -\frac{(2-2y)^3}{6}y^2 + 2y.(2-2y)^2 )| \limits^1_0\\\\ 0 - [ 7 (-\frac{8}{6} + 4 ) -\frac{7}{3} ( -\frac{16}{8} + 8 ) -7 ( 0 ) ] \\\\- [ \frac{56}{3} - 14 ] \\\\\int\int {F} \, dS = -\frac{14}{3}$