Respuesta :
Answer:
[tex]\frac{56}{15}[/tex]
Step-by-step explanation:
Given that [tex]\vec F(x, y, z) = x^2y \hat i + xy^2 \hat j + 3xyz \hat k[/tex]
F(x, y, z) = x^2y \hat i + xy^2 \hat j + 3xyz \hat k
The flux of \vec F across the surface S is given by
[tex]\int \int _s \vec F \cdot dS[/tex]
Where the survare S is the surface of the tetrahedron bounded by the planes x = 0, y = 0, z = 0, and x + 4y + z = 4.
Now, by using the Gauss Divergence theorem:
[tex]\int \int _s \vec{F} \cdot dS=\int \int \int _v \bigtriangledown \cdot \vec F dV[/tex]...(i)
where V is the volume enclosed by the surface S and [tex]dV=dxdydz[/tex]
\bigtriangledown \cdot \vec F = \hat i \frac {\partial \vec F}{\partial x}+\hat j \frac {\partial \vec F}{\partial y}+\hat k \frac {\partial \vec F}{\partial z}
=2xy+2xy+3xy=7xy
Putting this value in equation (i), we have
\int \int \int _v \bigtriangledown \cdot \vec F dV = \int \int \int _v (7xy)dxdydz
= [tex]7\int_{y=0}^{y=1} \int_{x=0}^{x=4-4y} \left( \int_{z=0}^{z=4-x-4y}xy\;dz\right)dxdy[/tex]
[tex]=7\int_{y=0}^{y=1} \int_{x=0}^{x=4-4y} xy(4-x-4y-0)dxdy \\\\=7\int_{y=0}^{y=1} \int_{x=0}^{x=4-4y}(4xy-x^2y-4xy^2)dxdy \\\\=7\int_{y=0}^{y=1} (2x^2y-\frac 1 3 x^3y-2 x^2y^2)|_{x=0}^{x=4-4y}dy \\\\[/tex]
[tex]=7\int_{y=0}^{y=1} (2(4-4y)^2y-\frac 1 3 (4-4y)^3y-2 (4-4y)^2y^2)dy[/tex]
[tex]=7\int_{y=0}^{y=1} (32(1-y)^2y-\frac {64}{3} (1-y)^3y-32(1-y)^2y^2)dy[/tex]
[tex]=7\int_{y=0}^{y=1} (32(1-2y+y^2)y-\frac {64}{3} (1-y^3-3y+3y^2)y-32(1-2y+y^2)y^2)dy[/tex]
[tex]=7\int_{y=0}^{y=1} \{32(y-2y^2+y^3)-\frac {64}{3} (y-y^4-3y^2+3y^3)-32(y^2-2y^3+y^4)\}dy[/tex]
[tex]=7 \{32(\frac 1 2 y^2- \frac 2 3 y^3+\frac 1 4 y^3)-\frac {64}{3} (\frac 1 2 y^2-\frac 1 5 y^5-y^3+\frac 3 4 y^4)-32(\frac 1 3 y^3-\frac 1 2 y^4+\frac 1 5 y^5)\} |_{y=0}^{y=1}[/tex]
[tex]=7 \{32(\frac 1 2 - \frac 2 3 +\frac 1 4 )-\frac {64}{3} (\frac 1 2 -\frac 1 5 -1+\frac 3 4 )-32(\frac 1 3 -\frac 1 2 +\frac 1 5 )\}[/tex]
[tex]=7\times \frac {8}{15} \\\\=\frac{56}{15}[/tex]
Hence, the flux of F across S is [tex]\frac{56}{15}[/tex].
