The flux is 9.
Flux is the presence of a force field in a specified physical medium, or the flow of energy through a surface.
Given:
2x+2y+z= 1
F = 2i+2j + 2k
Now,
r = xi + yj + z( 1-2x-2y) K
dr/dx= i - 2k
dr/dy = j-2k
dr/dx* dr/dy
= ( i - 2k) * (j-2k)
= 2i + 2j + k
F(x)= 2i+2j + 2k
F(x). da = 4 +4 +2 = 10 dxdy
Hence, flux
= [tex]\int\limits^1_0 {\int\limits^{1-2y}_0 {10 } \, dx dy } \,[/tex]
= [tex]\int\limits^1_0[/tex] 10(1-2y) dx
= [tex]\int\limits^1_0[/tex] 10-2y
= 10(1) - (1)²
=9
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The plane has intercepts (1/2, 0, 0), (0, 1/2, 0), and (0, 0, 1). Parameterize [tex]S[/tex] by the vector function
[tex]\vec s(u,v) = \dfrac{(1-u)(1-v)}2 \, \vec\imath + \dfrac{u(1-v)}2 \, \vec\jmath + v \,\vec k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. (More explicitly, we have the parameterization
[tex]\vec s(u,v) = (1-v)((1-u) p_1 + u p_2) + v p_3[/tex]
where [tex]p_i[/tex] denote the given points.)
The normal vector to [tex]S[/tex] is
[tex]\vec n = \dfrac{\partial\vec s}{\partial u} \times \dfrac{\partial\vec s}{\partial v} = \dfrac{1-v}2\,\vec\imath + \dfrac{1-v}2\,\vec\jmath + \dfrac{1-v}4\,\vec k[/tex]
Then the flux of [tex]\vec F = 2\,\vec\imath+2\,\vec\jmath+2\,\vec k[/tex] across [tex]S[/tex] is given by the surface integral,
[tex]\displaystyle \iint_S \vec F \cdot d\vec\sigma = \iint_S \vec F \cdot \vec n \, dA[/tex]
[tex]\displaystyle = \int_0^1 \int_0^1 \left(2\,\vec\imath+2\,\vec\jmath+2\,\vec k) \cdot \left(\frac{1-v}2\,\vec\imath + \frac{1-v}2\,\vec\jmath + \frac{1-v}4\,\vec k\right) \, du \, dv[/tex]
[tex]\displaystyle = \frac52 \int_0^1 \int_0^1 (1-v) \, du \, dv[/tex]
[tex]\displaystyle = \frac52 \int_0^1 (1-v) \, dv = \boxed{\frac54}[/tex]
