Parameterize the surface (call it [tex]S[/tex]) by
[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(1-u-v)\,\vec k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_v\times\vec s_u=-\vec\imath-\vec\jmath-\vec k[/tex]
Then the integral of [tex]\vec F[/tex] over [tex]S[/tex] with the given orientation is
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1(v^3\,\vec\imath+8\vec\jmath-u\,\vec k)\cdot(-\vec\imath-\vec\jmath-\vec k)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^1\int_0^1(-v^3-8+u)\,\mathrm du\,\mathrm dv=\boxed{-\frac{31}4}[/tex]
We are to find the surface integral of [tex]\mathbf{\int \int _S \ F.ds}[/tex]
where;
∴
The plane x + y + z = 1
z = 1 - x - y
[tex]\dfrac{\partial z}{\partial x} = -1[/tex]
[tex]\dfrac{\partial z}{\partial y} = -1[/tex]
Since the surface is oriented downward
∴
[tex]dS = \Big( \dfrac{\partial z}{\partial x}i + \dfrac{\partial z}{\partial y}j - k) dxdy[/tex]
[tex]dS = (-i-j-k) dxdy[/tex]
However,
Flux = [tex]\mathbf{\int \int _S \ F.ds}[/tex]
[tex]= \int \int_R F. \Big( \dfrac{\partial z}{\partial x}i + \dfrac{\partial z}{\partial y}j - k) dxdy \\ \\ \\ = \int^1_{ x=0} \int ^{1-x}_{y=0} ( y^3i + 8j -xk)*(-i-j-k) dx dy \\ \\ \\ =\int^1_{ x=0} \int ^{1-x}_{y=0} (-y^3 -8+x) dxdy \\ \\ \\ = \int^1_{ x=0} \Big( \int ^{1-x}_{y=0} \Big(x-y^3 -8\Big) dy \Big) dx \\ \\ \\ = \int^1_{ x=0} \Bigg [xy - \dfrac{y^4}{4}-8y \Bigg]^{1-x}_{y=0} \ dx \\ \\ \\[/tex]
[tex]= \int^1_{ x=0} \Bigg [x(1-x) - \dfrac{1}{4}(1-x)^{4} - 8(1-x) \Bigg ] dx \\ \\ \\ = \int^1_{ x=0} \Bigg [(x-x^2) - \dfrac{1}{4}(x-1)^4+8(x-1)\Bigg] dx[/tex]
[tex]= \Bigg [ \dfrac{x^2}{2}-\dfrac{x^3}{3} - \dfrac{1}{4} \dfrac{(x-1)^5}{5}+(x-1)^2\Bigg] ^1_0[/tex]
[tex]= \Bigg [ \dfrac{1}{2}-\dfrac{1}{3} -0+0-0+0+ \dfrac{1}{4}\times \dfrac{(-1)^5}{5}-(0-1)^2\Bigg][/tex]
[tex]= \Bigg [ \dfrac{1}{2}-\dfrac{1}{3} - \dfrac{1}{20}-1\Bigg][/tex]
[tex]= \Bigg [ \dfrac{30-20-3-60}{60}\Bigg][/tex]
[tex]= \Bigg [ \dfrac{-53}{60}\Bigg][/tex]
Therefore, we can conclude that the surface integral is [tex]\mathbf{-\dfrac{53}{60}}[/tex]
Learn more about surface integral here:
https://brainly.com/question/1423573?referrer=searchResults
