Answer :
Parameterize [tex]S[/tex] by
[tex]\vec r(u,v)=u\,\vec\imath+v\,\vec\jmath+(4-u^2-v^2)\,\vec k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then
[tex]\vec r_u\times\vec r_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k[/tex]
In the integral we get
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S[/tex]
[tex]\displaystyle=\int_0^1\int_0^1(uv\,\vec\imath+v(4-u^2-v^2)\,\vec\jmath+u(4-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv[/tex]
[tex]\displaystyle=\int_0^1\int_0^1(4u-u^3+2u^2v+8v^2-uv^2-2u^2v^2-2v^4)\,\mathrm du\,\mathrm dv=\frac{713}{180}[/tex]
The value of the surface integral S F · dS for the given vector field F and the oriented surface S is 713/180 and this can be determined by doing integration.
Given :
- F(x, y, z) = xy i + yz j + zx k
- S is the part of the paraboloid [tex]\rm z = 4-x^2-y^2[/tex] that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has an upward orientation.
First, S is parameterized in order to evaluate the surface integral.
[tex]\rm \bar{r}(u,v) = u\; i + v\; j + (4-u^2-v^2)\;k[/tex] where [tex]\rm 0 \leq u\leq1[/tex] and [tex]\rm 0 \leq v\leq1[/tex].
The expression for [tex]\rm ||r_u\times r_v||[/tex] is given by:
[tex]\rm ||r_u\times r_v||=2u\;i+2v\; j +k[/tex]
Now, the surface integral is given by:
[tex]\rm \int\int_S \bar{F}.d\bar{S} = \int^1_0\int^1_0(uv \;i+v(4-u^2-v^2)\;j+u(4-u^2-v^2)\;k).(2u\;i+2v\;j+k)\;dudv[/tex]
[tex]\rm \int\int_S \bar{F}.d\bar{S} = \int^1_0\int^1_0(4u-u^3+2u^2v+8v^2-2u^2v^2-2v^4)\;dudv[/tex]
[tex]\rm \int\int_S \bar{F}.d\bar{S} = \dfrac{713}{180}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/24308099