Respuesta :

LammettHash

Parameterize [tex]S[/tex] by

[tex]\vec s(u,v)=(1-u)v\,\vec\imath+8uv\,\vec\jmath+8(1-v)\,\vec k[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex].

Then the surface element is

[tex]\mathrm dS=\|\vec s_u\times\vec s_v\|\,\mathrm du\,\mathrm dv=8\sqrt{66}\,v\,\mathrm du\,\mathrm dv[/tex]

and the surface integral is

[tex]\displaystyle\iint_Sxy\,\mathrm dS=8\sqrt{66}\int_0^1\int_0^18(1-u)uv^3\,\mathrm du\,\mathrm dv=\boxed{8\sqrt{\dfrac{22}3}}[/tex]

lhmarianateixeira

In this exercise we have to use the knowledge of parameterization and integrals to calculate the area needed, so we will find that this corresponds to:

[tex]8\sqrt{\frac{22}{3} }[/tex]

Parameterizing the function given as:

[tex]s(u, v) = (1-u)v+8uv+8(1-v)[/tex]

Knowing that the integration limits will be given by:

[tex]0\leq u\leq 1 \ and \ 0\leq v\leq 1[/tex]

So in this way calculating the surface we will find that it will be:

[tex]dS= 8\sqrt{66} v du dv[/tex]

Now putting all this data together in the integral, we will find that it will correspond to:

[tex]\int\limits \, \int\limits_S {xy} \, dxS\\8\sqrt{66}\int\limits^1_0 \int\limits^1_0 {8(1-u)uv^3} \, dufv\\= 8\sqrt{\frac{22}{3} }[/tex]

See more about parameterizing at brainly.com/question/14770282