Evaluate the surface integral. s (x + y + z) ds, s is the parallelogram with parametric equations x = u + v, y = u − v, z = 1 + 2u + v, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2.

Respuesta :

[tex]S[/tex] is given to be parameterized by

[tex]\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\langle u+v,u-v,1+2u+v\rangle[/tex]

with [tex]0\le u\le3[/tex] and [tex]0\le v\le2[/tex]. We have

[tex]\mathbf r_u=\langle1,1,2\rangle[/tex]
[tex]\mathbf r_v=\langle1,-1,1\rangle[/tex]
[tex]\mathbf r_u\times\mathbf r_v=\langle3,1,-2\rangle[/tex]
[tex]\left\|\mathbf r_u\times\mathbf r_v\right\|=\sqrt{14}[/tex]

The surface integral is then

[tex]\displaystyle\iint_S(x+y+z)\,\mathrm dS=\iint_S(x(u,v)+y(u,v)+z(u,v))\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}((u+v)+(u-v)+(1+2u+v))\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}(4u+v+1)\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)[/tex]
[tex]=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)[/tex]
[tex]=48\sqrt{14}[/tex]