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aivan3 [116]
4 years ago
14

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.
Mathematics
1 answer:
Stolb23 [73]4 years ago
4 0
S is given to be parameterized by

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

with 0\le u\le3 and 0\le v\le2. We have

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

The surface integral is then

\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
=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}((u+v)+(u-v)+(1+2u+v))\,\mathrm dv\,\mathrm du
=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}(4u+v+1)\,\mathrm dv\,\mathrm du
=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)
=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)
=48\sqrt{14}
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