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1 answer:

VladimirAG [237]3 years ago

5 0

Where is the picture?

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

Stolb23 [73]

$S$ 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

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