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Yuliya22 [10]
3 years ago
6

Evaluate the surface integral. (give your answer correct to at least three decimal places.) s is the boundary of the region encl

osed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 2
Mathematics
1 answer:
telo118 [61]3 years ago
3 0
Split up the surface S into three main components S_1,S_2,S_3, where

S_1 is the region in the plane y=0 bounded by x^2+z^2=1;

S_2 is the piece of the cylinder bounded between the two planes y=0 and x+y=2;

and S_3 is the part of the plane x+y=2 bounded by the cylinder x^2+z^2=1.

These surfaces can be parameterized respectively by

S_1:~\mathbf s_1(u,v)=\langle u\cos v,0,u\sin v\rangle
where 0\le u\le1 and 0\le v\le2\pi,

S_2:~\mathbf s_2(u,v)=\langle\cos v,u,\sin v\rangle
where 0\le u\le2-\cos v and 0\le v\le2\pi,

S_3:~\mathbf s_3(u,v)=\langle u\cos v,2-u\cos v,u\sin v\rangle
where 0\le u\le1 and 0\le v\le2\pi.

The surface integral of a function f(x,y,z) along a surface R parameterized by \mathbf r(u,v) is given to be

\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\iint_Sf(\mathbf r(u,v))\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times\frac{\partial\mathbf r(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv

Assuming we're just finding the area of the total surface S, we take f(x,y,z)=1, and split up the total surface integral into integrals along each component surface. We have

\displaystyle\iint_{S_1}\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}u\,\mathrm dv\,\mathrm du
\displaystyle\iint_{S_1}\mathrm dS=\pi

\displaystyle\iint_{S_2}\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2-u\cos v}\mathrm du\,\mathrm dv
\displaystyle\iint_{S_2}\mathrm dS=4\pi

\displaystyle\iint_{S_3}\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\sqrt2u\,\mathrm dv\,\mathrm du
\displaystyle\iint_{S_3}\mathrm dS=\sqrt2\pi

Therefore

\displaystyle\iint_S\mathrm dS=\left\{\iint_{S_1}+\iint_{S_2}+\iint_{S_3}\right\}\mathrm dS=(5+\sqrt2)\pi\approx20.151
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