Question

Evaluate the surface integral. (give your answer correct to at least three decimal places.) s is the boundary of the region enclosed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 2

Answer 1

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\|(\partial\mathbf r(u,v))/(\partial u)*(\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`