Mathematics, 24.09.2019 11:30 kaileyy06

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

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Answer from: Arellano2373

Split up the surface

into three main components S_1,S_2,S_3

, where

is the region in the plane y=0

bounded by x^2+z^2=1

;

is the piece of the cylinder bounded between the two planes y=0

and

;

and

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

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

, we take

, 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

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

Answer from: Quest

A. is the answer in this question

Answer from: Quest

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