Question

Evaluate the surface integral int_SF*dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x,y,z)=xi+2yj+3zk S is the cube with vertices (+-1,+-1,+-1)

Integrals
ANSWERED
asked 2020-11-03
Evaluate the surface integral
\(\displaystyle\int_{{S}}{F}\cdot{d}{S}\)
for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x,y,z)=xi+2yj+3zk
S is the cube with vertices \(\displaystyle{\left(\pm{1},\pm{1},\pm{1}\right)}\)

Answers (1)

2020-11-04

Remember that \(\displaystyle{F}\cdot{d}{s}={F}\cdot\hat{{n}}{d}{S}\ {w}{h}{e}{r}{e}\ \hat{{n}}\) is the unit outward normal vector. Luckily that vector will be constant across each of the 6 faces of the cube. So we just need to break this integral into 6 parts:
\(\displaystyle\int_{{S}}{F}\cdot{d}{S}=\int_{{\top}}{F}\cdot\hat{{n}}_{{\top}}{d}{S}+\int_{{\le{f}{t}}}{F}\cdot\hat{{n}}_{{\le{f}{t}}}{d}{S}+\int_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{F}\cdot\hat{{n}}_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{d}{S}+\ldots\)
Let's start with the top side of the cube. Along that face, the outward normal points up. I.e.\(\displaystyle\hat{{n}}={k}\).
Also along the top face we have \(\displaystyle-{1}\le{x}\le{1},-{1}\le{y}\le{1},{\quad\text{and}\quad}{z}={1}\).Hence we have
\(\displaystyle\int_{\top}{F}\cdot\hat{{n}}_{{\top}}={\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left[{x}{i}+{2}{y}{j}+{3}{\left({1}\right)}{k}\right]}\cdot{k}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{3}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={3}{\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle{3}{\left({2}\right)}{\int_{{{y}=-{1}}}^{{{y}={1}}}}{\left.{d}{y}\right.}\)
=3(2)(2)=12 Along the left face (pretend you're sitting somewhat out on the positive\(\displaystyle\times{a}\xi{s}{l}\infty{k}\in{g}{a}{t}\ {t}{h}{e}\ {c}{u}{b}{e}{)},{t}{h}{e}\ {o}{u}{t}\ {w}{a}{r}{d}{\left\|{a}\right\|}{l}\ {i}{s}-{j}{\quad\text{and}\quad}{t}{h}{e}\ {v}{a}{l}{u}{e}\ {s}{o}{f}{x},{y},{\quad\text{and}\quad}{z}\ {a}{r}{e}-{1}\le{x}\le{1},{y}=-{1},{\quad\text{and}\quad}-{1}\le{z}\le{1}\).Thus
\(\displaystyle\int_{{\le{f}{t}}}{F}\cdot\hat{{n}}_{{\le{f}{t}}}{d}{S}={\int_{{{z}=-{1}}}^{{{z}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left[{x}{i}+{2}{\left(-{1}\right)}{j}+{3}{z}{k}\right]}\cdot-{j}{\left.{d}{x}\right.}{\left.{d}{z}\right.}={8}\)
At this point hopefully you get the idea so I'm just going to quickly list the rest of the integrals and their values:
\(\displaystyle\int_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{F}\cdot\hat{{n}}_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{d}{S}={4}\)
\(\displaystyle\int_{{()}}{F}\cdot\hat{{n}}_{{()}}{d}{S}={8}\)
\(\displaystyle\int{\left({b}{a}{c}{k}\right)}{F}\cdot\hat{{n}}_{{{b}{a}{c}{k}}}{d}{S}={4}\)
\(\displaystyle\int{\left(\bot\to{m}\right)}{F}\cdot\hat{{n}}_{{\bot\to{m}}}{d}{S}={12}\)
So if the flux through each face is 12, 8, 4, 8, 4, and 12, respectively, then the total flux from the cube is
12+8+4+8+4+12=48

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