# 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)

Question
Integrals
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 $$\displaystyle{\left(\pm{1},\pm{1},\pm{1}\right)}$$

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

### Relevant Questions

True or false?
Given an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve with positive orientation, and a vector field whose components have continuous partial derivatives on an open region in three-dimensional Euclidean space. Then the line integral of this vector field along the boundary curve equals the flux of the curl of the vector field across the surface.
Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
$$\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}$$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $$\displaystyle{y}=\frac{\pi}{{2}},{z}={0}$$, and z = x.
Let C be the ellipse contained in the xy plane whose equation is $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore W = 0.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that W = 12.56.
c) We can apply Stokes' Theorem and conclude that W = 6.28
d) We can apply Stokes' Theorem and conclude that W = 12.56.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}$$, S is the sphere with center the origin and radius 2.
Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}$$,
S is the surface of the solid bounded by the cylinder $$\displaystyle{y}^{{2}}+{z}^{{2}}={9}$$ and the planes x = −3 and x = 1.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
Use the Divergence Theorem to find the flux of $$\displaystyle{F}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}$$ outward through the surface of the region enclosed by the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ and the planes z = 1 and z =-1.
Divergence Theorem for more general regions Use the Divergence Theorem to compute the net outward flux of the following vector fields across the boundary of the given regions D.
$$\displaystyle{F}={\left\langle{z}-{x},{x}-{y},{2}{y}-{z}\right\rangle}$$, D is the region between the spheres of radius 2 and 4 centered at the origin.
Use Stokes' Theorem to evaluate $$\displaystyle\int\int_{{S}}{C}{U}{R}{L}{f}\cdot{d}{S}$$.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}$$,
S is the part of the cone $$\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}$$ that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.