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There is vector field $F = [x,y,-z]$. We need to find the flux of the vector field outward across the given surface $\sigma=x^2+y^2+z^2=1, x\ge0, y\ge0, z\ge0$ directly and by using Gauss theorem (i.e. $\iiint\limits_V\mathrm{div}\,\mathbf{F}\,dV=\iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\;\!\!\;\subset\!\!\supset\mathbf F\cdot\mathbf{n}\,dS$).

In this case the surface is just a sphere.

So, any explanation how to calculate the flux by two ways are highly welcomed. According to Andrei's answer below: $\int_{0}^{\pi}sin\theta d\theta \int_{0}^{2\pi}d\phi=2*2\pi=4\pi$, right?

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  • $\begingroup$ It's hard to know what you are asking. Do you know how to calculate divergence of $F$? There is a triple integral and a surface integral. Are you having trouble with one or both of these? In what way? $\endgroup$
    – saulspatz
    May 20, 2019 at 20:03
  • $\begingroup$ @saulspatz Thank you, I know how to calculate the divergence of $F$. First of all I need to calculate the flux through all parts of the surface. $\endgroup$
    – Mikhail Gaichenkov
    May 20, 2019 at 20:19

1 Answer 1

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The normal to the surface is $\mathbf n=[x,y,z]$, where $x^2+y^2+z^2=1$. Then $\mathbf F\cdot\mathbf n=x^2+y^2-z^2=1-2z^2$. Then I would use polar coordinates: $\mathbf F\cdot\mathbf n = 1-2\cos^2\theta$ and $ds=\sin\theta d\theta d\phi$. The integration over $\phi$ is from $0$ to $2\pi$, and the integration of $\theta$ is from $0$ to $\pi$. Can you finish this?

For the divergence part, the integrand is a constant, so you just need to multiply it with the volume of the sphere.

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  • $\begingroup$ Thanks, the integral is $4\pi$, right? What is the divergence part? And how did you get $1-2z^2$, could you explain please? $\endgroup$
    – Mikhail Gaichenkov
    May 22, 2019 at 21:04
  • $\begingroup$ Volume of a sphere is $4\pi/3$. The divergence part is the other method to solve the problem (other than the surface integral). $\mathrm{div} \mathbf F=1$. The $1-2z^2$ comes from $x^2+y^2-z^2=x^2+y^2+z^2-2z^2$. And you have $x^2+y^2+z^2=1$ on the surface. $\endgroup$
    – Andrei
    May 22, 2019 at 21:23
  • $\begingroup$ Well, so $1*4\pi/3=4\pi/3$ On the other hand the integration over $\phi$ and $\theta$ results in just $4\pi$. What to take into account else? $\endgroup$
    – Mikhail Gaichenkov
    May 23, 2019 at 19:52
  • $\begingroup$ I've forgot to put a square symbol for $\cos\theta$ term. It's $z^2$, so the expression to integrate is $1-2\cos^2\theta$. Integral over $1$ will give $4\pi$, and you have the integral $\int_0^\pi \cos^2\theta\sin\theta d\theta=2/3$. Integral over $\phi$ is $2\pi$, so the answer is $4\pi-2\cdot2\pi\cdot\frac 23=\frac 43\pi$ $\endgroup$
    – Andrei
    May 23, 2019 at 20:55

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