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Let $S$ be the oriented surface $x^{2} + y^{2} + z^{2} =1$ with the unit normal $\hat{n}$ pointing outward for the vector field $F = xi +yj+ zk$, the value of double integration of $\vec{F} . \hat{n} dS$ is -

We need to evaluate flux across the given surface, which can easily be done by gauss divergence formula and it comes out to be $4\pi$.

I am trying to solve this by below method-

$\int \int \vec{F} . \hat{n} dS$ $$= \int \int \vec{F} . \frac{grad S}{ | grad S| } \frac{|grad S|}{|grad S. \hat{p}|} dA$$ ( this integration is over R which is shadow region of surface S, $\hat{p}$ is the unit normal vector to R) So, flux = $$\int \int \frac{F.grad S}{|grad S. \hat{p}|} dA$$

$$= \int\int \frac{(xi + yj + zk).(2xi +2yj+2zk)dxdy}{|2z. \hat{k}|}$$

$$= \int\int \frac{2x^{2} + 2y^{2} + 2z^{2}}{2z} dxdy$$

$$= \int \int (1/z) dxdy$$ $$= \int \int 1/(1 - x^{2} -y^{2})^{1/2}dxdy $$

further, this integration leads to $2\pi$ which is wrong. So, where did I go wrong?

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    $\begingroup$ It's not wrong. $2\pi$ is the flux over the upper half of the sphere. $4\pi$ is the entire sphere. $\endgroup$
    – Dylan
    Dec 23, 2018 at 7:22
  • $\begingroup$ @Dylan why does the above integration find the flux over the upper half sphere only? $\endgroup$ Dec 23, 2018 at 7:27
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    $\begingroup$ Both the upper and the lower half have the same projection on the $\Bbb R^2$ plane. So the projection integral is equivalent to only one half of the sphere $\endgroup$
    – Dylan
    Dec 23, 2018 at 7:29
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    $\begingroup$ It's a little inconvenient to answer when you leave out the bounds of the integration when that's exactly what you're confused about. Probably putting them in would resolve your own confusion. $\endgroup$
    – zoidberg
    Dec 23, 2018 at 7:30
  • $\begingroup$ @Mathsaddict Why did you put a modulus around $\vec{\nabla S}\cdot\hat p$? $\endgroup$ Dec 23, 2018 at 10:49

1 Answer 1

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Note that $\vec F=x\hat i+y\hat j+z\hat k=\vec r,\hat n=\hat r\ \therefore\vec F\cdot\vec r=r=\sqrt{x^2+y^2+z^2}=1$

$dS=rd\theta\cdot r\sin\theta\ d\phi=r^2\sin\theta\ d\theta\ d\phi=\sin\theta\ d\theta\ d\phi$

$\displaystyle\therefore\int\int\vec F\cdot\hat n\ dS=\int_0^{2\pi}\int_0^{\pi}\sin\theta\ d\theta\ d\phi=4\pi$.

As for your method, the flux through the upper half is $2\pi$. Add to that the $2\pi$ of the lower half.

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