# Flux across the surface $x^{2} + y^{2} + z^{2} = 1$

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?

• It's not wrong. $2\pi$ is the flux over the upper half of the sphere. $4\pi$ is the entire sphere. Dec 23, 2018 at 7:22
• @Dylan why does the above integration find the flux over the upper half sphere only? Dec 23, 2018 at 7:27
• 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 Dec 23, 2018 at 7:29
• 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. Dec 23, 2018 at 7:30
• @Mathsaddict Why did you put a modulus around $\vec{\nabla S}\cdot\hat p$? Dec 23, 2018 at 10:49

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.