How Do We Find the Parametrization σ?

[mathmari]

Hey!

I want to show that $\iint_{\Sigma}(\nabla\times f)\cdot d\Sigma=0$ for the function $f(x,y,z)=(1,1,1)\times g(x,y,z)$ when $\Sigma$ is the surfcae that is defined by the relations $x^2+y^2+z^2=1$ and $x+y+z\geq 1$.
I have done the following:

Let $g(x,y,z)=(g_1, g_2, g_3)$. Then $f(x,y,z)=(1,1,1)\times (g_1, g_2, g_3)=(g_3-g_2, g_1-g_3, g_2-g_1)$.

From Stokes theorem we have that $$ \iint_{\Sigma}(\nabla\times f)\cdot d\Sigma=\oint_{\sigma}f\cdot d\sigma$$

So, we have to find the boundary of the surface, $\sigma$. For that do we have to set equal the relations $x^2+y^2+z^2=1$ and $x+y+z= 1$. I yes, we get the following:

$$x^2+y^2+z^2=x+y+z\\ \Rightarrow x^2-x+y^2-y+z^2-z=0 \\ \Rightarrow x^2-x+\frac{1}{4}+y^2-y+\frac{1}{4}+z^2-z+\frac{1}{4}=\frac{3}{4} \\ \Rightarrow \left (x-\frac{1}{2}\right )^2+\left (y-\frac{1}{2}\right )^2+\left (z-\frac{1}{2}\right )^2=\frac{3}{4} \\ \Rightarrow \frac{4}{3}\left (x-\frac{1}{2}\right )^2+\frac{4}{3}\left (y-\frac{1}{2}\right )^2+\frac{4}{3}\left (z-\frac{1}{2}\right )^2=1 \\ \Rightarrow \left (\frac{2}{\sqrt{3}}x-\frac{1}{\sqrt{3}}\right )^2+\left (\frac{2}{\sqrt{3}}y-\frac{1}{\sqrt{3}}\right )^2+\left (\frac{2}{\sqrt{3}}z-\frac{1}{\sqrt{3}}\right )^2=1 $$

How can we continue? How can we get $\sigma$ ?

 

[HallsofIvy]

Hey!

I want to show that $\iint_{\Sigma}(\nabla\times f)\cdot d\Sigma=0$ for the function $f(x,y,z)=(1,1,1)\times g(x,y,z)$ when $\Sigma$ is the surfcae that is defined by the relations $x^2+y^2+z^2=1$ and $x+y+z\geq 1$.

Doesn't that depend on exactly what the function, g, is?

 

[mathmari]

Doesn't that depend on exactly what the function, g, is?

We don't have any information about g. So, we cannot show that without an expression for g? (Wondering)

 

[HallsofIvy]

well, I expect that the key point here is that the vector (1, 1, 1) is perpendicular to the plane x+ y+ z= 1. And notice that, of course, the spherical surface and the plane have the same boundary. By Stokes theorem, as you say, \(\displaystyle \int\int_\Sigma (\nabla \times f)d\Sigma= \oint f\cdot d\sigma\). Applying exactly the same theorem to the plane, and writing \(\displaystyle \Pi\) for the region of the plane bounded by the circle of intersection, we have \(\displaystyle \int\int_\Pi (\nabla \times f)d\Pi= \oint f\cdot (\nabla\times f) d\sigma\).

Together, \(\displaystyle \int\int_\Sigma (\nabla \times f)d\Sigma= \int\int_\Pi (\nabla\times f)d\Pi\).