MHB 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, $$\int\int_\Sigma (\nabla \times f)d\Sigma= \oint f\cdot d\sigma$$. Applying exactly the same theorem to the plane, and writing [math]\Pi[/math] for the region of the plane bounded by the circle of intersection, we have $$\int\int_\Pi (\nabla \times f)d\Pi= \oint f\cdot (\nabla\times f) d\sigma$$.

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