stokes-theoremvector analysis
Evaluate $∫ydx+zdy+xdz$ where C is intersection of $x+z=a$ and $x^2+y^2+z^2=a^2$
I am using Stoke's Theorem to find the integral. I could solve the first half
$\operatorname{curl}\text{ }F = -\hat{\textbf{i}} -\hat{\textbf{j}} -\hat{\textbf{k}}$
Normal Vector N = $\frac{\hat{\textbf{i}}+\hat{\textbf{k}}}{\sqrt2}$
Ultimately I end up with having to find surface integral of $ds$. How do I proceed?
The vector identity you need is this: For $C^1$ vector fields $\vec F$ and $\vec G$ on $\Bbb R^3$, we have $$\text{div}(\vec F\times\vec G) = (\text{curl}\,\vec F,\vec G) - (\vec F,\text{curl}\,G).$$ Then the Divergence Theorem tells us that $$\iiint_V \text{div}(\vec F\times\vec G)\,dV = \iint_S (\vec F\times\vec G,\vec n)\,dS,$$ where $\vec n$ is the unit outward-pointing normal to $S=\partial V$. Now, if you assume that at each point of $S$, $\vec n$ is a linear combination of $\vec F$ and $\vec G$, then the integrand on the right-hand side vanishes. This gives what I believe is the result you were trying to establish.
Your surface $$ x^2+y^2+z^2-2ax-2ay=0 $$ is a sphere around $(a,a,0)$ with radius $\sqrt{2}\,a\,.$ The plane $$ x+y=2a $$ intersects that sphere in a great circle (of radius $\sqrt{2}\,a$). The unit normal to that plane is $$\tag{$\color{red}{red}$ arrow} \frac{\hat{i}+\hat{j}}{\sqrt{2}}\,. $$ The curve that starts at $(2a,0,0)$ and goes below the $xy$-plane is oriented counterclockwise when we look from outside down to the $\color{blue}{blue}$ disk $S_2$ that caps the hemispheres. It is correct that the flux of $\nabla \times F=(-1,-1,-1)$ through the hemisphere $S_1$ plus the $\color{green}{green}$ disk $S_2$ is zero but this seems to only confuse you. What you have to calculate is just (by Stokes, and without extra minus sign) $$ \int_{\partial S_2}F\cdot\,ds=\int_{S_2}(\nabla\times F)\cdot\frac{\hat{i}+\hat{j}}{\sqrt{2}}\,dS $$ because that has the right orientation indicated by the $\color{orange}{orange}$ counterclockwise arrows.
The labels in the picture assume $a=1\,:$
Best Answer
It's been a while since I've done stokes theorem, so apologies if I make any error or whatnot.
The boundary curve is given by
we can choose any surface that has this boundary. I'll go for the plane since it is most simple where $z=a-x$ with restrictions on the ellipse boundary.
You calculated the curl to be $\langle -1, -1, -1\rangle$ and the normal vector denotes positive orientation so we are good.
Thus we have $$\int_C\overrightarrow{\textbf{F}}\cdot\text{ d}\overrightarrow{\textbf{r}}=\iint_S\operatorname{curl }\overrightarrow{\textbf{F}}\cdot\text{ d}\overrightarrow{\textbf{S}}=\iint_D\langle -1, -1, -1\rangle\cdot\nabla f\text{ d}A$$ where $D$ is the 2D shadow of the ellipse boundary on the $x-y$ plane and $$\nabla f=\nabla\left(z-a+x\right)=\langle 1, 0, 1 \rangle$$
Now we have $$\iint_D\langle -1, -1, -1\rangle\cdot\langle 1, 0, 1 \rangle\text{ d}A=-2\iint_D\text{ d}A$$
We just need to calculate the area of the ellipse given by the equation $$\frac{x^{2}}{\left(\frac{a}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{a}{\sqrt{2}}\right)^{2}}\le1$$ so this is just $$\pi\cdot \frac{a}{2}\cdot \frac{a}{\sqrt{2}}$$
So if I havent made a mistake the answer should just be $$-2\cdot\frac{a^2\pi}{2\sqrt2}=-\frac{a^2\pi}{\sqrt2}$$