Apply Stokes theorem to prove that $\int_{c} ydx+zdy+xdz =-2\sqrt{2}\pi a^2$
Where C is the curve given by $x^2+y^2+z^2-2ax-2ay=0, x+y=2a$ ; and begins at the point (2a,0,0) and it goes first below the z plane.
My work
I move clockwise on the boundary of the sphere cut by the plane since i want the surface to be on the left side upon movement on the curve inline with stoke's theorem.
The normal on the curve is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$
Curl of F $y\hat{i} + z\hat{j} + x\hat{k}$ is $-\hat{i} -\hat{j} -\hat{k}$
Now taking surface of the sphere as $S_1$ and surface of the great circle as $S_2$ we have using gauss divergence theorem
$$
\int_{S_1 + S_2} (\nabla \times F) \hat{n} dS = 0 \\
\int_{S_1} (\nabla \times F) \hat{n} dS = – \int_{S_2} (\nabla \times F) \hat{n} dS \\
\int_{S_1} (\nabla \times F) \hat{n} dS = – \int_{S_2} (-\hat{i} -\hat{j} -\hat{k}) * \frac{\hat{i} + \hat{j}}{\sqrt{2}} dS \\
\int_{S_1} (\nabla \times F) \hat{n} dS = \int_{S_2} \sqrt{2} dS \\
\int_{S_1} (\nabla \times F) \hat{n} dS = 2 \sqrt{2} \pi a^2
$$
The above calculation is off by a factor of – sign
Also I am unable to understand how does goes below the z axis affects the calculation.
Best Answer
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 $a=1\,:$