Evaluate $\int ydx + zdy + xdz$ where $C $ is intersection of $x+y=2$
and $x^2+y^2+z^2=2(x+y) $ traversed counterclockwise as viewed from
origin
I am using Stokes' theorem to solve this question so
We want $\int \int curl F.N \; dS$ where $N$ is the normal unit vector to surface S, where S is a surface bounded by $C$
$F = \langle y,z,x\rangle$
$curl F = \langle -1,-1,-1 \rangle $
I take $S $ on the plane $x+y = 2$
$\nabla (x+y) = \langle 1,1,0 \rangle = A(say)$
Then unit normal vector $N = \langle -1/\sqrt2,-1/\sqrt2, 0\rangle $ {Multiplied by $-1$ because we are viewing it from origin }
$curl F.N= \sqrt(2)$
Now intersection of $x+y=2 $ and $x^2+y^2+z^2 = 2(x+y) $ gives
$x^2 + y^2 + z^2 = 4$
To get projection onto $xy$ plane $z=0$ we get
$x^2 +y^2 =4$
Now I am stuck in finding $dS$
How do I get dS = $\sqrt{z_x^2 +z_y^2 +1}dA$ where $A:x^2 +y^2 =4$
This is because my $S:x+y=2$ has no $z$ term
Best Answer
Since $\nabla\times F\cdot \vec{N}=\sqrt{2}$, you only need to evaluate $$\sqrt{2}\iint_S dS$$ which is $\sqrt{2}$ multiplied by the area of the disk. (Note that the intersection is a disk, and it passes through the center of the original ball.) The original ball can be written as $$(x-1)^2+(y-1)^2+z^2=2$$ whose center is $(1,1,0)$. The plane $x+y=2$ also passes through the center $(1,1,0)$. Hence the area of the disk is $\pi r^2=2\pi$, and the answer for the integral is $2\sqrt{2}\pi$.
Alternatively, if we project this surface to $xz$-plane, (we cannot project it to $xy$-plane since it is perpendicular to the $xy$-plane) we substituting $y=2-x$ into the other equation to obtain $$x^2+(2-x)^2+z^2=4\implies (x-1)^2+\frac{z^2}{2}=1.$$ Now setting $y=2-x$, we have $$\sqrt{1+y_x^2+y_z^2}=\sqrt{2}.$$ Computing $$\sqrt{2}\cdot\sqrt{2}\iint_A dA$$ where $A$ is region enclosed by the ellipse, we obtain $2\sqrt{2}\pi$, same as above.