Stokes' theorem on intersection of $ x^2+y^2+z^2=1$ and $x+y=0$

Question

Using Stokes' theorem, evaluate the line integral $$\int_L f\cdot dr $$ where $L$ is the intersection of $ x^2+y^2+z^2=1$ and $x+y=0$ traversed in counter clockwise direction when viewed from $(1,1,0)$. $f=yi+zj+xk.$

I evaluated curl $f=-(î+j+k)$ and the normal vector to the plane $x+y=0$ is $ \frac{-(î+j)}{\sqrt{2}}$. Since the curve lies in $x+y=0$ plane so I am having difficulty in further proceedings.

Answer 1

The normal vector is $\frac{i+j}{\sqrt{2}}$, since the circle is traversed in counter clockwise direction.

Stoke's theorem then gives you $$\iint (-1,-1,-1)\cdot(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0) dS$$ where the surface is a circle with radius $1$. I think you can go from there.