Evaluate $$\int_L (y i + zj + xk) \cdot dr$$ where $L$ is the intersection of the unit sphere and $x+y = 0 $ traversed in the clockwise direction when viewed from $(1,1,0)$.
My attempt:
$∇ \times A = - i -j -k$
S: Surface of intersection of the sphere and the plane. Then definitely S lies on the plane and is bounded by a level curve of the sphere. So, S is given by $x+y = 0$, $x^2 + y^2 = 1$.
Now, a unit normal to S is given by n = $(i + j)/√2$
So, $(∇ \times A)\cdot n = -\sqrt 2$
implies $\int\int ((∇ \times A) \cdot n) \ dS$ reduces to $-\sqrt2 \int\int dS$. Since the plane passes through the origin, the circle it defines in the unit sphere has radius 1. So $\int\int dS = π$.
Hence, the final answer is $-\sqrt{2π}$. I have doubts regarding two things:
(a) Is the above solution correct? I came across this problem in two books. One lists the answer as $-\sqrt{2π}$ while the other says it is $\pi+2$.
(b) How do I ascertain that the unit normal is outward, so that the final answer is positive? Also, I haven't used the statement "traversed in the clockwise direction when viewed from $(1,1,0)$" while solving the question. How do I apply it?
(a) The answer is √2π.
(b) Outward normal should be interpreted as per the orientation mentioned in the question.
The unit sphere centred at origin and the plane through origin intersect to form the circle L. S is the surface bounded by this curve L. So S is a disc of unit radius.