Stokes theorem on calculating the integral

Question

Using Stokes theorem calculate the integral:$\int_C x^2y^2 dx+dy+zdz$ where C is curve bounded with $x^2+y^2=2x, y\ge 0, z=1$ oriented from $A(2,0,1)$ to $B(0,0,1)$. So using Stokes formula I should bring this down to solving surface integral od second kind. After finding partial derivates od $P=x^2y^2, Q=1, R=z$, I need to calculate normal vector, and then scalar product of field F and normal n. But not sure how to use given orientation, what is it telling me?

Answer 1

Vector field is $\vec{F} = (x^2y^2,1,z)$. The curve is bounded by $x^2+y^2=2x, y\ge 0, z=1$ and you are going from point $A(2,0,1)$ to $B(0,0,1)$.

You can rewrite the surface as $(x-1)^2+y^2 = 1$ which is a cylinder of radius $1$ with center at $(1,0,z)$. Now coming to the curve, it is a semicircle on this surface from point $A$ to point $B$ at $z=1$. Please note you must go anti-clockwise from $A$ to $B$ as $y \geq 0$.

You can parametrize your curve as
$r(t) = (2\cos^2 t, 2\cos t \sin t, 1) $ or $(1+\cos2t, \sin2t,1), 0 \leq t \leq \frac{\pi}{2}$.

$r'(t) = (-2\sin2t, 2\cos 2t,0)$

$\vec{F} \cdot r'(t) = (\sin^2 2t (1+\cos 2t)^2,1,1) \cdot (-2\sin2t, 2\cos2t,0)$

$ = -2\sin^32t(1+\cos 2t)^2 + 2\cos2t$

Line integral is $ \ \displaystyle \int_0^{\pi/2} \vec{F} \cdot r'(t) \ dt$