If $\textbf{F}(x,y,z)=(xy-z,x+y^2,xyz)$ is a force field acting on a particle traversing the intersection of the cones $z=\sqrt{x^2+y^2},z=2-\sqrt{x^2+y^2}$ and the semispace $y\geq0$, compute the work done through the point $(-1,0,1)$ to the point $(1,0,1)$.
Is my strategy more or less on track?
I "closed" the curve $C=C_1 \cup C_2$ to apply Stokes' theorem. $C_1$ is the semicircle and $C_2$ the line from $(1,0,1)$ to the point $(-1,0,1)$.
We know that
$$\int_{C_1} \textbf{F} d\textbf{s} + \int_{C_2} \textbf{F} d\textbf{s} = \int_C \textbf{F} d\textbf{s} \stackrel{\text{Stokes}}{=} \iint_S \text{curl} \textbf{ F} d\textbf{S} \iff \\ \int_{C_1} \textbf{F} d\textbf{s} = \iint_S \text{curl} \textbf{ F} d\textbf{S} – \int_{C_2} \textbf{F} d\textbf{s}$$
Given that, for the parametrization $\sigma:[0,1]\to \mathbb{R}^3,\sigma(t)=(1-2t,0,1)$,
$$\int_{C_2} \textbf{F} d\textbf{s} = \int_0^1 \textbf{F}((1-2t,0,1)) \cdot (-2,0,0) dt =2$$
And that, for the parametrization $\gamma:[0,1]\times [0,\pi]\to\mathbb{R}^3,\gamma(u,v)=(u\cos v,u\sin v, 1)$,
$$\iint_S \text{curl} \textbf{ F} d\textbf{S} = \int_0^{\pi}\int_0^1 (u-u^2\cos v) du dv= \frac{\pi}{2}$$
Then $W=\int_{C_1} \textbf{F} d\textbf{s} = \frac{\pi}{2}-2$?
Best Answer
You can move on the curve and compute it directly. The curve is semicircle that parametrize as follows: $$W = \int_0^\pi (\cos t \sin t - 1 , \cos t + \sin^2 t , \cos t \sin t ) (\cos(2t) , -\sin t + \sin(2t) , \cos(2t)) dt$$ $$\int_0^\pi [(\frac{\sin(4t)}{4} - \cos(2t)] + [\frac{\sin(2t)}{-2}-\sin^3t+\cos t \sin^2t+ \sin^2 t\sin(2t)] + [\frac{\sin(4t)}{4}] dt$$ $$= \int_0^\pi -\sin^3 t dt = \int_0^\pi \frac{3\sin t - \sin(3t)}{4}dt = \left[\frac{-3\cos t}{4} + \frac{\cos(3t)}{12}\right]_0^\pi = \frac{4}{3}$$