$\int_C F dr $ where $F(x,y) = (y, -x)$ where $C$ is the curve $x^2+y^2=1$

Question

I have the following problem:

Calculate $$\int_C F dr $$

where $F(x,y) = (y, -x)$ where $C$ is the curve $x^2+y^2=1$ that connects the points $(1,0)$ with $(0,1)$ counterclockwise.

What I get is that I need to parametrize $C$. But how can I parametrize it in terms of a $t$ and not with cosine or sine. Or is using trigonometric functions the way to go? And then, the counterclockwise part also confuses me. Does this implies I need to use $\int_{-C} Fdr=-\int_C Fdr $?

Thanks.

Answer 1

Trig functions work perfectly fine: $$C=(\cos t,\sin t),0\le t\le\pi/2$$ $$\int_CF\,dr=\int_0^{\pi/2}(\sin t,-\cos t)\cdot(-\sin t,\cos t)\,dt=\int_0^{\pi/2}-1\,dt=-\frac\pi2$$ The statement that the curve runs counterclockwise around the unit circle tells us that the parametrisation above is correct – no need to use a sign reversal.

Answer 2

As $t$ moves from $0$ to $\pi/2$, the point $(\cos(t),\sin(t))$ moves from where to where? Is it moving clockwise or counterclockwise?

"Counterclockwise" is actually the default orientation in parametrizing the unit circle. The parametrization of the circle with trigonometric functions is also the most natural one, given that they provide a parametrization of constant speed. There are other parametrizations, such as the rational one $\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$, but they don't have this nice constant-speed feature.