Evaluate the line integral over the vector field given this semi-circle

Question

So here is the problem on the practice test verbatim

Evaluate $ \int_{C}^{ } \textbf{F} \cdot d\textbf{r} \text{ where } C = \{ (x,y) \in \mathbb{R} \mid (x-1)^2 + y^2 = 1 \text{ and } y \ge 0 \} $ oriented counter clockwise and $\textbf{F}(x,y) = <-y,x> $

I've considered doing it a few ways but keep getting stuck. At first I thought I'd parameterize it but then I realized I didn't know what to do with $\textbf{F}(x,y) = <-y,x> $ so then I thought converting it to curl would make sense but then I get confused because if I dot with the gradient vector then both x and y go to zero since the x component isn't in terms of x and the y component isn't in terms of y.

Answer 1

Brute force way

Parameterise the curve by some suitable $r(t)=(x(t), y(t))$ where $t$ goes from $t_0$ to $t_1$. The line integral is then given by $$ \int_{t_0}^{t_1} F(x(t),y(t))\cdot r'(t) dt $$ For instance, setting $r(t) = (\cos(t) + 1, \sin(t))$ with $t\in [0, \pi]$, we get $$ \int_0^\pi(-\sin(t), \cos(t) + 1)\cdot (-\sin(t), \cos(t)) dt = \int_0^\pi(1 + \cos(t))dt $$ which isn't too hard to calculate.

Using theorems

The curl theorem states that the integral of a vector field $F$ along a closed (nice enough) curve is equal to the integral of the curl of $F$ over the interior of the curve. Our curve isn't closed yet, so we need to fix that.

Add a straight line segment to our curve, from $(0,0)$ to $(2,0)$. The vector field is orthogonal to this line, so the integral over the line is $0$, meaning the value of the integral won't change from this addition. This makes our curve into a closed semicircle, and we may apply the theorem.

The curl of $F$ is constantly equal to $2$, so integrating the curl over the region simply means multiplying the area of the semicircle by $2$, and we're done.

Answer 2

We parametrise the semi-circle as $$\mathbf r(t)=\langle2\cos^2t,\sin2t\rangle\qquad0\le t\le\pi/2$$ $$\mathbf r'(t)=\langle-2\sin2t,2\cos2t\rangle$$ where the first expression can be derived from the polar form of the curve $r=2\cos\theta$. The line integral then becomes $$\int_0^{\pi/2}\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt$$ $$=\int_0^{\pi/2}\langle-\sin2t,2\cos^2t\rangle\cdot\langle-2\sin2t,2\cos2t\rangle\,dt$$ $$=\int_0^{\pi/2}(2\sin^22t+4\cos ^2t\cos2t)\,dt$$ $$=\int_0^{\pi/2}(1-\cos4t+2\cos ^22t+2\cos2t)\,dt$$ $$=\int_0^{\pi/2}(1-\cos4t+\cos4t+1+2\cos2t)\,dt$$ $$=2\int_0^{\pi/2}(1+\cos2t)\,dt$$ $$=2\left[t+\frac12\sin2t\right]_0^{\pi/2}=\pi$$