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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.

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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.

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  • $\begingroup$ I'm almost there... For the function we're integrating over, <-y,x>, why did x go to cos(t) + 1? I thought it would just be cos(t) which interestingly enough will yield the same answer $\endgroup$ – financial_physician Aug 14 '19 at 18:56
  • $\begingroup$ @financial_physician The equation for the circle you're integrating over is $(x-1)^2+y^2=1$, not $x^2+y^2=1$, so parameterising it with $x(t)=\cos t$ just doesn't give the right curve. As to why it gives the same answer, well, the theorem-y approach is better at answering that. $\endgroup$ – Arthur Aug 14 '19 at 19:24
  • $\begingroup$ Sorry, I don't think I'm doing a very good job explaining what I'm trying to ask. 𝑟(𝑡)=(cos(𝑡)+1, sin(𝑡)) makes sense to me so dr = (-sin(t), cos(t)) makes sense. What I'm trying to figure out in the the equation you got for F(x(t),y(t)) which appears to be (−sin(𝑡),cos(𝑡)+1). I'm assuming you got that from our original function F(x,y) = <-y,x>. Here, you got <-sin(t), cos(t) + 1> but I thought y = sin(t) and x = cos(t) so wouldn't it be <-sin(t), cos(t)> for F(x,y)? $\endgroup$ – financial_physician Aug 14 '19 at 19:35
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    $\begingroup$ @financial_physician No, you insert the parameterised $x(t)$ and $y(t)$ into $F$. You are, after all, looking for the vector field at points on the curve. Points on the curve have the form $(x(t), y(t))=(\cos t+1, \sin t)$, and inserting the $x$ and $y$ coordinates of points into $F$ tells you what the vector field looks like at those points. $\endgroup$ – Arthur Aug 14 '19 at 19:39
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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 $$\int_0^{\pi/2}\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt$$ $$=\int_0^{\pi/2}\langle-\sin2t,2\cos^2t\rangle\cdot\langle-2\sin2t,2\cos2t\rangle\,dt$$ $$=\int_0^{\pi/2}(2\sin^22t+4\cos ^2t\cos2t)\,dt$$ $$=\int_0^{\pi/2}(1-\cos4t+2\cos ^22t+2\cos2t)\,dt$$ $$=\int_0^{\pi/2}(1-\cos4t+\cos4t+1+2\cos2t)\,dt$$ $$=2\int_0^{\pi/2}(1+\cos2t)\,dt$$ $$=2\left[t+\frac12\sin2t\right]_0^{\pi/2}=\pi$$

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