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.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $$\int_0^{\pi/2}\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt$$ $$=\int_0^{\pi/2}\langle-\sin2t,2\cos^2t\rangle\cdot\langle-2\sin2t,2\cos2t\rangle\,dt$$ $$=\int_0^{\pi/2}(2\sin^22t+4\cos ^2t\cos2t)\,dt$$ $$=\int_0^{\pi/2}(1-\cos4t+2\cos ^22t+2\cos2t)\,dt$$ $$=\int_0^{\pi/2}(1-\cos4t+\cos4t+1+2\cos2t)\,dt$$ $$=2\int_0^{\pi/2}(1+\cos2t)\,dt$$ $$=2\left[t+\frac12\sin2t\right]_0^{\pi/2}=\pi$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.