I have the following problem:

Given a vector field in polar coordinates $$ \mathbf{F}(r,\theta) = -4 \sin \theta\ \mathbf{i}\ +\ 4 \sin \theta\ \mathbf{j},$$ calculate the work done when a particle is moved from point $(1,0)$ to the origin, following the spiral whose polar equation is $r = e^{-\theta}.$

My attempt was to write the equation of the spiral like so $$\mathbf{\alpha}(t) = e^{-\theta} \cos \theta\ \mathbf{i} + e^{-\theta} \sin \theta\ \mathbf{j} \\ \mathbf{\alpha}'(t) = -e^{-\theta}(\cos \theta + \sin \theta\ \mathbf{i}\ + \sin \theta - \cos \theta\ \mathbf{j}),$$ so the line integral would become $$\int_C \mathbf{F}\cdot\mathbf{\alpha'}(t) = \int_C 8 e^{-\theta}\sin\theta\cos\theta\ d\theta.$$

But this doesn't give me the right answer, what am I doing wrong?

NOTE: I know this question was asked before, but it doesn't have an accepted answer, and what I read from there wasn't very helpful.

  • 3
    $\begingroup$ Did you integrate from $0$ to $\infty$? What is the correct answer according to the book? $\endgroup$ – Ofek Gillon Mar 10 '17 at 15:33
  • $\begingroup$ No, I haven't tried integrating from $0$ to $\infty$, I'll try that right away. The book says that the answer is $\frac{8}{5}.$ $\endgroup$ – lorenzattractor Mar 11 '17 at 0:08
  • $\begingroup$ It worked! I integrated from $0$ to $\infty$ and got the correct answer, but why? I really don't get it why it worked. $\endgroup$ – lorenzattractor Mar 11 '17 at 0:48
  • $\begingroup$ Because you need to integrate from $(1,0)$ to $(0,0)$. We see that for $\theta = 0 $ we get the point $(1,0)$, but in which $\theta$ do we approach $(0,0)$? You can see that you need to solve the equation $e^{-\theta} = 0 \Rightarrow \theta= \infty$ , so you need to integrate from $\theta=0$ to $\theta=\infty$ $\endgroup$ – Ofek Gillon Mar 11 '17 at 7:50

As Ofek pointed out, one should integrate from $0$ to $\infty$ given that solving the equation $e^{- \theta} = 0$ gives $ \theta = \infty$, and this is true because we are using the polar equation given in the exercise.

So the answer to this problem is simply taking $\mathbf{F}(r,\theta)$, compute the dot product with the derivative of the parametric curve $\alpha ' (t),$ and finally compute the following integral using integration by parts $$\int_0^{\infty} 4 e^{-\theta} \sin 2\theta\ d \theta,$$ where I have used the trigonometric identity $2 \sin \theta \cos \theta = \sin (2 \theta).$ This will yield the correct answer, and I credit Ofek for it.


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