Line integral with vector field in polar coordinates 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.
 A: 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.
A: Alternatively, one may work with differential forms to do the full computation in polar coordinates. Pulling back the form to be integrated we get $$\omega = 4\sin \theta (-dx + dy) = 4\sin \theta \big[ (\sin\theta - \cos\theta )dr + r(\sin\theta + \cos\theta)d\theta\big].$$ We can parametrize the spiral in the polar plane via $$\gamma(t) = (r(t),\theta(t)) = (e^{-t},t), \quad t \in [0,\infty).$$ This yields $$\dot\gamma(t) = -e^{-t}\partial_r + \partial_\theta,$$ and now we are ready to integrate:
$$\int_\gamma \omega = \int_0^\infty \omega_{\gamma(t)}(\dot\gamma(t))\, dt = \int_0^\infty 8e^{-t}\sin t \cos t \, dt = \frac 85.$$
