Line integral in polar coordinates vs change of variables

Question

Assume we have a two-dimensional force field: $$F(r, \theta) = -4\sin(\theta)i + 4\sin(\theta)j$$

Compute the work done in moving a particle from the point $(1, 0)$ to the origin along the spiral whose polar equation is $r = e^{-\theta}$.

I know how to compute the answer in general. We need to represent the path $\alpha(\theta) = (e^{-\theta}\cos(\theta), e^{-\theta}\sin(\theta))$ as such, and then take a line integral to infinity.

However, I also first made a mistake by trying the following integral:

$$\int_0^{+\infty} F(r, \theta) \cdot \frac{\partial (e^{-\theta}, \theta)}{\partial \theta} d\theta$$

which of course gives an incorrect result.

The problem I cannot figure is how do we do the variable change here? In particular, where do the respective coordinate systems (Cartesian vs polar) come along and which should be transformed into which?

I justify the correct answer by noticing, that $F(r, \theta)$ is a vector in $R^2$, while the path $\alpha(\theta) = (e^{-\theta}, \theta)$ is a path in some space which is NOT an $R^2$. As such, no inner product can be formed.

1. Why the incorrect path not working? In short, why do we need to use $\alpha(\theta) = (e^{-\theta}\cos(\theta), e^{-\theta}\sin(\theta))$ instead of $\alpha(\theta) = (e^{-\theta}, \theta)$

2. If the vector field $F$ is already parameterized by $\theta$, where does $\sin(\theta)$ and $\cos(\theta)$ come from in its definition, given that the path should be described as $(e^{-\theta}\cos(\theta), e^{-\theta}\sin(\theta))$

3. How can I use the path $\alpha(\theta) = (e^{-\theta}, \theta)$ in whatever space it is defined?

Answer 1

Your formula is correct, but the interpretation is wrong. Let's start with the "correct" way, as in point 1. $$(x,y)=(e^{-\theta}\cos\theta,e^{-\theta}\sin\theta)=e^{-\theta}\cos\theta\hat i+e^{-\theta}\sin\theta\hat j$$ Then $$\vec{dl}=\{[-e^{-\theta}\cos\theta-e^{-\theta}\sin\theta]\hat i+[-e^{-\theta}\sin\theta+e^{-\theta}\cos\theta]\hat j\}d\theta$$ And then the line integral is $$\int_\alpha\vec F\cdot\vec{dl}=\int_0^\infty4e^{-\theta}(\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta)d\theta=\int_0^\infty4e^{-\theta}d\theta$$ Now the "incorrect" way. You write the path in polar coordinates $r$ and $\theta$. For dot product to work, you need to have two orthogonal directions, $\hat r$ and $\hat \theta$. It is easy to see that when translated to Cartesian coordinates $$\begin{align}\hat r&=\cos\theta\hat i+\sin\theta\hat j\\\hat\theta&=-\sin\theta\hat i+\cos\theta\hat j\end{align}$$ The next step is to write $$\vec F(r,\theta)=4\hat\theta$$ and $$\vec{dl}=dr\hat r+rd\theta\hat\theta$$ With this, your line integral becomes $$\int_\alpha\vec F\cdot\vec{dl}=\int_0^\infty 4rd\theta=\int_0^\infty4e^{-\theta}d\theta$$ So the two forms are identical. Your mistake was to use different coordinate system for the path and the force.

Note As mentioned in the comment, I've solved a slightly different problem. But the method is the same. With the correct force, in Cartesian coordinates we have: $$\int_0^\infty \vec F\cdot\vec{dl}=\int_0^\infty(\sin\theta\cos\theta+\sin^2\theta-\sin^2\theta+\sin\theta\cos\theta)d\theta=4\int_0^\infty e^{-\theta}\sin(2\theta) d\theta$$ The last integral can be done by integrating by parts twice, and the value for that is $2/5$, so the final answer is $8/5$.

Similarly, we want to write $\vec F$ in the polar coordinate system: $$\vec F=a\hat r+b\hat\theta=a\cos\theta\hat i+a\sin\theta\hat j-b\sin\theta\hat i+b\cos\theta\hat j$$ Identifying coefficients of $\hat i$ and $\hat j$ you get: $$\begin{align}a\cos\theta-b\sin\theta&=-4\sin\theta\\a\sin\theta+b\cos\theta&=4\sin\theta\end{align}$$ To find $a$, you multiply the first equation by $\cos\theta$, the second by $\sin\theta$ and you add them together: $$a(\cos^2\theta+\sin^2\theta)=4\sin^2\theta-4\sin\theta\cos\theta$$ Similarly, to find $b$, you multiply the first equation by $-\sin\theta$ and the second by $\cos\theta$, and add them: $$b(\sin^2\theta+\cos^2\theta)=4\sin^2\theta+4\sin\theta\cos\theta$$ Then the integral becomes $$\int_\alpha\vec F\cdot\vec{dl}=\int_0^\infty (a\frac{dr}{d\theta}+r b) d\theta=\int_0^\infty\left[(4\sin^2\theta-4\sin\theta\cos\theta)(-e^{-\theta})+e^{-\theta}(4\sin^2\theta+4\sin\theta\cos\theta)\right]d\theta=4\int_0^\infty e^{-\theta}\sin(2\theta)d\theta$$ Once again, this is the same answer as with the Cartesian coordinate system.