# How to write line integral to describe a force given in polar coordinates?

I have a force given by $$\vec{F}=200N\cos\theta \hat{i}+200N\sin\theta \hat{j}=200N \hat{R}$$ and I'm trying to figure out how to set up two line integrals to calculate work as a function of initial and final positions in terms of both$$(x,y)$$ and $$(R,\theta)$$, and also determine whether or not this force is consevative.

For $$(R,\theta)$$ I have \begin{align} x&=200N\cos\theta \hat{i} & y& =200N\sin\theta \hat{j} \\ dx&=-200N\sin\theta d\theta & dy&=200N\cos\theta d\theta \end{align}

\begin{align} \int_{C}^{}\vec{F}\cdot d\vec{s} & =\int_{C}^{}\left \langle x,y \right \rangle\cdot \left \langle dx,dy \right \rangle \\ & =\int_{C}^{}200N\cos\theta (-200N\sin\theta d\theta )+(200N\sin\theta )(200N\cos\theta d\theta ) \\ & =0 \end{align}

I don't know if this is right or how I should go about writing the line integral in terms of x and y. Can someone help out?

## 3 Answers

So a conservative force can be expressed as the gradient of a scale function: $$V(r, \theta)$$. Your force is:

$$\vec F(r, \theta) = k\hat r$$

The gradient in polar coordinates is:

$$\vec \nabla V(r, \theta) = \frac{\partial V}{\partial r}\hat r + \frac 1 r \frac{\partial V}{\partial \theta}\hat{\theta}$$

so (1) your potential can't be a function of $$\theta$$. (2) Let's guess a power of $$r$$:

$$V(r) = ar^n$$ $$\vec \nabla V(r, \theta) = anr^{n-1}\hat r$$

which is solved with $$n=1$$ and $$a=k$$:

$$V(r) = kr$$

The line integral along any curve $$C$$ satisfies

$$\int_C \vec F(r)\cdot d\vec s = V(r_1, \theta_1)-V(r_0, \theta_0) = k(r_1 - r_0)$$

where the $$(r_i, \theta_i)$$ are the endpoints of $$C$$.

• Thanks for this answer, but I'm still having a tad bit of trouble. I'm trying to find the work done by the force as a function of initial and final positions in both the $x,y$ and $(R, \theta)$ coordinate systems. – CalebWilliamsUIC Dec 2 '20 at 20:38

To do this in polar coordinates, you'd need to express your $$\hat R$$ vector and $$r$$ as functions of some parameter. But you'd most likely end up expressing the $$\hat R$$ vector in terms of $$x$$ and $$y$$ positions anyway.

The force is conservative if, as you know, it can be described as the gradient of some potential. In polar coordinates, the gradient operator applied to some potential $$U$$ is: $$\nabla U =\hat{\pmb e}_{r}\frac{\partial U}{\partial r}+\hat{\pmb e}_{\theta}\frac{1}{r}\frac{\partial U}{\partial \theta}.$$

At any given point you have:

$$\hat{r}=\cos{\theta}\hat{i}+\sin{\theta}\hat{j}$$

$$\hat{\theta}= -\sin{\theta} \hat{i}+ \cos{\theta}\hat{j}$$

Note the dot product of either vector with itself is 1 and the product of one with the other is zero, they are orthogonal unit vectors.

The infinitesimal line element is: $$d\vec{s}=dr \ \hat{r}+rd\theta \ \hat{\theta}$$

From above, $$\vec{F}=F_0\hat{r}$$. By definition of work, Work is the line integral of the force through a distance:$$W=\int {\vec{F}\cdot d\vec{s}}$$.

$$\vec{F}\cdot d\vec{s}=F_0\hat{r}\cdot(dr\hat{r}+rd\theta\hat{\theta})=F_0dr$$. Here $$d\vec{s}$$ is the linear differential vector.

It follows that $$W=\int F_0 dr=F_0\Delta r$$. The integral only depends on the net change in $$r$$.

We then have two ways to prove $$\vec{F}$$ is conservative. You can't leave a given point, then return without there being a zero net change in r, but a zero net change in r implies no work is done along a closed path, so we have from this that $$\vec{F}$$ is conservative.

We also have that $$\vec{F}=F_0\nabla{r}$$. $$\vec{F}$$ is a conservative force because it can be expressed as a gradient.

• I don't really understand this. How can I use this to solve the problem with the line integrals? – CalebWilliamsUIC Dec 2 '20 at 21:25