# Line integral off by a factor of 2

Perform the closed line integral $$\int \vec{F}\cdot d\vec{s}$$ for the given field along the specifcied path. Let $$\hat F =r\hat{\theta}$$ starting and finishing at $$(2,0)$$ along the counter-clockwise circle $$r=2$$.

$$\int_CF \cdot d\vec{s}=\int_0^{2\pi}F(\vec{r}(t))\cdot r'(t) dt = \int_0^{2\pi}2 \vec{\theta}\cdot \theta dt = 4\pi.$$

Let $$\vec{r}(t)=2\hat{r} + t\hat{\theta}$$, $$t\in [0,2\pi]$$. Then $$\vec{r}'(t) = (0,1)$$.

However in my book, the answer is $$8\pi$$. What have I done wrong?

EDIT: This was very instructive! I didn't do the change of coordinates correctly. Funnily enough, I just learned how to do change of tangent basis coordinates in my geometry class but didn't see the application here.

• I thought the parameterization formula went $\int F(r(t))\cdot r'(t)dt$? Does this not apply here? Feb 25 at 4:51
• Your $\vec{r}(t)$ is incorrect (plug in your points and you'll see that it doesn't map correctly). It is simply $$\vec{r}(t) = 2\hat{r} \implies \vec{r}'(t) = 2\hat{\theta}$$ In general you are asking for problems when you try to do vector integrals with noncartesian unit vectors, because they are not constants unlike $\hat{x}$, $\hat{y}$, etc which were constant values everywhere. Feb 25 at 4:52
• @NinadMunshi but shouldn't the angle move as t moves? there should be some deppendence on t Feb 25 at 4:54
• You have the incorrect assumption that $(r,\theta)$ written in vector form would be $r\hat{r} + \theta\hat{\theta}$. That is not true, it would simply be $(r,\theta)\mapsto r\hat{r}(\theta)$, using $\hat{r} = \cos\theta \hat{x} + \sin\theta \hat{y}$ Feb 25 at 4:56
• As an exercise to see the difference try computing $$\int_0^{2\pi}\hat{x}\:d\theta$$ and then compare it to the value of $$\int_0^{2\pi}\hat{r}\:d\theta$$ Feb 25 at 4:59

To disentangle the notation I write $$\boldsymbol{\gamma}(t)=\gamma_x(t)\,\hat{\boldsymbol{x}}+\gamma_y(t)\,\hat{\boldsymbol{y}}$$ for the curve over which we integrate and start with Cartesian coordinates. The tangent vector at this curve is $$\boldsymbol{\gamma}'(t)=\gamma_x'(t)\,\hat{\boldsymbol{x}}+\gamma_y'(t)\,\hat{\boldsymbol{y}}=\underbrace{\gamma_x'(t)\,\partial_x+\gamma_y'(t)\,\partial_y}\,.$$ The notation in the underbraced term has the huge advantage that the chain rule leads directly to the expressions in polar coordinates: $$\boldsymbol{\gamma}'(t)=\gamma_r'(t)\,\partial_r+\gamma_\theta'(t)\,\partial_\theta\,,$$ where \begin{align}\tag{1} \partial_x&=\cos\theta\,\partial_r-\color{red}{\tfrac1r}\sin\theta\,\partial_\theta\,,& \partial_y&=\sin\theta\,\partial_r+\color{red}{\tfrac1r}\cos\theta\,\partial_\theta\,,\\[2mm] \gamma_x'&=\cos\theta\,\gamma_r'-\color{red}{r}\sin\theta\,\gamma_\theta'\,,& \gamma_y'&=\sin\theta\,\gamma_r'+\color{red}{r}\cos\theta\,\gamma_\theta'\,.\tag{2} \end{align} The common physics convention is that the normalized unit basis vectors are \begin{align} \hat{\boldsymbol{r}}&=\partial_r\,,& \hat{\boldsymbol{\theta}}&=\color{red}{\tfrac1r}\partial_\theta\,. \end{align} By this normalization (1) and (2) become orthonormal transformations. The transformation (2) holds for any vector field. When it is made clear that its polar components are to be understood w.r.t. the coordinate basis $$\{\partial_r,\partial_\theta\}=\{\hat{\boldsymbol{r}},\color{red}{r}\hat{\boldsymbol{\theta}}\}$$ this becomes \begin{align}\tag{3} F_x&=\cos\theta\,F_r-\color{red}{r}\sin\theta\,F_\theta\,,& F_y&=\sin\theta\,F_r+\color{red}{r}\cos\theta\,F_\theta\,. \end{align} The vector field that we integrate along $$\gamma$$ is \begin{align} \boldsymbol{F}=F_x\,\hat{\boldsymbol{x}}+F_y\,\hat{\boldsymbol{y}}= F_x\,\partial_x+F_y\,\partial_y= F_r\,\partial_r+F_\theta\,\partial_\theta=F_r\,\hat{\boldsymbol{r}}+\color{red}{r}F_\theta\,\hat{\boldsymbol{\theta}}\,, \end{align} where the polar notation is compatible with (1) and (2) and the fact that we express $$F_r,F_\theta$$ in the coordinate basis. In Cartesian coordinates, the line integral is \begin{align} \int_C\boldsymbol{F}\cdot ds=\int_a^bF_x\gamma_x'+F_y\gamma_y'\,dt\,.%=\int_a^bF_r\,\gamma_r'+F_\theta\gamma_\theta'\,dt\, \end{align} where $$\boldsymbol{\gamma}(t)=\pmatrix{\gamma_x\\\gamma_y}=\pmatrix{r\cos t\\r\sin t}$$ (these are Cartesian coordinates!). Therefore, $$\boldsymbol{\gamma}'(t)=\pmatrix{\gamma_x'\\\gamma_y'}=\pmatrix{-r\sin t\\r\cos t}\,.$$ In your case $$\boldsymbol{F}=r\hat{\boldsymbol{\theta}}=\partial_\theta$$ so that $$F_r=0$$ and $$F_\theta=\color{red}{1}\,.$$ Using (3) we get $$F_x=-\color{red}{r}\sin\theta\,,\quad F_y=\color{red}{r}\cos\theta\,.$$ Choosing $$t=\theta$$ in $$[0,2\pi]$$ we get \begin{align} \int_C\boldsymbol{F}\cdot ds=\int_0^{2\pi}\color{red}{r^2}\sin^2\theta+\color{red}{r^2}\cos^2\theta=2\pi\color{red}{r^2}\,. \end{align} Perhaps it is worth mentioning that $$F_x=-r\sin\theta\,,\quad F_y=r\cos\theta\,,\quad \boldsymbol{F}=\pmatrix{-y\\x}$$ so that the line integral around the circle coincides with that of the one-form $$-y\,dx+x\,dy\,.$$ Since the radius is $$r$$ this form coincides on $$C$$ with $$\boldsymbol{\omega}=r^2\frac{-y\,dx+x\,dy}{x^2+y^2}$$ which is known to equal $$r^2\,d\theta$$ and has a very simple integral. No headaches whasoever.