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I have this line integral $$\oint_{\partial D}(f\nabla f)\cdot\hat{n}\,ds$$ which I would like to "change variables" so that the final result is in terms of a line integral of $g$ on $[0,2\pi]$ where $g(\theta)=f(e^{i\theta})$.

Background info: $D=\{(x,y)\in\mathbb{R}^2: x^2+y^2<1\}$, so $\partial D$ is a unit circle.

My working (which is definitely wrong, but I am not sure where it goes wrong.): Parametrize the circle $\partial D$ by $x=\cos\theta$, $y=\sin\theta$. We may take $\hat n\,ds=(dy,-dx)$.

$\begin{align*} &\oint_{\partial D}(f\nabla f)\cdot(dy,-dx)\\ &=\oint_{\partial D}(ff_x,ff_y)\cdot(dy,-dx)\\ &=\oint_{\partial D} (ff_x\,dy-ff_y\,dx) \end{align*}$

Making the change of variable $g(\theta)=f(e^{i\theta})$, then $f_x=\frac{dg}{d\theta}\frac{d\theta}{dx}=g'\cdot\frac{1}{-\sin\theta}$, $f_y=\frac{dg}{d\theta}\frac{d\theta}{dy}=g'\frac{1}{\cos\theta}$, $dx=-\sin\theta\, d\theta$, $dy=\cos\theta\,d\theta$.

Substituting all of these in, the eventual integral becomes $$\int_0^{2\pi} gg'(\tan\theta-\cot\theta)\,d\theta.$$

I am certain this is wrong since $\tan\theta=\infty$ at $\theta=\pi/2$. Which part of my working is wrong?

Thanks for any help. Very curious to know the error.

yoyostein
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    You don't need to add "multivariable calculus expert needed" in the title. It can be safely assumed that the vast majority people on this website are proficient with calculus, and the ones that aren't probably won't try to answer a question about a line integral. – anomaly May 31 '16 at 06:36

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You are just applying the 2 variable chain rule incorrectly (or rather, at selected points pretending that there is only one variable). Consider the following calculation that basically forgets the special form of your integration path.

You define $$ g(r,\theta)=f(r\cos\theta, r\sin\theta). $$

The 2-variable chain rule then gives us the formulas $$ \frac{\partial g}{\partial \theta}=-f_x\cdot r\sin\theta+f_y\cdot r\cos\theta $$ and $$ \frac{\partial g}{\partial r}=f_x\cdot \cos\theta+f_y\cdot \sin\theta. $$ Multiplying the former by $-\sin\theta$ and the latter by $r\cos\theta$ and adding the resulting equations together eliminates $f_y$, and leads to $$ f_x=\frac1r\left(r\cos\theta \frac{\partial g}{\partial r}-\sin\theta \frac{\partial g}{\partial \theta}\right). $$ Something very different from what you gave. It is easy to similarly solve for $f_y$ in terms of $g_r$ and $g_\theta$.

So $g_r$ affects $f_x$ everywhere (why wouldn't it). Even along the unit circle.

Jyrki Lahtonen
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    In other words, $\theta$ is a function of both $x$ and $y$, so you cannot just take the reciprocal of $\partial x/\partial\theta$. – Jyrki Lahtonen May 31 '16 at 07:02
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    I am somewhat unhappy with this. While it gets to the point, I think I failed to find the conceptual error. I need to commute next, so leaving the floor for somebody to give a better answer. – Jyrki Lahtonen May 31 '16 at 07:08
  • Your answer has enlightened me. I see the error now. – yoyostein May 31 '16 at 07:31