calculus - Find the net flux of temperature given the temperature distribution $u(z)= - ln left| z right|$

Given the logarithmic temperature distribution $u(z)= - \ln \left| z \right| = - \frac{1}{2} \ln (x^2+y^2),$ verify that the net flux of temperature across the unit circle is $-2 \pi$.

The net flux of temperature is given by $\int_{\Gamma} (\partial u/ \partial \eta) ds$ where $\partial u/ \partial \eta$ is the outward normal derivative. The outward normal derivative is given by $\nabla u(z) \cdot N(z)$ where $\nabla u(z)$ is the gradient of $u$ and $N(z)$ is the normal vector at $z$. So $$\partial u/ \partial \eta= (u_x,u_y) \cdot N(x,y) = \left(-\frac{x}{x^2+y^2},-\frac{y}{x^2+y^2} \right) \cdot (x,y) = -\frac{x^2+y^2}{x^2+y^2}=-1.$$

Thus $\int_{\Gamma} (\partial u/ \partial \eta) ds = -\int_{\Gamma} 1 ds= -\int_{0}^{2 \pi} 1 ds = -s]_{0}^{2 \pi} = -2 \pi$.

Is this correct? I'm not entirely sure that my work is valid.

Answer

The work in the OP is solid. But I thought it might be instructive to see how use of polar coordinates facilitates the ensuing analysis.

To that end, let $\vec \rho =\hat xx+\hat yy$ be the $2$-dimensional position vector and let the vector $\vec F(\vec \rho)$ be defined as

$$\vec F(\vec \rho) =-\nabla \log(|\vec \rho|).$$

Then, net outward flux $\Psi$ of the unit circle $|\vec \rho|=1$ of

$$\begin{align} \Psi&=\oint_{|\vec \rho|=1} \vec F(\vec \rho)\,\cdot \hat n\,d\ell\\\\ &=\oint_{|\vec \rho|=1} \left(-\nabla\log(|\vec \rho|)\right)\,\cdot \hat n\,d\ell \tag 1 \end{align}$$

Working in polar coordinates $(\rho,\phi)$, the outward unit normal is $\hat n=\hat \rho$, where $\hat \rho =\frac{\vec \rho}{\rho}$ and the differential length is $\vec d\ell= \rho\,d\phi. Moreover, the integrand can be written

$$\begin{align}\vec F(\vec \rho)&=-\frac{d}{d \rho}\left(\log(\rho)\right)\,\nabla \left(\rho\right)\\\\ &=-\frac{\hat \rho}{\rho} \end{align}$$

Putting it all together, $(1)$ becomes

$$\begin{align} \Psi&=\int_0^{2\pi}\left(-\frac{\hat \rho}{\rho}\right)\cdot \hat \rho \,\rho\,d\phi\\\\ &=-2\pi \end{align}$$

as was to be shown!