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.
 A: 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!
