Computation surface integral $\int_{\partial B_R}g\frac{\partial u}{\partial \nu}\,d\sigma$.

Question

I have the following computation written in my notes. Let $u$ be a function such that, in polar coordinates, $u(r,\theta)=r^7(1/98+1/90\cdot \cos(2\theta))$. Let $g(x,y)=\log\sqrt{(x-1)^2+y^2}$. I want to understand the computation of $$\int_{\partial B_R}g\frac{\partial u}{\partial \nu}\,d\sigma,$$ where $B_R$ is the ball of radius $R$ centred at $0$ and $\nu$ is the outer unit vector of $B_R$. In my notes I have: $$\int_{\partial B_R}g\frac{\partial u}{\partial \nu}\,d\sigma=7R^7\int_0^{2\pi}g(R\cos\theta,R\sin\theta)\left(\frac{1}{98}+\frac{1}{90}\cos(2\theta)\right)\,d\theta. \quad (*)$$ I do not understand that equality. For me, if I have the vector field $F=(P,Q)=g\,\nabla u=(g\,u_x,g\,u_y)$ and $\gamma$ is the curve representing $\partial B_R$ in a counterclockwise sense, then $$\int_{\partial B_R}F\cdot\nu\,d\sigma=\int_{\gamma}P\,dy-Q\,dx=\int_{\gamma}(g\,u_x\,dy-g\,u_y\,dx)=$$ $$=R\int_0^{2\pi}g(R\cos\theta,R\sin\theta)[u_x(R\cos\theta,R\sin\theta)\cos\theta+u_y(R\cos\theta,R\sin\theta)\sin\theta]\,d\theta.$$ Where are the partial derivatives of $u$ in $(*)$? Could you explain this to me in detail?

Answer 1

Given a curve $\gamma$ of class $C^{1}$ in $\mathbb{R}^{2}$ parametrized by a function $h:[a,b]\rightarrow\mathbb{R}^{2}$, the integral of a function $f$ along $\gamma$ is given by $$ \int_{\gamma}f\,d\sigma=\int_{a}^{b}f(h(t))\sqrt{(h_{1}^{\prime}% (t))^{2}+(h_{2}^{\prime}(t))^{2}}dt. $$ In your case, you can take $h(\theta)=(R\cos\theta,R\sin\theta)$ and so $h^{\prime}(\theta)=(-R\sin\theta,R\cos\theta)$ which gives \begin{align*} \int_{\partial B_{R}}f\,d\sigma & =\int_{0}^{2\pi}f(R\cos\theta,R\sin \theta)\sqrt{(-R\sin\theta)^{2}+(R\cos\theta)^{2}}d\theta\\ & =\int_{0}^{2\pi}f(R\cos\theta,R\sin\theta)R\,d\theta. \end{align*} The outer normal $\nu$ to $\partial B_{R}$ at $(x,y)$ is given by $\frac{(x,y)}{\sqrt{x^{2}+y^{2}}}$, so in polar coordinate $\nu(\theta )=(\cos\theta,\sin\theta)$ and so \begin{align*} \frac{\partial u}{\partial\nu}(R\cos\theta,R\sin\theta) & =\nabla u(R\cos\theta,R\sin\theta)\cdot\nu(\theta)\\ & =\partial_{x}u(R\cos\theta,R\sin\theta)\cos\theta+\partial_{y}u(R\cos \theta,R\sin\theta)\sin\theta\\ & =\frac{\partial u}{\partial r}(R\cos\theta,R\sin\theta)=7R^{6}\left( \frac{1}{98}+\frac{1}{90}\cos2\theta\right) . \end{align*} It follows that \begin{align*} \int_{\partial B_{R}}g\frac{\partial u}{\partial\nu}\,d\theta & =\int% _{0}^{2\pi}g(R\cos\theta,R\sin\theta)\frac{\partial u}{\partial r}(R\cos \theta,R\sin\theta)R\,d\theta\\ & =\int_{0}^{2\pi}g(R\cos\theta,R\sin\theta)7R^{6}\left( \frac{1}{98}% +\frac{1}{90}\cos2\theta\right) R\,d\theta, \end{align*} which is what you have in your notes.

Answer 2

In polar coordinates the gradient is $$\nabla u=\hat{r}\frac{\partial u}{\partial r} + \hat{\theta}\frac{1}{r}\frac{\partial u}{\partial\theta}$$ Dotting with the outward normal $\hat{r}$ yields $$\hat{r}\cdot\nabla u=\frac{\partial u}{\partial r}$$ Likewise, in polar coordinates, the differential of a circular arc is $$d\sigma = r\,d\theta$$

Substituting these 2 results into the integral $$\eqalign{ \oint g\,\frac{\partial u}{\partial\nu}\,d\sigma &= \oint g\,\frac{\partial u}{\partial r}\,\,r\,d\theta \cr &= \oint g\,7\,r^6\,\Big(\frac{1}{98}+\frac{\cos(2\theta)}{90}\Big)\,\,r\,d\theta \cr &= \oint g\,7\,r^7\,\Big(\frac{1}{98}+\frac{\cos(2\theta)}{90}\Big)\,\,d\theta \cr }$$ The path of this integral is the surface of the ball, where $r$ is a constant. The effect of this is to evaluate any $r$-dependent functions at the boundary value, e.g. $\,f(r)$ becomes $f(R)$.

Pulling the constant terms out of the integral, we are left with the result from your notes $$\eqalign{ \oint g\,\frac{\partial u}{\partial\nu}\,d\sigma &= 7\,R^7\,\int_0^{2\pi} \Big(\frac{1}{98}+\frac{\cos(2\theta)}{90}\Big)\,g\,d\theta \cr \cr }$$