Let f(x, y) be twice continuously differentiable on the unit disc D = {x^2 + y^2 <= 1} with f_xx + f_yy = x^2 + y^2. Evaluate the double integral over D of (x/sqrt(x^2+y^2)) f_x + (y/sqrt(x^2+y^2)) f_y.

Step-by-step solution

1. 1

   Recognise the integrand as a radial derivative. In polar coordinates $x=r\cos\theta$, $y=r\sin\theta$, so $\tfrac{x}{\sqrt{x^2+y^2}}=\cos\theta$ and $\tfrac{y}{\sqrt{x^2+y^2}}=\sin\theta$. Therefore

   $\cos\theta\,f_x+\sin\theta\,f_y=\frac{\partial f}{\partial r}$

   and the problem becomes $I=\displaystyle\iint_{D}\frac{\partial f}{\partial r}\,dA$. Note that $f$ itself is never given — so the answer must be forced by the Laplacian condition alone.

2. 2

   Introduce the angular average. Define

   $g(r)=\frac{1}{2\pi}\int_{0}^{2\pi}f(r,\theta)\,d\theta$

   In polar form the Laplacian is $\nabla^{2}f=\tfrac1r\big(rf_r\big)_r+\tfrac{1}{r^{2}}f_{\theta\theta}$. Averaging over $\theta$ kills the $f_{\theta\theta}$ term (it integrates to zero over a full period) and leaves

   $\frac1r\big(rg'(r)\big)'=r^{2}$

   because the right-hand side $x^2+y^2=r^2$ does not depend on $\theta$.

3. 3

   Solve the resulting ODE. Multiplying by $r$ and integrating,

   $\big(rg'\big)'=r^{3}\quad\Longrightarrow\quad rg'=\frac{r^{4}}{4}+C$

   Smoothness of $f$ at the origin forces $rg'\to 0$ as $r\to 0$, hence $C=0$ and

   $g'(r)=\frac{r^{3}}{4}$

   This is the key point: the arbitrary harmonic part of $f$ has a constant angular average (mean value property), so it contributes nothing to $g'$.

4. 4

   Convert the integral into g'. With $dA=r\,dr\,d\theta$,

   $I=\int_{0}^{1}\!\!\int_{0}^{2\pi}\frac{\partial f}{\partial r}\,r\,d\theta\,dr=\int_{0}^{1}r\left(\int_{0}^{2\pi}f_r\,d\theta\right)dr=\int_{0}^{1}r\cdot 2\pi g'(r)\,dr$

5. 5

   Evaluate.

   $I=2\pi\int_{0}^{1}r\cdot\frac{r^{3}}{4}\,dr=\frac{\pi}{2}\int_{0}^{1}r^{4}\,dr=\frac{\pi}{2}\cdot\frac{1}{5}=\frac{\pi}{10}$

   $\boxed{\dfrac{\pi}{10}}$

6. 6

   Check with an explicit solution. Take $f=\tfrac{r^{4}}{16}=\tfrac{(x^2+y^2)^2}{16}$. Then $\nabla^{2}f=\tfrac1r\left(r\cdot\tfrac{4r^{3}}{16}\right)'=\tfrac1r\left(\tfrac{r^{4}}{4}\right)'=r^{2}$ ✓, and $f_r=\tfrac{r^{3}}{4}$, so

   $I=\int_{0}^{2\pi}\!\!\int_{0}^{1}\frac{r^{3}}{4}\,r\,dr\,d\theta=2\pi\cdot\frac{1}{20}=\frac{\pi}{10}$

   matching the general argument. Adding any harmonic function to this $f$ leaves both the hypothesis and the answer unchanged.