Show a limit of a bounded function is 0 then solving the integral $B=\{x^2+y^2\le1\}$, & for all $\delta>0$, there is $B_\delta=\{x^2+y^2\le\delta\}$.  $f$ is a continuous function and $\|\nabla f\|\le1$ on $B$, and suppose $\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=e^{x^2+y^2}$ on $B$ not including $0$.
Use boundedness of $\nabla f$ to show $\lim_{\delta\to 0}\int_{\partial B_\delta}f_y \, dx-f_x \, dy=0$, then solve $\lim_{\delta\to 0}\int_{\partial B}f_y \, dx-f_x \, dy=0$
This is the problem and I'm not sure what to do.  I've been trying to use the Green's Theorem to solve this, and by changing to polar coodinates and I'm getting nowhere.  Any hints or help would be appreciated.  Thanks in advance
 A: We can't use Green's theorem yet at we do not know that the partials exist at zero. However, the direct approach is straightforward. Assuming $|\nabla f|\leq M$ and using the parameterisation $\gamma(t)=\delta(\cos t,\sin t)$, we have
$$\left|\int_{\partial B_\delta}f_y\ dx-f_x dy\right|=\delta\left|\int_0^{2\pi}-(f_y\circ\gamma)(t)\sin t-(f_x\circ\gamma)(t)\cos t\ dt\right|\leq4\pi M\delta\to0$$
as $\delta\to0$. Now assuming the second order partials are continuous where they exist, let $A_\delta=B\setminus B_\delta$ and we can use Green's Theorem to get
$$\int_{\partial B}f_y\ dx-f_x dy=\left(\int_{\partial A}+\int_{\partial B_\delta}\right)f_y\ dx-f_x dy=-\int_{A_\delta} e^{x^2+y^2}\ dA+\left(\int_{\partial B_\delta}f_y\ dx-f_x dy\right).$$
The second integral tends to zero as $\delta\to0$ as discussed above. Hence
$$\int_{\partial B}f_y\ dx-f_x dy=-\lim_{\delta\to0}\int_\delta^1\int_0^{2\pi}re^{r^2}\ d\theta dr=-\pi\int_0^12re^{r^2}\ dr=\pi(1-e).$$
A: Set $[0,2\pi]\ni t\mapsto \gamma_\delta(t)=(x(t),y(t))\in \partial B_\delta$ with $x(t)=\delta\cos(t)$ and $y(t)=\delta\sin(t)$.
\begin{align}
\lim_{\delta\to 0}\left|\int_{\partial B_\delta}f_y \, dx-f_x \, dy\right|
=
&
\lim_{\delta\to 0}\left|\int_{\gamma_\delta}f_y(x(t),y(t))\cdot y'(t) \, dt-f_x(x(t),y(t))\cdot x'(t) \, dt\right|
\\
=
&
\lim_{\delta\to 0}\left|\int_{0}^{2\pi}
\Big\langle \nabla f\big(x(t),y(t)\big)\,,\,\big(y'(t), x'(t)\big)\Big\rangle \, dt\right|
\\
\leq
&
\lim_{\delta\to 0}\int_{0}^{2\pi}
\left|\Big\langle \nabla f\big(x(t),y(t)\big)\,,\,\big(y'(t), x'(t)\big)\Big\rangle \right| \, dt
\\
\leq
&
\lim_{\delta\to 0}\int_{0}^{2\pi}
\left\| \nabla f\big(x(t),y(t)\big)\right\|\left\|\big(y'(t), x'(t)\big) \right\| \, dt
\\
\leq
&
\lim_{\delta\to 0}\int_{0}^{2\pi}
1\cdot\delta \, dt
\\
\leq 
&
\lim_{\delta \to 0}2\pi\delta=0
\end{align}
