Evaluation of $\iint_{D}\left(x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y$ Suppose that $f(x,y)$ is defined on $D=\{(x,y)\mid x^2+y^2\le1\}$ and has continuous second-order partial derivatives in $D$. If
$$\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\mathrm{e}^{-(x^2+y^2)}$$
then find $\iint_{D}\left(x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y$.
I have tried the polar coordinate form, obtained
$$
\frac{\partial^2f}{\partial x^2}  =\cos^2\theta\frac{\partial^2f}{\partial r^2}-\frac{2\sin\theta\cos\theta}{r}\cdot\frac{\partial^2f}{\partial r\partial \theta}+\frac{\sin^2\theta}{r^2}\frac{\partial^2f}{\partial\theta^2}+\frac{\sin^2\theta}{r}\frac{\partial f}{\partial r}+\frac{2\sin\theta\cos\theta}{r^2}\frac{\partial f}{\partial\theta}\\
\frac{\partial^2f}{\partial y^2}  =\sin^2\theta\frac{\partial^2f}{\partial r^2}+\frac{2\sin\theta\cos\theta}{r}\cdot\frac{\partial^2f}{\partial r\partial \theta}+\frac{\cos^2\theta}{r^2}\frac{\partial^2f}{\partial\theta^2}+\frac{\cos^2\theta}{r}\frac{\partial f}{\partial r}-\frac{2\sin\theta\cos\theta}{r^2}\frac{\partial f}{\partial\theta}\\
\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2} =\frac{\partial^2f}{\partial r^2}+\frac1r\frac{\partial f}{\partial r}+\frac1{r^2}\frac{\partial^2f}{\partial\theta^2}=\mathrm{e}^{-r^2}
$$
and 
\begin{equation*}
x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=r\frac{\partial f}{\partial r}
\end{equation*}
so, the double integral becomes
$$\iint_D r\frac{\partial f}{\partial r}r\mathrm{d}r\mathrm{d}\theta$$
how to solve this? 
 A: First, note this identity
$$\psi \nabla ^2 \phi=\nabla \cdot (\psi \nabla \phi)-\nabla \psi \cdot \nabla \phi \tag{1}$$
Now, choose $\phi=f$ and $\psi=\frac{1}{2}{\bf{x}}\cdot{\bf{x}}=\frac{1}{2}|{\bf{x}}|^2$ where $\bf{x}$ is the position vector. Then equation $(1)$ becomes
$$\frac{1}{2}|{\bf{x}}|^2 \nabla ^2 f=\nabla \cdot (\frac{1}{2}|{\bf{x}}|^2 \nabla f)-{\bf{x}} \cdot \nabla f \ \\
{\bf{x}} \cdot \nabla f=\nabla \cdot (\frac{1}{2}|{\bf{x}}|^2 \nabla f)- \frac{1}{2}|{\bf{x}}|^2 \nabla ^2 f \\
\tag{2}$$
Next, we integrate over the domain $D$, use the divergence theorem, and $\nabla^2f=e^{-|{\bf{x}}|^2}$ to get
$$\int_{D}{\bf{x}} \cdot \nabla f d_{A}=\int_{\partial D} \frac{1}{2}|{\bf{x}}|^2 \nabla f \cdot {\bf{n}} d_{L} - \int_{D} \frac{1}{2}|{\bf{x}}|^2 \text{e}^{-|{\bf{x}}|^2}d_{A}\tag{3}$$
and noting that $|{\bf{x}}|=1$ on $\partial D$ will lead to
$$2\int_{D}{\bf{x}} \cdot \nabla fd_{A}=\int_{\partial D}  \nabla f \cdot {\bf{n}} d_{L} - \int_{D} |{\bf{x}}|^2 \text{e}^{-|{\bf{x}}|^2}d_{A}\tag{4}$$
Again, use the Divergence theorem for the first integral on the RHS of $(4)$
$$\begin{align}
2\int_{D}{\bf{x}} \cdot \nabla fd_{A} &= \int_{D}  \nabla \cdot \nabla f d_{A} - \int_{D} |{\bf{x}}|^2 \text{e}^{-|{\bf{x}}|^2}d_{A} \\
2\int_{D}{\bf{x}} \cdot \nabla fd_{A} &= \int_{D}  \nabla^2 f d_{A} - \int_{D} |{\bf{x}}|^2 \text{e}^{-|{\bf{x}}|^2}d_{A} \\
2\int_{D}{\bf{x}} \cdot \nabla fd_{A} &= \int_{D}  \text{e}^{-|{\bf{x}}|^2} d_{A} - \int_{D} |{\bf{x}}|^2 \text{e}^{-|{\bf{x}}|^2}d_{A}\\
\tag{5}
\end{align}$$
and the final result will be
$$\boxed{I=\int_{D}{\bf{x}} \cdot \nabla f= \frac{1}{2}\int_{D} \left(1-|{\bf{x}}|^2\right) \text{e}^{-|{\bf{x}}|^2}d_{A}}\tag{6}$$
You can evaluate the integral in RHS of $(6)$ using polar coordinates
$$\begin{align}
I &= \frac{1}{2}\int_{0}^{2\pi}\int_{0}^{1} \left(1-r^2\right) e^{-r^2}rdrd\theta \\
&= \pi \int_{0}^{1} r \left(1-r^2\right) e^{-r^2} dr \\
&=\color{blue}{\boxed{\frac{\pi}{2\text{e}}}}\tag{7}
\end{align}$$
A: @H.R. posted a solid solution that used vector analysis.  I thought it would be useful to some readers to see a solution that relies on scalar analysis only.  To that end we proceed.
Let $D$ be the unit disk, centered at the origin, and let $I$ be the integral of interest defined as
$$\begin{align}
I&=\int_{D}\left(x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial x}\right)\,dS\\\\
&=\int_{-1}^1 \left(\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}x\frac{\partial f}{\partial x} \,dx\right)\,dy+\int_{-1}^1 \left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}y\frac{\partial f}{\partial y} \,dy\right)\,dx \tag 1
\end{align}$$
We integrate by parts the inner integral of the first integral on the right-hand side of $(1)$.  We let $u= \frac{\partial f(x,y)}{\partial x}$ and $v=\frac12\left(x^2+y^2\right)$ to reveal
$$\begin{align}
\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}x\frac{\partial f}{\partial x} \,dx&=\left .\left(\frac12\left(x^2+y^2\right)\frac{\partial f}{\partial x}\right)\right|_{x=-\sqrt{1-y^2}}^{\sqrt{1-y^2}}-\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \left(\frac12\left(x^2+y^2\right)\frac{\partial^2 f}{\partial x^2}\right)dx\\\\
&=\frac12\left .\left(\frac{\partial f}{\partial x}\right)\right|_{x=-\sqrt{1-y^2}}^{\sqrt{1-y^2}}-\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \left(\frac12\left(x^2+y^2\right)\frac{\partial^2 f}{\partial x^2}\right)dx \tag 2
\end{align}$$
An analogous development of the second integral on the right-hand side of $(2)$ reveals
$$\begin{align}
\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}y\frac{\partial f}{\partial y} \,dy&=\frac12\left .\left(\frac{\partial f}{\partial y}\right)\right|_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}-\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(\frac12\left(x^2+y^2\right)\frac{\partial^2 f}{\partial y^2}\right)dy \tag 3
\end{align}$$
Combining the results of $(2)$ and $(3)$ yields
$$\begin{align}
I&=\frac12\oint_{x^2+y^2=1}\left(\frac{\partial f}{\partial x}dy-\frac{\partial f}{\partial y}dx\right)-\frac12\int_D \left(x^2+y^2\right)\left(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\right)\,dS\\\\
&=\frac12 \int_D \left(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\right)\,dS-\frac12\int_D \left(x^2+y^2\right)\left(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\right)\,dS\\\\
&=\frac12\int_0^{2\pi}\int_0^1 \left(1-\rho^2\right)e^{-\rho^2}\,\rho\,d\rho\,d\phi\\\\
&=\frac12 \pi e^{-1}
\end{align}$$
