# how to show $\iint\limits_D(x\frac{\partial f }{\partial x}+ y \frac{\partial f}{\partial y}) \, dx \, dy= \frac{\pi}{2e}$?

Assume $$f\in C^2(D)$$, where $$D=\{(x,y)\in \mathbb R^2: x^2+ y^2 \le 1\}$$, if $$\frac{\partial^2 f }{\partial x^2} + \frac{\partial^2 f }{\partial y^2} =e^{-x^2 -y^2}$$ how do I show $$\iint\limits_D(x\frac{\partial f }{\partial x}+ y \frac{\partial f}{\partial y})\,dx\,dy= \frac{\pi}{2e}$$

I think I should use Green's Theorem, but I wasn't able to make progress.

Notice that the integral can be rewritten as

$$\iint_D (x,y)\cdot \nabla f dA = \iint_D \nabla \left( \frac{x^2 + y^2}{2}\right) \cdot \nabla f dA$$

then use integration by parts (really the divergence theorem in higher dimensions) with $$u=\nabla f$$ and $$dv = (x,y)$$:

$$\iint_D \nabla \left( \frac{x^2 + y^2}{2}\right) \cdot \nabla f dA = \int_{\partial D} \left(\frac{x^2+y^2}{2}\right) \nabla f \cdot \mathbf{n} ds - \iint_D \left(\frac{x^2+y^2}{2}\right) \Delta f dA$$

$$= \frac{1}{2} \int_{\partial D} \nabla f \cdot \mathbf{n} ds - \iint_D \left(\frac{x^2+y^2}{2}\right) \Delta f dA$$

We can use the divergence theorem on the first integral

$$\frac{1}{2}\iint_{D} \Delta f dA - \iint_D \left(\frac{x^2+y^2}{2}\right) \Delta f dA$$

then plug in $$\Delta f = e^{-x^2-y^2}$$ and convert to polar coordinates

$$\frac{1}{2}\int_0^1 \int_0^{2\pi}re^{-r^2}d\theta dr-\frac{1}{2} \int_0^1 \int_0^{2\pi} r^3e^{-r^2}d\theta dr=\pi\int_0^1 re^{-r^2}dr-\pi \int_0^1 r^3e^{-r^2}dr = \frac{\pi}{2e}$$