Double Integral and PDE

Stokes' Theorem in the plane tells us that, for smooth functions $A,B$ $\iint_D (\partial_xA + \partial_yB)\,dx\,dy \; = \; \oint_C(A\,dy - B\,dx)$ where $C$ is the positively oriented unit circle. Thus $\begin{aligned} \iint_D \Big(\partial_x(r^2\partial_xf) + \partial_y(r^2\partial_yf)\Big]\,dx\,dy & = \; \oint_Cr^2 (\partial_xf\,dy - \partial_yf\,dx) \\ & = \; \oint_C (\partial_xf\,dy - \partial_yf\,dx) \\ & = \; \iint_D \Big(\partial_x(\partial_xf) + \partial_y(\partial_yf)\Big)\,dx\,dy \; = \; \iint_D \nabla^2f\,dx\,dy \; = \; \iint_D \,dx\,dy \end{aligned}$ so that $\iint_D\Big(r^2\nabla^2f + 2\big(x\partial_xf + y\partial_yf\big)\Big)\,dx\,dy \; =\; \iint_D\,dx\,dy$ and hence $\iint_D \big(x\partial_xf + y\partial_yf\big)\,dx\,dy \; = \; \tfrac12\iint_D(1-r^2)\,dx\,dy \; = \; \pi\int_0^1 (1-r^2)r\,dr \; = \; \boxed{\tfrac14\pi}$

Note: My assumed form for the solution is somewhat lacking in generality, as @Mark Hennings has pointed out in his solution

Taking double derivatives and ending up with constants means that we're talking about quadratics. The general form should be:

$f = A x^2 + B x + C y^2 + D y + E \\ 2 A + 2 C = 1$

Evaluating the integrand:

$x \, \frac{\partial{f}}{\partial{x}} + y \, \frac{\partial{f}}{\partial{y}} = x (2 A x + B) + y (2 C y + D) \\ = 2 A x^2 + 2 C y^2 + B x + D y \\ = 2 A \, r^2 \, cos^2 \theta + 2 C \, r^2 \, sin^2 \theta + B r \, cos \theta + D r \, sin \theta$

Integrating sine or cosine over the unit disk results in zero overall, so the lower order terms drop out, and we're effectively left with:

$x \, \frac{\partial{f}}{\partial{x}} + y \, \frac{\partial{f}}{\partial{y}} = 2 A \, r^2 \, cos^2 \theta + 2 C \, r^2 \, sin^2 \theta$

The integral becomes:

$\int_0^{2 \pi} \int_0^1 (2 A \, r^2 \, cos^2 \theta + 2 C \, r^2 \, sin^2 \theta ) r \, d r \, d \theta \\ = \int_0^{2 \pi} \int_0^1 (2 A \, r^3 \, cos^2 \theta + 2 C \, r^3 \, sin^2 \theta ) d r \, d \theta$

Substituting in for C in terms of A yields:

$\int_0^{2 \pi} \int_0^1 (2 A \, r^3 \, cos^2 \theta + (1 - 2 A) \, r^3 \, sin^2 \theta ) d r \, d \theta$

The integrals of sine squared and cosine squared over the unit disk are both equal, so these components now cancel, and we're left with:

$\int_0^{2 \pi} \int_0^1 r^3 \, sin^2 \theta \, d r \, d \theta = \frac{\pi}{4}$