Problem: Let $D=\{x^{2}+y^{2} \leq1\}$ and $ f_{xx}+f_{yy}=1$ for $(x,y)\in D$. Prove that $$\iint_{D}(xf_{x}+yf_{y})dxdy=\frac{\pi}{4}$$
My solution: $\iint_{D}(xf_{x}+yf_{y})dxdy=\int_{0}^{1}dr\int_{0}^{2\pi}[rcos\theta f_{x}(rcos\theta,rsin\theta)+rsin\theta f_{y}(rcos\theta,rsin\theta)]rd\theta\=\int_{0}^{1}dr[\int_{L_{r}}(xf_{x}+yf_{y})ds]=\int_{0}^{1}dr[\int_{L_{r}}(f_{x}dy-f_{y}dx)ds]=\int_{0}^{1}dr[\iint_{D_{r}}(f_{xx}+f_{yy}dx)dS]=\int_{0}^{1}dr[\pi r^{2}]=\frac{\pi}{3}\quad???$
where $L_{r} $ represents a circle with radius $r$ in $D$ and $D_{r}$ is the interior of $L_{r}$.
However, I don't know what's wrong with my solution...Could anyone help me? Please..
In the following, I take
$D(\rho) = \{ (x, y) \in \Bbb R^2 \mid x^2 + y^2 \le \rho^2 \}; \tag 0$
Let $r$ and $\theta$ denote the usual polar coordinates and set
$\vec r = (x, y), \tag 1$
and
$\vec n = (\cos \theta, \sin \theta); \tag 2$
then
$\vec r = (x, y) = (r\cos \theta, r \sin \theta) = r \vec n; \tag 3$
thus
$xf_x(x, y) + yf_y(x, y) = \vec r \cdot \nabla f(x, y) = r \vec n \cdot \nabla f(x, y); \tag 4$
we have, expressing the integral in polars:
$\displaystyle \int \int_{D(\rho)} (x f_x(x, y) + y f_y(x, y)) \; dx dy = \int \int_{D(\rho)} (r \vec n \cdot \nabla f(x, y)) \; dx dy$ $= \displaystyle \int \int_{D(\rho)} (r \vec n \cdot \nabla f(r\cos \theta, r \sin \theta)) \; r dr d\theta = \displaystyle \int_0^\rho \left [ \int_0^{2\pi} (r \vec n \cdot \nabla f(r\cos \theta, r \sin \theta)) \; r d\theta \right ] \; dr$ $= \displaystyle \int_0^\rho \left [ r \int_0^{2\pi} (\vec n \cdot \nabla f(r\cos \theta, r \sin \theta)) \; r d\theta \right ] \; dr; \tag 5$
we evaluate the inner integral on the right of (5), taking note of the fact that it is taken over the circle of radius $r$ centered at $(0, 0)$ in the $xy$-plane, which is the boundary of the disk $D(r)$; the line element on this circle is in fact $rd\theta$, and its unit normal is $\vec n$; now, by the divergence theorem,
$\displaystyle \int_0^{2\pi} (\vec n \cdot \nabla f(r\cos \theta, r \sin \theta)) \; r d\theta = \int \int_{D(r)} \nabla \cdot \nabla f(s\cos \theta, s \sin \theta) \; s ds d\theta$ $= \displaystyle \int \int_{D(r)} \nabla^2 f(s\cos \theta, s \sin \theta) \; s ds d\theta; \tag 6$
furthermore, since
$\nabla^2 f(s\cos \theta, s \sin \theta) = 1 \tag 7$
for $s \le 1$, we have, for $r \le 1$,
$\displaystyle \int \int_{D(r)} \nabla^2 f(s\cos \theta, s \sin \theta) \; s ds d\theta = \int \int_{D(r)} 1 \; s ds d\theta = \pi r^2; \tag 8$
if we combine (6) and (8) and insert the result into (5) we find
$\displaystyle \int \int_{D(\rho)} (x f_x + y f_y) \; dx dy = \int_0^\rho r (\pi r^2) \; dr = \int_0^\rho \pi r^3 \; dr = \dfrac{\pi \rho^4}{4}; \tag 9$
taking $\rho = 1$ yields the requisite result.
I'm not exactly sure where our OP user450201 veered off of the correct path, but my suspicions are centered on
$\displaystyle \int_{0}^{1}dr \left [\int_{L_{r}}(xf_{x}+yf_{y})ds \right ]=\int_{0}^{1}dr \left [\int_{L_{r}}(f_{x}dy-f_{y}dx)ds \right], \tag{10}$
which occurs in the equation he presents under "my solution"; I don't fully comprehend why $xf_{x}+yf_{y}$ is replaced by $f_{x}dy-f_{y}dx$; this doesn't really make sense to me since the number of "differentials" $dx, dy, ds$ is different from side-to-side, whereas each inner integral is taken over $L_r$. As I said, though I'm not positive where the error lies this is where my suspicions are primarily directed.