I have a problem set with the following problem:
“Prove that, if $D_1={(x,y) | x^2+y^2 \leq r^2}$, then $\ lim_{r\to 0} \frac{1}{\pi r^2} \iint_{D_1} f(x,y) dA =f(0,0)$.
I am just stuck, as I can imagine it visually, but I have no idea what to do since it’s an arbitrary function. Should I switch to polar? Any help or hints is appreciated.
On
Here is my attempt:
If $f$ is not continuous, then the statement does not necessarily hold. As Kawi pointed out in comments, a counter example is that $f(x,y) = 0$ for all $(x,y)\ne 0$ but $f(x,y)=1$.
So let’s assume that $f(x,y)\in C(\mathbb{R}^2)$. Write the integral in polar form:
$$ \iint_{D_1} f(x,y)\mathrm{d}A=\int_0^r \mathrm{d}R\int_0^{2\pi}f(R\cos\theta, R\sin\theta)R \mathrm{d}\theta$$
Consider $f(R\cos\theta, R\sin\theta) $ as a funtion depending on $\theta$, and apply the first mean value theorem of numerical integration, the preceding formula became
$$\int_0^r \mathrm{d}R\cdot f(R\cos\theta^*, R\sin\theta^*)\int_0^{2\pi}R\mathrm{d}\theta $$
where $\theta^*\in[0,2\pi]$ is a variable depending on $R$. Apply the mean value theorem again, we have
$$\int_0^r f(R\cos\theta^*, R\sin\theta^*)2\pi R \mathrm{d}R = f(R^*\cos\theta^*, R^*\sin\theta^*)\int_0^r 2\pi R \mathrm{d}R $$
where $0\le R^*\le r$. Multiply the integral by $\frac 1 {\pi r^2}$, take the limit $r\to 0$, and from the continuity of $f$ we see that the original formula is equal to $f(0,0)$.
EDIT: Actually, there is a more straightforward proof. Apply the mean value theorem directly to the original integral, and the result follows immediately.
If $f$ is continuous at $(0,0)$ this result is very easy to prove. Note that $\frac 1 {\pi r^{2} } \int\int_D 1 dA=1$. Hence $\frac 1 {\pi r^{2} } \int\int_D f(x,y) dA -f(0,0)=\frac 1 {\pi r^{2} } \int\int_D [f(x,y)-f(0,0)] dA$. If $r$ is small enough then $\frac 1 {\pi r^{2} } |\int\int_D [f(x,y)-f(0,0)] dA | <\epsilon \frac 1 {\pi r^{2} } \int\int_D 1 dA=\epsilon$.