A question about estimating an area of overlapping circles I need to show $$\text{Area}((B-B') \cup (B'-B))\leq 4R\delta$$ where B and B' are circles with the same radius R, B is centered at the origin and B' is centered at $(\delta,0)$. Since $B-B'$ and $B'-B$ have the same area and are disjoint, it suffices to show $$\text{Area}(B-B')\leq 2R\delta$$ I have tried looking at triangles, using the implicit equations for the circles, and using the fact that $$\text{Area}(B-B')=\text{Area}(B)-\text{Area}(B\cap B')$$ I have found that $$\text{Area}(B\cap B')\leq 2R(2R-\delta),\quad \text{Area}(B)\leq4R^2$$ I found the first part by overestimating the area of the intersection by a rectangle. If I subtract these two quantities I reach the result, but does not seem like it is exactly allowed the way these inequalities are set up. This is supposed to be part of a proof that a bounded harmonic function is constant for my Analysis of PDE class. 
 A: The following figure shows the upper right quarter of the figure. If you add the red triangular piece to the area of $\>B'\setminus B\>$ you obtain an area which is obviously $=R\cdot\delta$. It follows that the area we are interested in is $<R\delta$, resp. $<4R\delta$ in total.

A: The required area 
$$A = 4.\left(\int_{-R}^{\frac{\delta}{2}}\sqrt{R^2-t^2}dt - \int_{\frac{\delta}{2}}^{R}\sqrt{R^2-t^2}dt\right)$$
$$A\le 4.\left(\int_{-R}^{\frac{\delta}{2}}Rdt - \int_{\frac{\delta}{2}}^{R} Rdt\right)$$
$$A\le 4.R\left(\frac{\delta}{2}+R - R+\frac{\delta}{2}\right)$$
$$A\le4R\delta$$
A: We have that the area of $(B-B') \cup (B'-B)$ is four times the following 
$$\begin{align}\int_{\delta/2}^{R+\delta}\sqrt{R^2-(t-\delta)^2}dt-\int_{\delta/2}^{R}\sqrt{R^2-t^2}dt
&=
\int_{-\delta/2}^{R}\sqrt{R^2-t^2}dt-\int_{\delta/2}^{R}\sqrt{R^2-t^2}dt\\
&=\int_{-\delta/2}^{\delta/2}\sqrt{R^2-t^2}dt\leq R\delta\end{align}$$
where the circles $B$ and $B'$ are given by  $x^2+y^2=R^2$ and $(x-\delta)^2+y^2=R^2$.
See also the related question Show that $\int^{R+\delta}_{\delta/2} \sqrt{R^2-(x-\delta)^2}\,dx-\int^{R}_{\delta/2} \sqrt{R^2-x^2}\,dx\leq R\delta$, for $0<\delta<R.$
