Area of a leaf

The area between two overlapping circles is $A = r^2\cos^{-1}(\frac{d^2 + r^2 - R^2}{2dr}) + R^2\cos^{-1}(\frac{d^2 + R^2 - r^2}{2dR}) - \frac{1}{2}\sqrt{(-d + r + R)(d + r - R)(d - r + R)(d + r + R)}$ where $d$ is the distance between the centers of the two circles and $R$ and $r$ are the radii of the two circles.

In this case, $R = 2$ , $r = 1$ , and $d = 2\sqrt{2}$ and so the area $A \approx \boxed{0.108}$ .

We will tackle this with coordinate geometry and calculus. Let $A$ be the origin, $AB$ the positive x-axis, and $AD$ the negative y-axis. The equation for the circle with center $A$ is $y = -\sqrt{1-x^{2}}$ . The equation for the circle with center $C$ is $y = \sqrt{4-(x-2)^{2}}-2$ . First we must find where they intersect by equating the two equations for the circles. $\begin{aligned} -\sqrt{1-x^{2}} + 2 & = \sqrt{4-(x-2)^{2}} \\ 1-x^{2} - 4\sqrt{1-x^{2}}+4 & = -x^{2} + 4x \\ -4\sqrt{1-x^{2}} & = 4x-5 \\ 32x^{2} -40x + 9 & = 0 \end{aligned}$ Solving this quadratic results in our two x-coordinates of intersection, which are $\frac{5 + \sqrt{7}}{8}$ and $\frac{5 - \sqrt{7}}{8}$ . Now we set up the integral for solving for the area of the leaf where $I$ is the desired integral. $\begin{aligned} I & = \int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \sqrt{4- (x-2)^{2}}-2 + \sqrt{1-x^{2}} \;\mathrm{d}x \\ & = \int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \sqrt{4- (x-2)^{2}}\;\mathrm{d}x -2\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}}\;\mathrm{d}x + \int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \sqrt{1-x^{2}}\; \mathrm{d}x \end{aligned}$ Now we will focus on solving the first portion of the integral. $\begin{aligned} &\large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \sqrt{4- (x-2)^{2}}\;\mathrm{d}x \qquad u = x-2,\; \mathrm{d}u = \mathrm{d}x \\ = & \large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \sqrt{4- u^{2}}\;\mathrm{d}u \qquad u = 2\sin(v), \; \mathrm{d}u = 2\cos(v)\; \mathrm{d}v \\ = & 4\large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \cos(v)\sqrt{1-\sin^{2}(v)} \; \mathrm{d}v \\ = & 4\large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \cos^{2}(v) \; \mathrm{d}v \\ = & 2\large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}}1 + \cos(2v) \; \mathrm{d}v \\ = & 2(v + \frac{1}{2}\sin(2v)) \\ = & 2\arcsin(\frac{u}{2}) + \sin(2\arcsin(\frac{u}{2})) \\ = & 2\arcsin(\frac{x-2}{2}) + \sin(2\arcsin(\frac{x-2}{2})) \Bigg\rvert_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \\ \approx & 0.944 \end{aligned}$ The second, or trivial, portion of the integral is equal to $-\frac{\sqrt{7}}{2} = -1.323$ . We will now focus on the third portion of the integral. $\begin{aligned} & \large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \sqrt{1-x^{2}}\; \mathrm{d}x \qquad x = \sin(u), \; \mathrm{d}x = \cos(u) \;\mathrm{d}u \\ = & \large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \cos^{2}(u) \; \mathrm{d}u \\ = & \frac{1}{2}\large\int_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} 1 + \cos(2u) \; \mathrm{d}u \\ = & \frac{1}{2}u + \frac{1}{4}\sin(2u) \\ = & \frac{1}{2}\arcsin(x) + \frac{1}{4}\sin(2\arcsin(x)) \Bigg\rvert_{\frac{5 - \sqrt{7}}{8}} ^{\frac{5 + \sqrt{7}}{8}} \\ \approx & 0.487 \end{aligned}$ Finally, adding all portions of the original integral, $I$ , we get $0.944 + 0.487 - 1.323 = \boxed{0.108}$

After connecting the points as above in the figure, we know that $AC=2\sqrt2$ since the side length of the square is 2. Also, $AX=1$ and $CX=2$ (radii). Now let $AZ=x$ , making $CZ=2\sqrt2-x$ and applying the Pythagorean theorem, we get $XZ=\sqrt{(AX)^2-(AZ)^2}=\sqrt{(CX)^2-(CZ)^2}$ and plugging in values $\sqrt{(1)^2-(x)^2}=\sqrt{(2)^2-(2\sqrt2-x)^2}$ and solving further $1-x^2=4-(8-4x\sqrt{2}+x^2)$ $1-x^2=4-8+4x\sqrt{2}-x^2$ $1=-4+4x\sqrt{2}$ $\frac{5}{4\sqrt{2}}=x$ which simplifies to $x=AZ=\frac{5\sqrt2}{8}$ and $CZ=\frac{11\sqrt2}{8}$ .

$1)$ For the circle with radius 1, we first find the area of $\widehat {AXY}$ and subtract the area of $\triangle AXY$ from it. To find the area of the sector, first we calculate its angle $\cos {\angle XAZ}= \frac{AZ}{AX}$ or $\angle {XAZ}= \cos^{-1} \left(\frac{\frac{5\sqrt{2}}{8}}{1} \right)\approx27.88^\circ$ and multiply it by two to find the whole angle $\angle XAZ \cdot 2=\angle XAY\approx55.77^\circ$

Now we'll find the area of the sector as $\frac{55.77^\circ}{360^\circ}\cdot \pi \cdot (1)^2\approx 0.4867$

To find the area of the triangle, first we know $AX=1$ and $AZ=\frac{5\sqrt2}{8}$ and we find $XZ$ using Pythagorean Theorem $XZ=\sqrt{AX^2-AZ^2}$ $XZ=\sqrt{1^2-\left({\frac{5\sqrt2}{8}} \right)^2}$ as $XZ=\frac{\sqrt{14}}{8}$ hence the area of $\triangle AXY = AZ \cdot XZ$ $\triangle AXY=\frac{5\sqrt2}{8}\ \cdot \frac{\sqrt{14}}{8}$ $\triangle AXY=\frac{5\sqrt{28}}{64}\approx{0.4134}$ Now the area of the red portion of the green leaf is just $0.4867-0.4134=0.0733 \implies A$

$2)$ For the circle with radius 2, we first find the area of $\widehat {CXY}$ and subtract the area of $\triangle CXY$ from it as we did above. To find the area of the sector, we calculate $\cos {\angle XCZ}= \frac{CZ}{CX}$ $\angle {XCZ}= \cos^{-1} \left(\frac{\frac{11\sqrt{2}}{8}}{2} \right)$ $\cos^{-1} \left({\frac{11\sqrt2}{16}}\right)\approx13.52^\circ$ and multiply it by two to find the whole angle $\angle XCZ \cdot 2=\angle XCY\approx 27.04^\circ$ Now similar to what we did above, we are going to find the area of the sector by $\frac{27.04^\circ}{360^\circ}\cdot \pi \cdot (2)^2\approx 0.9442$ To find the area of the triangle we will do $\triangle CXY = CZ \cdot XZ$ $\triangle CXY=\frac{11\sqrt2}{8}\ \cdot \frac{\sqrt{14}}{8}$ $\triangle CXY=\frac{11\sqrt{28}}{64}\approx{0.9095}$ The area of the green portion of the leaf is just $0.9442-0.9095=0.0347 \implies B$

The final answer is $A+B=0.0733+0.0347\approx \boxed{0.108}$ .