The key here is that the area of each "leaf" is the area of the quarter circle minus the area of the triangle defined by the secant and radii. In the illustration, I added labels A-E to points on the circles.
Attachment:
overlapping_arcs.png [ 48.4 KiB | Viewed 411 times ]
First, let's find the area of the bigger leaf. The circle's center is C with a radius of 5. Thus, the area of the quarter circle is \(\frac{25\pi}{4}\).
Sides AC and BC are radii, so they are both 5. The area of triangle ABC is \(\frac{25}{2}\).
To isolate just the shaded arc's area, we subtract the area of the right triangle ABC. That gets us the area of half of the bigger leaf, \(\frac{25\pi}{4} - \frac{25}{2}\)
Now we double that to get the area of the full leaf, \(\frac{25\pi}{2} - 25\).
Now, we can do the same process to find the area of the smaller leaf. The circle's center is E with a radius of 3, and the area of quarter circle is \(\frac{9\pi}{4}\). The area of triangle DBE is \(\frac{9}{2}\).
The area of the smaller leaf is \(\frac{9\pi}{2} -9\).
Finally, we subtract the area of the smaller leaf from the bigger leaf to find the area of the shaded region.
\((\frac{25\pi}{2} - 25) - (\frac{9\pi}{2} -9) = 8 \pi-16\)