Find the shaded area, which is enclosed by a square, semi-circle and a quarter of circle.
An alternative is to use calculus. The quarter circle has radius $\displaystyle \displaystyle \begin{align*} 7\,\textrm{cm} \end{align*}$ and is centred at $\displaystyle \displaystyle \begin{align*} (0, 0) \end{align*}$, so has the equation $\displaystyle \displaystyle \begin{align*} x^2 + y^2 = 49 \end{align*}$.
The semicircle has radius $\displaystyle \displaystyle \begin{align*} \frac{7}{2}\,\textrm{cm} \end{align*}$ and is centred at $\displaystyle \displaystyle \begin{align*} \left(7, \frac{7}{2}\right) \end{align*}$, so has the equation $\displaystyle \displaystyle \begin{align*} \left(x - 7\right)^2 + \left(y - \frac{7}{2}\right)^2 = \frac{49}{4} \end{align*}$.
So if we are trying to find the area of the enclosed region, we use $\displaystyle \displaystyle \begin{align*} \int{\int_R{}\,dA} \end{align*}$, where $\displaystyle \displaystyle \begin{align*} A \end{align*}$ represents an infinitesimal element of area. Since you are dealing with circular bounds, it will be easiest to convert to polars, so each element of area can be written as $\displaystyle \displaystyle \begin{align*} r\,dr\,d\theta \end{align*}$.
Now as for the bounds, it can clearly be seen that since the semicircle has a radius of $\displaystyle \displaystyle \begin{align*} \frac{7}{2}\,\textrm{cm} \end{align*}$ and the quarter circle has a radius of $\displaystyle \displaystyle \begin{align*} 7\,\textrm{cm} \end{align*}$, the region of integration extends over $\displaystyle \displaystyle \begin{align*} \frac{7}{2} \leq r \leq 7 \end{align*}$.
If we solve the two equations simultanously, we can see that they intersect at $\displaystyle \displaystyle \begin{align*} (x, y) = \left(\frac{21}{5}, \frac{28}{5}\right) \end{align*}$. This means at the point of intersection:
$\displaystyle \displaystyle \begin{align*} x &= r\cos{\theta} \\ \frac{21}{5} &= 7\cos{\theta} \\ \frac{3}{5} &= \cos{\theta} \\ \theta &= \arccos{\frac{3}{5}} \end{align*}$
So the region of integration is swept out over $\displaystyle \displaystyle \begin{align*} 0 \leq \theta \leq \arccos{\frac{3}{5}} \end{align*}$.
Therefore our double integral is
$\displaystyle \displaystyle \begin{align*} A &= \int_0^{\arccos{\frac{3}{5}}}{\int_{\frac{7}{2}}^7 {r\,dr}\,d\theta} \\ &= \int_0^{\arccos{\frac{3}{5}}}{\left[\frac{r^2}{2}\right]_{\frac{7}{2}}^7\,d\theta} \\ &= \int_0^{\arccos{\frac{3}{5}}}{\frac{7^2}{2} - \frac{\left(\frac{7}{2}\right)^2}{2}\,d\theta} \\ &= \int_0^{\arccos{\frac{3}{5}}}{ \frac{147}{8} \, d\theta } \\ &= \left[\frac{147\,\theta}{8}\right]_0^{\arccos{\frac{3}{5}}} \\ &= \frac{147\arccos{\frac{3}{5}}}{8} - \frac{147 \cdot 0}{8} \\ &= \frac{147\arccos{\frac{3}{5}}}{8} \\ &\approx 17.039\,\textrm{cm}^2 \end{align*}$
That's two different answers, (11.781 and 17.039).
When I first saw this problem I automatically thought calculus. On reflection, after bjhopper's response, I think I prefer the geometric method. However, I already had a result, (using Cartesian co-ordinates rather than polar).
I'd changed the orientation so that the quarter circle had its centre at (0,7) and the half circle had its centre at (7/2,0) so as to have simple upper and lower boundary curves.
That leads to the integral
$\displaystyle \int ^{28/5}_{0} \sqrt{7x-x^{2}}-7+\sqrt{49-x^{2}}\quad dx$
which evaluates to approximately 11.7813.