# Math Help - Area enclosed by a semi-circle and a quarter circle

1. ## Area enclosed by a semi-circle and a quarter circle

Find the shaded area, which is enclosed by a square, semi-circle and a quarter of circle.

2. ## Re: Area enclosed by a semi-circle and a quarter circle

Draw the common chord of the two circles.Area of shaded region is the sum of the two segments.Central angles of these are readily calculable

3. ## Re: Area enclosed by a semi-circle and a quarter circle

Got it. Thanks a lot, bjhopper!

4. ## Re: Area enclosed by a semi-circle and a quarter circle

Could you explain this in a little bit more depth? I am not seeing how to get this. Thank you.

6. ## Re: Area enclosed by a semi-circle and a quarter circle

An alternative is to use calculus. The quarter circle has radius \displaystyle \begin{align*} 7\,\textrm{cm} \end{align*} and is centred at \displaystyle \begin{align*} (0, 0) \end{align*}, so has the equation \displaystyle \begin{align*} x^2 + y^2 = 49 \end{align*}.

The semicircle has radius \displaystyle \begin{align*} \frac{7}{2}\,\textrm{cm} \end{align*} and is centred at \displaystyle \begin{align*} \left(7, \frac{7}{2}\right) \end{align*}, so has the equation \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 \begin{align*} \int{\int_R{}\,dA} \end{align*}, where \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 \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 \begin{align*} \frac{7}{2}\,\textrm{cm} \end{align*} and the quarter circle has a radius of \displaystyle \begin{align*} 7\,\textrm{cm} \end{align*}, the region of integration extends over \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 \begin{align*} (x, y) = \left(\frac{21}{5}, \frac{28}{5}\right) \end{align*}. This means at the point of intersection:

\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 \begin{align*} 0 \leq \theta \leq \arccos{\frac{3}{5}} \end{align*}.

Therefore our double integral is

\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*}

7. ## Re: Area enclosed by a semi-circle and a quarter circle

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.

$\int ^{28/5}_{0} \sqrt{7x-x^{2}}-7+\sqrt{49-x^{2}}\quad dx$

which evaluates to approximately 11.7813.

8. ## Re: Area enclosed by a semi-circle and a quarter circle

Nice work Yeoky

Thanks.

10. ## Re: Area enclosed by a semi-circle and a quarter circle

May I know how you integrate? Thanks.

11. ## Re: Area enclosed by a semi-circle and a quarter circle

Thank a lot, Prove it.

May I know the answer should be 17.039 or 11.781?