LinkBack URL
About LinkBacksFind 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 radiusand is centred at
, so has the equation
.
The semicircle has radiusand is centred at
, so has the equation
.
So if we are trying to find the area of the enclosed region, we use, where
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
.
Now as for the bounds, it can clearly be seen that since the semicircle has a radius of.
If we solve the two equations simultanously, we can see that they intersect at. This means at the point of intersection:
So the region of integration is swept out over.
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*}](http://latex.codecogs.com/png.latex?\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
which evaluates to approximately 11.7813.
  |
