Question

Find the area of the minor segment of the circle \[x^2 + y^2 = a^2\] cut off by the line \[x = \frac{a}{2}\]

Answer 1

The equation of the circle is \[x^2 + y^2 = a^2\]

Centre of the circle = (0, 0) and radius = a.

The line \[x = \frac{a}{2}\] is parallel to y-axis and intersects the x-axis at \[\left( \frac{a}{2}, 0 \right)\]

Required area = Area of the shaded region
                      = 2 × Area of the region ABDA

\[= 2 \times \int_\frac{a}{2}^a y_{\text{ circle }} dx\]
\[ = 2 \int_\frac{a}{2}^a \sqrt{a^2 - x^2}dx\]
\[ = \left.2 \left( \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right)\right|_\frac{a}{2}^a \]
\[ = 2\left[ \left( 0 + \frac{a^2}{2} \sin^{- 1} 1 \right) - \left( \frac{a}{4} \times \sqrt{a^2 - \frac{a^2}{4}} + \frac{a^2}{2} \sin^{- 1} \frac{1}{2} \right) \right]\]
\[= 2\left( \frac{a^2}{2} \times \frac{\pi}{2} - \frac{a}{4} \times \frac{\sqrt{3}a}{2} - \frac{a^2}{2} \times \frac{\pi}{6} \right)\]

\[ = \frac{a^2 \pi}{2} - \frac{\sqrt{3} a^2}{4} - \frac{a^2 \pi}{6}\]
\[ = \frac{6 a^2 \pi - 3\sqrt{3} a^2 - 2 a^2 \pi}{12}\]
\[ = \frac{a^2}{12}\left( 4\pi - 3\sqrt{3} \right)\text{ square units }\]