Question

Find the area of the shaded region of the circle of radius $\mathrm{a}$ when the chord is $\mathrm{h}$ units $(0<\mathrm{h}<\mathrm{a})$ from the center of the circle (see figure).

Answer 1

Finding the area of the shaded region of the given circle:

The diagram represents a circle with center $\left(0,0\right)$ and radius $\mathrm{a}$.

Therefore, the equation of ye given circle is,

${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$

Hence, the area of the circle can be expressed as,

$$\begin{gathered} \mathrm{A}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{y}\mathrm{dy} \\ =2{\int }_{\mathrm{h}}^{\mathrm{a}}\sqrt{{\mathrm{a}}^2-{\mathrm{h}}^2}d\mathrm{y}\left[\because \mathrm{x}=\mathrm{h}\right] \end{gathered}$$

Integrating the above Equation we can write,

$$\begin{gathered} A={\left[a^2{\sin }^{-1}\left(\frac{y}{a}\right)-y\sqrt{a^2-y^2}\right]}_h^a \\ A=a^2{\sin }^{-1}\left(\frac{a}{a}\right)-a^2{\sin }^{-1}\left(\frac{h}{a}\right)+h\sqrt{a^2-h^2} \\ A=a^2{\sin }^{-1}\left(1\right)-a^2{\sin }^{-1}\left(\frac{h}{a}\right)+h\sqrt{a^2-h^2} \\ A=\frac{a^2\mathrm{\pi }}{2}-a^2{\sin }^{-1}\left(\frac{h}{a}\right)+h\sqrt{a^2-h^2}\left[\because {\sin }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$

Therefore, the area of the shaded region of the circle is $A=\frac{a^2\mathrm{\pi }}{2}-a^2{\sin }^{-1}\left(\frac{h}{a}\right)+h\sqrt{a^2-h^2}$.