Question

Using definite integrals, find the area of the circle x² + y² = a².

Answer 1

Area of the circle x² + y² = a² will be the 4 times the area enclosed between x = 0 and x = a in the first quadrant which is shaded.

$$\begin{gathered} A=4{\int }_0^a\left|y\right|dx \\ =4{\int }_0^a\left(\sqrt{a^2-x^2}\right)dx \\ =4{\left[\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\frac{x}{a}\right]}_0^a \\ =4\left[0+\frac{1}{2}{a}^2{\sin }^{-1}1\right] \\ =4\left[\frac{1}{2}{a}^2\frac{\mathrm{\pi }}{2}\right]\left(\because {\sin }^{-1}1=\frac{\mathrm{\pi }}{2}\right) \\ =a^2\mathrm{\pi }\mathrm{sq}\mathrm{units} \end{gathered}$$