Question
Find the area enclosed by the circle $x^2 + y^2 = a^2$
✓
Answer
The equation of circle is $x^2 + y^2 = a^2 .... (i)$
Its centre is origin and radius $a.$
We know that circle $x^{2 }+ y^2 = a^2$ is symmetrical about both axes.
$\therefore$ required area = $4 \int \limits_{0}^{2} y d x=4 \int \limits_{0}^{2} \sqrt{a^{2}-x^{2}} d x$
$=4\left[\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a}$
$=4\left[\left(0+\frac{\mathrm{a}^{2}}{2} \sin ^{-1} 1\right)-\left(0+\frac{\mathrm{a}^{2}}{2} \sin ^{-1} 0\right)\right]$
$=4\left[\left(0+\frac{a^{2}}{2} \sin ^{-1} 1\right)-\left(0+\frac{a^{2}}{2} \sin ^{-1} 0\right)\right]$
$=4\left[\frac{a^{2}}{2} \times \frac{\pi}{2}\right]$ $\left[\because \sin ^{-1} 0=0\right]$
$=\pi a$
Its centre is origin and radius $a.$
We know that circle $x^{2 }+ y^2 = a^2$ is symmetrical about both axes.
$\therefore$ required area = $4 \int \limits_{0}^{2} y d x=4 \int \limits_{0}^{2} \sqrt{a^{2}-x^{2}} d x$
$=4\left[\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a}$
$=4\left[\left(0+\frac{\mathrm{a}^{2}}{2} \sin ^{-1} 1\right)-\left(0+\frac{\mathrm{a}^{2}}{2} \sin ^{-1} 0\right)\right]$
$=4\left[\left(0+\frac{a^{2}}{2} \sin ^{-1} 1\right)-\left(0+\frac{a^{2}}{2} \sin ^{-1} 0\right)\right]$
$=4\left[\frac{a^{2}}{2} \times \frac{\pi}{2}\right]$ $\left[\because \sin ^{-1} 0=0\right]$
$=\pi a$
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