Question
Find the area of the circle $x^2 + y^2 = 6^2$
✓
Answer
By the symmetry of the circle, required area of the circle is $4$ times the area of the region $OPQO.$
For the region $OPQO$,
The limits of integration are $x = 0$ and $x = 6.$
Given equation of the circle is $x^2 + y^2 = 6^2$
$\therefore y^2 = 6^2– x^2$
$\therefore y= \pm \sqrt{6^2-x^2}$
$\therefore y=\sqrt{6^2-x^2} \quad \ldots \ldots .[\because$ In first quadrant, $y>0]$
$\therefore$ Required area $=4$ (area of the region OPQO)
$ =4 \times \int_0^6 y \cdot d x$
$=4 \times \int_0^6 \sqrt{6^2-x^2} d x$
$=4\left[\frac{x}{2} \sqrt{(6)^2-x^2}+\frac{(6)^2}{2} \sin ^{-1}\left(\frac{x}{6}\right)\right]_0^6$
$=4\left\{\left[\frac{6}{2} \sqrt{(6)^2-(6)^2}+\frac{(6)^2}{2} \sin ^{-1}\left(\frac{6}{6}\right)\right]-\left[\frac{0}{2} \sqrt{(6)^2-(0)^2}+\frac{(6)^2}{2} \sin ^{-1}\left(\frac{0}{6}\right)\right]\right\}$
$=4\left\{\left[0+\frac{36}{2} \sin ^{-1}(1)\right]-[0+0]\right\} $
$=4\left(\frac{36}{2} \times \frac{\pi}{2}\right)$
$=36 \pi \text { sq.units }$
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