Question
Find the area of the circle $x^2 + y^2 = 25.$
✓
Answer
By the symmetry of the circle, its area is equal to $4$ times the area of the region $OABO$.
Clearly, for this region, the limits of integration are $0$ and $5$ .
From the equation of the circle, $y^2=25-x^2$.
In the first quadrant $y>0$
$
\therefore y =\sqrt{25-x^2}
$
$\therefore$ area of the circle $=4($ area of region $OABO )$
$
\begin{aligned}
=4 & \int_0^5 y d x=4 \int_0^5 \sqrt{25-x^2} d x \\
=4 & {\left[\frac{x}{2} \sqrt{25-x^2}+\frac{25}{2} \sin ^{-1}\left(\frac{x}{5}\right)\right]_0^5 } \\
& \ldots\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]
\end{aligned}
$
$
\begin{array}{r}
=4\left[\left\{\frac{5}{2} \sqrt{25-25}+\frac{25}{2} \sin ^{-1}(1)\right\}-\right. \\
\left.\left\{\frac{0}{2} \sqrt{25-0}+\frac{25}{2} \sin ^{-1}(0)\right\}\right]
\end{array}
$
$=4 \cdot \frac{25}{2} \cdot \frac{\pi}{2}=25 \pi$ sq units.
$
\cdots\left[\because \sin ^{-1}(1)=\frac{\pi}{2}, \sin ^{-1}(0)=0 .\right]
$
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