The area bounded by the circle $x^{2}+y^{2}=16$ and lines x=0 and x=2 is
The area bounded by the circle $x^{2}+y^{2}=16$ and lines x=0 and x=2 is
- $\left[4 \sqrt{3}+\frac{8 \pi}{3}\right]$ sq. units
- $\frac{1}{2}\left[4 \sqrt{3}+\frac{8 \pi}{3}\right]$ sq. units
- $\left[4 \sqrt{3}-\frac{8 \pi}{3}\right]$ sq. units
- $\frac{1}{2}\left[4 \sqrt{3}-\frac{8 \pi}{3}\right]$ sq. units
Solution
Given equation of circle is $x^{2}+y^{2}=16$
$\therefore y^{2}=16-x^{2} \therefore y=\sqrt{16-x^{2}}$
Required area is shaded.
$A=2 \mathrm{~A}(\mathrm{OABCO})$
$\begin{aligned} \text { Area } &=2 \int_{0}^{2} \sqrt{16-x^{2}} \mathrm{dx} \\ &=2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{0}^{2} \\ &=2\left\{\frac{2}{2} \sqrt{12}+8 \sin ^{-1} \frac{1}{2}-0\right\}=2\left[2 \sqrt{3}+8\left(\frac{\pi}{6}\right)\right]=4 \sqrt{3}+\frac{8 \pi}{3} \end{aligned}$
Asked in: MHT CET 2020 (14 Oct Shift 2)
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