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Question:
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Area lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is
(a) $\pi$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{4}$
(a) $\pi$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{4}$
Solution:
2566 Upvotes
Verified Answer
(a) $x^2+y^2=4$. It is a circle at the centre $(0,0)$ and $r=2$
$$
\begin{aligned}
&\text { Area }=\int_0^2 \sqrt{4-x^2} d x \\
&\text { Let } x=2 \sin \theta \\
&\Rightarrow d x=2 \cos \theta \mathrm{d} \theta \\
&\text { When } x=0, \theta=0 \\
&\left.=4 \int_0^{\pi / 2}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta=2\left(\theta+\frac{\sin 2 \theta}{2}\right)\right]_0^{\pi / 2}=\pi
\end{aligned}
$$
$$
\begin{aligned}
&\text { Area }=\int_0^2 \sqrt{4-x^2} d x \\
&\text { Let } x=2 \sin \theta \\
&\Rightarrow d x=2 \cos \theta \mathrm{d} \theta \\
&\text { When } x=0, \theta=0 \\
&\left.=4 \int_0^{\pi / 2}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta=2\left(\theta+\frac{\sin 2 \theta}{2}\right)\right]_0^{\pi / 2}=\pi
\end{aligned}
$$
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