MCQ
The area bounded by the curve $x^2+y^2=1$ in first quadrant is
- ✓$\frac{\pi}{4}$ sq. units
- B$\frac{\pi}{2}$ sq. units
- C$\frac{\pi}{3}$ sq. units
- D$\frac{\pi}{6}$ sq. units
✓
Answer
Correct option: A.
$\frac{\pi}{4}$ sq. units
(a): We have, $x^2+y^2=1$, which is a circle with centre $(0,0)$ and radius $=1$.
Required area
$
\begin{array}{l}
=\int_0^1 \sqrt{1-x^2} d x \\
=\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} \frac{x}{1}\right]_0^1 \\
=\left[\frac{1}{2} \sin ^{-1} 1\right]=\left(\frac{1}{2} \times \frac{\pi}{2}\right)=\frac{\pi}{4} \text { sq. units }
\end{array}
$
Required area
$
\begin{array}{l}
=\int_0^1 \sqrt{1-x^2} d x \\
=\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} \frac{x}{1}\right]_0^1 \\
=\left[\frac{1}{2} \sin ^{-1} 1\right]=\left(\frac{1}{2} \times \frac{\pi}{2}\right)=\frac{\pi}{4} \text { sq. units }
\end{array}
$
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