Question

The area of smaller part between the circle $x^2 + y^2 = 4$ and the line $x = 1$ is ______ sq. units.

• A. $\frac{4\pi}{3} - \sqrt{3}$
• B. $\frac{8\pi}{3} - \sqrt{3}$
• C. $\frac{4\pi}{3} + \sqrt{3}$
• D. $\frac{5\pi}{3} + \sqrt{3}$

Hint:

Geometric Shortcut! Area = (Area of Sector $OAB$) - (Area of Triangle $OAB$).
Angle of sector: $\cos \theta = 1/2 \implies \theta = 60^\circ$ (so total sector angle $120^\circ$ or $2\pi/3$).
Sector Area = $\frac{1}{2} r^2 (2\pi/3) = \frac{4\pi}{3}$. Triangle Area = $\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} (2) (2\sqrt{3}/2) = \sqrt{3}$.
Result: $\frac{4\pi}{3} - \sqrt{3}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The area is bounded by the circle $y = \pm \sqrt{4-x^2}$ from $x = 1$ to the radius $x = 2$. The total area is twice the area above the x-axis.

Step 2: Formula Application:
Area $= 2 \int_{1}^{2} \sqrt{2^2 - x^2} \, dx$. Using $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$.

Step 3: Explanation:
Area $= 2 \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}(\frac{x}{2}) \right]_1^2$ $= 2 \left[ (0 + 2 \cdot \frac{\pi}{2}) - (\frac{1}{2}\sqrt{3} + 2\sin^{-1}\frac{1}{2}) \right]$ $= 2 \left[ \pi - \frac{\sqrt{3}}{2} - 2(\frac{\pi}{6}) \right] = 2 \left[ \pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3} \right]$ $= 2 \left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right] = \frac{4\pi}{3} - \sqrt{3}$.

Step 4: Final Answer:
The area is $\frac{4\pi}{3} - \sqrt{3}$ sq. units.