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 Question:
We must find the area of the smaller region enclosed between a circle of radius 2 centered at the origin, and a vertical chord cut at $x = 1$.

Step 2: Detailed Explanation:
The circle equation is $x^2 + y^2 = 4$, which means $y = \pm\sqrt{4 - x^2}$.
The vertical line is $x = 1$. The "smaller part" lies to the right of this line, from $x = 1$ to the edge of the circle at $x = 2$.
Because the circle is perfectly symmetric across the X-axis, the total area is exactly twice the area located in the first quadrant:
$\text{Area} = 2 \int_{1}^{2} y \, dx = 2 \int_{1}^{2} \sqrt{4 - x^2} \, dx$
Use the standard integration formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:
Here, $a = 2$.
Area = $2 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_1^2$
Area = $\left[ x\sqrt{4 - x^2} + 4\sin^{-1}\left(\frac{x}{2}\right) \right]_1^2$
Evaluate at the upper limit ($x = 2$):
Upper = $2\sqrt{4 - 4} + 4\sin^{-1}\left(\frac{2}{2}\right) = 0 + 4\sin^{-1}(1) = 4\left(\frac{\pi}{2}\right) = 2\pi$
Evaluate at the lower limit ($x = 1$):
Lower = $1\sqrt{4 - 1} + 4\sin^{-1}\left(\frac{1}{2}\right) = \sqrt{3} + 4\left(\frac{\pi}{6}\right) = \sqrt{3} + \frac{2\pi}{3}$
Subtract the lower limit from the upper limit:
Area = Upper - Lower
Area = $2\pi - \left(\sqrt{3} + \frac{2\pi}{3}\right)$
Area = $2\pi - \frac{2\pi}{3} - \sqrt{3}$
Find a common denominator for the $\pi$ terms:
Area = $\frac{6\pi}{3} - \frac{2\pi}{3} - \sqrt{3}$
Area = $\frac{4\pi}{3} - \sqrt{3}$

Step 3: Final Answer:
The area is $\frac{4\pi}{3} - \sqrt{3}$ sq. units, matching option (a).