Question

Find the area bounded by the curves \[ x^2 + y^2 = 4 \quad \text{and} \quad x^2 + (y-2)^2 = 4. \]

• A. \( \dfrac{8\pi}{3} - 2\sqrt{3} \) Sq. units
• B. \( \dfrac{8\pi}{3} + \sqrt{3} \) Sq. units
• C. \( \dfrac{4\pi}{3} - 2\sqrt{3} \) Sq. units
• D. \( \dfrac{4\pi}{3} + 2\sqrt{3} \) Sq. units

Hint:

For overlapping circles of equal radii, memorizing the standard common-area formula saves time.

Answer 1

The Correct Option is A

To find the area bounded by the curves \( x^2 + y^2 = 4 \) and \( x^2 + (y-2)^2 = 4 \), we first identify what these equations represent:

• The equation \( x^2 + y^2 = 4 \) describes a circle centered at the origin (0,0) with a radius of 2.
• The equation \( x^2 + (y-2)^2 = 4 \) describes a circle centered at (0,2) with a radius of 2.

To find the region bounded between these two circles, we will determine the overlapping region, known as the lens-shaped area.

Step 1: Find Points of Intersection

The points of intersection satisfy both equations simultaneously:

• From the first equation: \( y = \pm \sqrt{4 - x^2} \)
• From the second equation: \( y = 2 \pm \sqrt{4 - x^2} \)

Equating the expressions for \( y \) from both equations gives:

• \( \sqrt{4 - x^2} = 2 - \sqrt{4 - x^2} \)
• Solving \( \sqrt{4 - x^2} = 1 \), we square both sides to solve:
• \( 4 - x^2 = 1 \)
• \( x^2 = 3 \)
• \( x = \pm \sqrt{3} \)

Both circles intersect at points \((\sqrt{3}, 1)\) and \((- \sqrt{3}, 1)\).

Step 2: Calculate the Area of Intersection

The area of the lens is computed as twice the area bounded by each circle from \(-\sqrt{3}\) to \( \sqrt{3}\) and the line \( y = 1 \).

Using Integration for the Upper Half:

The limits for the integration are from \(-\sqrt{3}\) to \(\sqrt{3}\).

The integral becomes:

\[ 2 \int_{0}^{\sqrt{3}} \left( \sqrt{4-x^2} - (2 - \sqrt{4-x^2}) \right) \, dx = 2 \int_{0}^{\sqrt{3}} \sqrt{4-x^2} + \sqrt{4-x^2} - 2 \, dx \]

Applying the identity \( \int \sqrt{a^2-x^2} \, dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C \):

This simplifies to the area formula for this symmetric lens:

\[ \text{Area} = 2 \times \left(\frac{1}{2} \times \left( 2 \times 2 \times \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \right) - (\sqrt{3}) \right) \]

\(\text{Area of Each Circular Segment with Radius } 2\):

\[ = 2 \times (2 \times \pi) \frac{60}{360} - 2\sqrt{3} = \frac{8\pi}{3} - 2\sqrt{3} \]

Thus, the area bounded by the curves is \( \frac{8\pi}{3} - 2\sqrt{3} \) square units, which corresponds to the correct option given.