Question

Calculate the area of the intersection between the disk \(x^2 + y^2 \leq 1\) and the region defined by \(y^2 \leq 1 - x\).

• A. \( \frac{\pi}{2} - \frac{2}{3} \)
• B. \( \frac{\pi}{2} + \frac{2}{3} \)
• C. \( \frac{\pi}{2} + \frac{4}{3} \)
• D. \( \frac{\pi}{2} - \frac{4}{3} \)

Hint:

When solving area problems involving curves and circles, break the region into simpler sub-regions to calculate their areas more efficiently.

Answer 1

The Correct Option is C

The specified region is defined by the inequalities:
The circle \( x^2 + y^2 \leq 1 \), 
The curve \( y^2 \leq 1 - x \).

 Step 1: Calculate the total area of the region.
The total area \( A \) is the sum of two components: 1. The area below the semicircle, and 2. The area below the curve \( y = \sqrt{1 - x} \). Consequently, the total area \( A \) is expressed as: \[ A = 2 \int_{0}^{1} \sqrt{1 - x^2} \, dx + \int_{0}^{1} \sqrt{1 - x} \, dx. \]

 Step 2: Evaluate the first integral (semicircular area).
The first integral represents the area of a quarter-circle and is a known value: \[ \int_{0}^{1} \sqrt{1 - x^2} \, dx = \frac{\pi}{4}. \] 

Step 3: Evaluate the second integral (area under the curve).
For the second integral, we have: \[ \int_{0}^{1} \sqrt{1 - x} \, dx. \] Employ the substitution \( 1 - x = t^2 \). This implies: \[ dx = -2t \, dt, \quad \text{with limits: } x = 0 \implies t = 1, \quad x = 1 \implies t = 0. \] Substituting into the integral yields: \[ \int_{0}^{1} \sqrt{1 - x} \, dx = \int_{1}^{0} t \cdot (-2t) \, dt = 2 \int_{0}^{1} t^2 \, dt. \] Now, we solve this integral: \[ 2 \int_{0}^{1} t^2 \, dt = 2 \left[ \frac{t^3}{3} \right]_{0}^{1} = \frac{2}{3}. \] 

Step 4: Aggregate the results.
Combining the areas computed from both integrals: \[ A = 2 \cdot \left( \frac{\pi}{4} \right) + \frac{2}{3} = \frac{\pi}{2} + \frac{4}{3}. \] 

Final Answer: \[ \boxed{\frac{\pi}{2} + \frac{4}{3}}. \]