Question

The area of the region described by \( \{(x, y) \, | \, x^2 + y^2 \leq 1 \, \text{and \, y^2 \leq 1 - x \} \) is:}

• 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 given region is bounded by:
The circle \( x^2 + y^2 \leq 1 \),
The line \( y^2 \leq 1 - x \). Step 1: Determine the total area of the region
The total area \( A \) is the sum of two parts: 1. The area under the semicircle, and 2. The area under the curve defined by \( y = \sqrt{1 - x} \). Thus, the total area \( A \) is: \[ A = 2 \int_{0}^{1} \sqrt{1 - x^2} \, dx + \int_{0}^{1} \sqrt{1 - x} \, dx. \] Step 2: Solve the first integral (semicircular area)
The first integral represents the area of a quarter-circle, and is known: \[ \int_{0}^{1} \sqrt{1 - x^2} \, dx = \frac{\pi}{4}. \] Step 3: Solve the second integral (area under the curve)
For the second integral, we have: \[ \int_{0}^{1} \sqrt{1 - x} \, dx. \] Let’s use the substitution \( 1 - x = t^2 \), then: \[ dx = -2t \, dt, \quad \text{when } x = 0 \implies t = 1, \quad x = 1 \implies t = 0. \] Substitute into the integral: \[ \int_{0}^{1} \sqrt{1 - x} \, dx = \int_{1}^{0} t \cdot (-2t) \, dt = 2 \int_{0}^{1} t^2 \, dt. \] Now solve: \[ 2 \int_{0}^{1} t^2 \, dt = 2 \left[ \frac{t^3}{3} \right]_{0}^{1} = \frac{2}{3}. \] Step 4: Combine the results
Now, combining the areas from both integrals: \[ A = 2 \cdot \left( \frac{\pi}{4} \right) + \frac{2}{3} = \frac{\pi}{2} + \frac{4}{3}. \] Final Answer: \[ \boxed{\frac{\pi}{2} + \frac{4}{3}}. \]