Question

Using integration, find the area of the region enclosed between the curve \( y = \sqrt{4 - x^2} \), and the lines \( x = -1 \), \( x = 1 \), and the \( x \)-axis.

Hint:

When solving problems involving semicircles, simplify using symmetry and appropriate substitutions like \( x = r\sin\theta \).

Answer 1

Step 1: Understand the problem
The curve \( y = \sqrt{4 - x^2} \) represents the upper semicircle with radius \( 2 \) and center at the origin. We are tasked to find the area of the region enclosed between this curve, the lines \( x = -1 \), \( x = 1 \), and the \( x \)-axis. 
Step 2: Set up the integral
The required area can be calculated as twice the area between the curve and the \( x \)-axis for \( x \in [0, 1] \) (using symmetry): \[ \text{Required area} = 2 \int_{0}^{1} \sqrt{4 - x^2} \, dx. \] Step 3: Evaluate the integral using substitution
Let \( x = 2 \sin \theta \), so \( dx = 2 \cos \theta \, d\theta \), and for \( x = 0 \), \( \theta = 0 \), and for \( x = 1 \), \( \sin \theta = \frac{1}{2} \), giving \( \theta = \frac{\pi}{6} \). Substituting: \[ \int \sqrt{4 - x^2} \, dx = \int \sqrt{4 - 4\sin^2\theta} \cdot 2\cos\theta \, d\theta = \int 2\cos^2\theta \cdot 2\cos\theta \, d\theta. \] Simplify: \[ \int_{0}^{\pi/6} 2\cos^2\theta \cdot 2 \, d\theta = \int_{0}^{\pi/6} 4\cos^2\theta \, d\theta. \] Use the double-angle identity \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \): \[ \int_{0}^{\pi/6} 4\cos^2\theta \, d\theta = \int_{0}^{\pi/6} 2(1 + \cos 2\theta) \, d\theta. \] Split into two integrals: \[ \int_{0}^{\pi/6} 2 \, d\theta + \int_{0}^{\pi/6} 2\cos 2\theta \, d\theta. \] Evaluate each term: 1. For \( \int_{0}^{\pi/6} 2 \, d\theta \): \[ \int_{0}^{\pi/6} 2 \, d\theta = 2\left[\theta\right]_{0}^{\pi/6} = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3}. \] 2. For \( \int_{0}^{\pi/6} 2\cos 2\theta \, d\theta \): \[ \int_{0}^{\pi/6} 2\cos 2\theta \, d\theta = \sin 2\theta \Big|_{0}^{\pi/6}. \] Using \( \sin 2\theta = 2\sin\theta\cos\theta \): \[ \sin 2\theta \Big|_{0}^{\pi/6} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}. \] Step 4: Combine results
The total area is: \[ \text{Required area} = 2 \left[ \frac{\pi}{3} + \frac{\sqrt{3}}{2} \right] = \frac{2\pi}{3} + \sqrt{3}. \] Conclusion:
The area of the region is: \[ \boxed{\frac{2\pi}{3} + \sqrt{3}}. \]