Question

Using integration, find the area of the region enclosed between the circle \( x^2 + y^2 = 16 \) and the lines \( x = -2 \) and \( x = 2 \).

Hint:

Use trigonometric substitution for integrals involving square roots of quadratic expressions.

Answer 1

The given circle is: \[ x^2 + y^2 = 16 \quad \Rightarrow \quad y = \pm \sqrt{16 - x^2}. \] The lines \( x = -2 \) and \( x = 2 \) are vertical lines. We need to find the area of the region between these vertical lines and the circle. 1. Area under the curve \( y = \sqrt{16 - x^2} \): The area between \( x = -2 \) and \( x = 2 \) above the \( x \)-axis is: \[ A_1 = \int_{-2}^{2} \sqrt{16 - x^2} \, dx. \] 2. Use symmetry: The total area enclosed is twice the area above the \( x \)-axis: \[ {Total Area} = 2A_1 = 2 \int_{-2}^{2} \sqrt{16 - x^2} \, dx. \] 3. Solve the integral: Substitute \( x = 4 \sin \theta \), so \( dx = 4 \cos \theta \, d\theta \) and \( \sqrt{16 - x^2} = 4 \cos \theta \): \[ \int_{-2}^{2} \sqrt{16 - x^2} \, dx = \int_{-\pi/6}^{\pi/6} 4 \cos \theta \cdot 4 \cos \theta \, d\theta = 16 \int_{-\pi/6}^{\pi/6} \cos^2 \theta \, d\theta. \] 4. Simplify \( \cos^2 \theta \): Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \): \[ 16 \int_{-\pi/6}^{\pi/6} \cos^2 \theta \, d\theta = 16 \int_{-\pi/6}^{\pi/6} \frac{1 + \cos 2\theta}{2} \, d\theta. \] Separate the terms: \[ 16 \int_{-\pi/6}^{\pi/6} \frac{1}{2} \, d\theta + 16 \int_{-\pi/6}^{\pi/6} \frac{\cos 2\theta}{2} \, d\theta. \] 5. Integrate: - First term: \[ 16 \int_{-\pi/6}^{\pi/6} \frac{1}{2} \, d\theta = 16 \cdot \frac{1}{2} \cdot \left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right) = 16 \cdot \frac{\pi}{6}. \] - Second term: \[ 16 \int_{-\pi/6}^{\pi/6} \frac{\cos 2\theta}{2} \, d\theta = 16 \cdot \frac{1}{2} \cdot \left[\frac{\sin 2\theta}{2}\right]_{-\pi/6}^{\pi/6} = 0. \] Total area above the \( x \)-axis: \[ A_1 = \frac{16\pi}{6} = \frac{8\pi}{3}. \] Total enclosed area: \[ {Total Area} = 2A_1 = 2 \cdot \frac{8\pi}{3} = \frac{16\pi}{3}. \] Final Answer: The area is \( \boxed{\frac{16\pi}{3}} \).