What is the area of the region bounded by the curve y = $$\sqrt{32-\text{x}^2}$$​ and x-axis in the first quadrant?

Correct option is C

​$$\begin{aligned}&\text{Use trigonometric substitution:} \\&\text{Let } x = 4\sqrt{2} \sin \theta \Rightarrow dx = 4\sqrt{2} \cos \theta \, d\theta \\&\text{When } x = 0, \, \theta = 0, \quad \text{and when } x = \sqrt{32} = 4\sqrt{2}, \, \theta = \frac{\pi}{2} \\[10pt]&\text{Now substitute into the integral:} \\&\int_{0}^{\sqrt{32}} \sqrt{32 - x^2} \, dx = \int_{0}^{\pi/2} \sqrt{32 - (4\sqrt{2} \sin\theta)^2} \cdot 4\sqrt{2} \cos\theta \, d\theta \\&= \int_{0}^{\pi/2} \sqrt{32(1 - \sin^2\theta)} \cdot 4\sqrt{2} \cos\theta \, d\theta = \int_{0}^{\pi/2} \sqrt{32\cos^2\theta} \cdot 4\sqrt{2} \cos\theta \, d\theta \\&= \int_{0}^{\pi/2} \sqrt{32} \cos\theta \cdot 4\sqrt{2} \cos\theta \, d\theta = 4\sqrt{2} \cdot \sqrt{32} \int_{0}^{\pi/2} \cos^2\theta \, d\theta = 32 \int_{0}^{\pi/2} \cos^2\theta \, d\theta \\[10pt]&\text{Use the identity } \cos^2\theta = \frac{1 + \cos(2\theta)}{2}: \\&32 \int_{0}^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta = 16 \int_{0}^{\pi/2} (1 + \cos(2\theta)) \, d\theta = 16 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{0}^{\pi/2} \\&= 16 \left[ \frac{\pi}{2} + 0 - (0 + 0) \right] = 16 \cdot \frac{\pi}{2} = {8\pi}\end{aligned}$$​​