Example 8 | Area bounded by arcs of quarter circles

$\displaystyle A = 4\int_{10}^{10\sqrt{3}} \Big( \sqrt{400 - x^2} - 10 \Big) \, dx$

Let
x = 20 sin θ
dx = 20 cos θ dθ

When x = 10, θ = π/6
When x = 10sqrt(3), θ = π/3

 

$\displaystyle A = 4\int_{\pi/6}^{\pi/3} \Big( \sqrt{400 - 400 \sin^2 \theta} - 10 \Big)(20 \cos \theta \, d\theta)$

$\displaystyle A = 80\int_{\pi/6}^{\pi/3} \cos \theta \Big( 20\sqrt{1 - \sin^2 \theta} - 10 \Big) \, d\theta$

$\displaystyle A = 80\int_{\pi/6}^{\pi/3} 10 \cos \theta \Big( 2\sqrt{\cos^2 \theta} - 1 \Big) \, d\theta$

$\displaystyle A = 800\int_{\pi/6}^{\pi/3} \cos \theta (2\cos \theta - 1) \, d\theta$

$\displaystyle A = 800\int_{\pi/6}^{\pi/3} (2\cos^2 \theta - \cos \theta) \, d\theta$

From
cos 2θ = 2cos² θ - 1
2cos² θ = cos 2θ + 1

 

$\displaystyle A = 800\int_{\pi/6}^{\pi/3} \Big[ (\cos 2\theta + 1) - \cos \theta \Big] \, d\theta$

$A = 800\left[ \dfrac{\sin 2\theta}{2} + \theta - \sin \theta \right]_{\pi/6}^{\pi/3}$

$A = 800\left[ \dfrac{\sin (2\pi/3)}{2} + \dfrac{\pi}{3} - \sin (\pi/3) \right] - 800\left[ \dfrac{\sin (2\pi/6)}{2} + \dfrac{\pi}{6} - \sin (\pi/6) \right]$

$A = 126.06 ~ \text{unit}^2$           answer