A problem involving the same area was carried out by geometry without involving any integration. See the solution for the area of CED in the following link: Solution by plane geometry.
$\displaystyle A_{ABC} = \int_{y_1}^{y_2} (x_R - x_L) \, dy$
Where
$y_1 = 0$
$y_2 = 10$
$x_R = 20$
$x_L = \sqrt{20^2 - y^2}$
Thus,
$\displaystyle A_{ABC} = \int_0^{10} (20 - \sqrt{20^2 - y^2}) \, dy$
At this point, you can use your scientific calculator to solve for the area of region ABC. From calculator.
$A_{ABC} = 8.68 \, \text{ cm}^2$
Required area,
$A = 8A_{ABC} = 8(8.68)$
$A = 69.42 \, \text{ cm}^2$ answer
For the sake of discussion, integration is carried further step by step below.
$\displaystyle A_{ABC} = 20\int_0^{10} dy - \int_{0}^{10} \sqrt{20^2 - y^2} \, dy$
For
$\displaystyle \int_0^{10} \sqrt{20^2 - y^2} \, dy$
Let
$y = 20 \sin \theta$
$dy = 20 \cos \theta ~ d\theta$
When y = 0, θ = 0
When y = 10, θ = 30° = π/6
$\displaystyle \int_0^{10} \sqrt{20^2 - y^2} \, dy = \int_0^{\pi/6} \sqrt{20^2 - 20^2 \sin^2 \theta} \, (20 \cos \theta) \, d\theta$
$\displaystyle \int_0^{10} \sqrt{20^2 - y^2} \, dy = \int_0^{\pi/6} \sqrt{20^2(1 - \sin^2 \theta)} \, (20 \cos \theta) \, d\theta$
$\displaystyle \int_0^{10} \sqrt{20^2 - y^2} \, dy = \int_0^{\pi/6} \sqrt{20^2 \cos^2 \theta} \, (20 \cos \theta) \, d\theta$
$\displaystyle \int_0^{10} \sqrt{20^2 - y^2} \, dy = \int_0^{\pi/6} (20 \cos \theta)(20 \cos \theta) \, d\theta$
$\displaystyle \int_0^{10} \sqrt{20^2 - y^2} \, dy = \int_0^{\pi/6} 200(2\cos^2 \theta) \, d\theta$
Thus,
$\displaystyle A_{ABC} = 20\bigg[ y \bigg]_0^{10} - 200\int_0^{\pi/6} (2\cos^2 \theta) \, d\theta$
$\displaystyle A_{ABC} = 20(10 - 0) - 200\int_0^{\pi/6} (\cos 2\theta + 1) \, d\theta$
$A_{ABC} = 200 - 200 \bigg[ \frac{1}{2}\sin 2\theta + \theta \bigg]_0^{\pi/6}$
$A_{ABC} = 200 - 200 \bigg[ (\frac{1}{2}\sin \frac{1}{3}\pi + \frac{1}{6}\pi) - (\frac{1}{2}\sin 0 + 0) \bigg]$
$A_{ABC} = 8.68 \, \text{ cm}^2$
Required area,
$A = 8A_{ABC} = 8(8.68)$
$A = 69.42 \, \text{ cm}^2$ answer
