# Example 7 | Area inside the square not common to the quarter circles

**Problem**

The figure shown below is composed of arc of circles with centers at each corner of the square 20 cm by 20 cm. Find the area inside the square but outside the region commonly bounded by the quarter circles. The required area is shaded as shown in the figure below.

**Solution**

**Solution by plane geometry**.

$\displaystyle A_{ABC} = \int_{y_1}^{y_2} (x_R - x_L) \, dy$

$y_1 = 0$

$y_2 = 10$

$x_R = 20$

$x_L = \sqrt{20^2 - y^2}$

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.

$\displaystyle A_{ABC} = 20\int_0^{10} dy - \int_{0}^{10} \sqrt{20^2 - y^2} \, dy$

$\displaystyle \int_0^{10} \sqrt{20^2 - y^2} \, dy$

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*