The area of the shadow part is calculated mathematically in the first volume of the sixth grade of Beijing Normal University.

You didn't tell R in the left semicircle about this question, which is not easy to do;

Area of half circle:1/2π R2 =1/2 * (8/2) 2? *π=8π

The area of the left semicircle: let the radius be x,

1/2? *? πx^2

The area of the right semicircle: the radius is 4-x,

1/2 *? π? *(4-x)^2= 1/2？ π? *( 16-8x+x^2)

Shaded area:

8π- 1/2 ? * ? πx^2 - 1/2？ π ? *( 16-8x+x^2)

=4πx-x^2 * π

Brother, I can only help you here!