Aregion r consists of a square bounded by the lines x = −5, x = 5, y = 0, and y = −10 and a half disk bounded by the semicircle y = 25 − x2 and the line y = 0. find the center of gravity, (x, y), of r.

Exactly (0, -100/(24 + 3*pi)) 
Approximately (0, -2.991792499)  
I am assuming that there's a typo in the formula for the semicircle. The equation given is: 
y = 25 - x^2 
which does NOT describe a semicircle. I assume that the correct equation is: 
y = sqrt(25 - x^2) 
which would describe a semi-circle with a radius of 5.  
First, determine the mass and centroid of the square and the semicircle. For the square, the centroid is the exact center of the square and the mass is the area. So 
Mass = (5 - -5)*(0 - -10) = 10*10 = 100 
Centroid = ((-5 + 5)/2, (0+ -10)/2) = (0/2, -10/2) = (0,-5)  
The mass for the semi-circle is also its area, and the centroid of a semicircle who's center is at the origin (0,0) is:
 (0, 4r/(3*pi) ) where r = radius of semicircle. So
 (0, 4*5/(3*pi))
 (0, 20/(3pi))
 Mass = (1/2)pi*r^2 = (1/2)pi*5^2 = (1/2)pi*25 = 12.5*pi  
 Since the centroid of both the square and semi-circle have a x-coordinate of 0, the x-coordinate of the center of gravity will also be 0. For the y coordinate, we need to use the following formula:
 (m1 + m2 + m3 + ... + mn)*Xcm = m1x1+m2x2+m3x3+...+mn*xn
 where
 m1,m2,m3,...,mn = masses of objects.
 x1,x2,x3,...,xn = distance from datum of objects.
 Xcm = distance from datum of center of mass. 
 I'll use the origin as the datum. So
 (m1 + m2)*Xcm = m1x1+m2x2
 Xcm = (m1x1+m2x2)/(m1 + m2)
 Xcm = (100(-5)+(12.5*pi)(20/(3pi)))/(100 + 12.5*pi)
 Xcm = (-500+250/3)/(100 + 12.5*pi)
 Xcm = (-1500/3+250/3)/(100 + 12.5*pi)
 Xcm = (-1250/3)/(100 + 12.5*pi)
 Xcm = -1250/(300 + 37.5*pi)
 Xcm = -2500/(600 + 75*pi)
 Xcm = -100/(24 + 3*pi) 
 So the center of gravity of r = (0, -100/(24 + 3*pi)) exactly, or approximately (0, -2.991792499)

Less force and energy was applied

Letter a, because when there are unlike charges they will always attract to each other.