Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 48

Given the curves are
y=sqrt(r^2-x^2) (this is a circle equation actually, but lets compute the centroid in a standard way) , y=0
let f(x) =sqrt(r^2-x^2)

and g(x)=0
In order to find the Centroid of the region bounded by the curves,
first we have to find the area bounded by the curves.
So now in order to find the area , we have to find the intersecting points of the curves. This can be obtained by equating f(x) and g(x) .
=> f(x) = g(x)
=> sqrt(r^2-x^2) =0
=> (r^2-x^2) =0
=> x^2 = r^2
=> x=+-r --------(1)
so the curves f(x)>=g(x) , -r<=x<=r
so the area = int _-r ^r [sqrt(r^2 - x^2) -0]dx
let x=rsin(theta) ------(2)
so , dx = rcos(theta) d theta
but from (1) and (2) we get
x=+-r , x= rsin(theta)
So,
sin(theta) = +-1
=> theta = sin^(-1) (+-1)
so theta = +-(pi/2)
so, now with the new integrals we get
area = int _-r ^r [sqrt(r^2 - x^2) -0] dx
= int_(-pi/2) ^(pi/2) sqrt(r^2 - r^2sin^2 (theta)) rcos(theta) d theta
= int_(-pi/2) ^(pi/2) sqrt(r^2(1 - sin^2 (theta))) rcos(theta) d theta
=int_(-pi/2) ^(pi/2) sqrt(r^2( cos^2 (theta))) rcos(theta) d theta
=int_(-pi/2) ^(pi/2) (rcos (theta)) rcos(theta) d theta
=int_(-pi/2) ^(pi/2) (rcos (theta))^2 d(theta)
=(r^2)int_(-pi/2) ^(pi/2) (cos^2 (theta)) d(theta)
we can right the above integral as
=(2) (r^2)int_0 ^(pi/2) (cos^2 (theta)) d(theta)
as we know that int cos^2(x) dx = (1/2)(x+(1/2)sin (2x))
now ,
area = (2) (r^2)int_0 ^(pi/2) (cos^2 (theta)) d(theta)
=(2) (r^2) [(1/2)(x+(1/2)sin (2x))]_0 ^(pi/2)
= 2(r^2)[((1/2)((pi/2)+(1/2)sin (2(pi/2))))-(1/2)((0)+(1/2)sin (2(0)))]
=2(r^2)[((1/2)((pi/2)+(0)))-0]
=2 r^2[pi/4]
= (pi r^2)/2
Now the centroid of the region bounded by the curves is given as
let (x_1,y_1) be the coordinates of the centroid so ,
x_1 is given as
x_1 = (1/(area)) int _a^b x[f(x)-g(x)] dx
where the a= -r , b= r
= (1/((pi r^2)/2)) int _-r ^r x[sqrt(r^2 -x^2)-(0)] dx
=(1/((pi r^2)/2)) int _-r ^r x[sqrt(r^2 -x^2)] dx
= (2/(pi r^2)) int _-r ^r [sqrt(r^2 -x^2)] xdx
let u = r^2 -x^2 => du = -2x dx
(-1/2)du = xdx
So the bounds of integration are from u=0 to u=0 (plug in x=-r and x=r to u=r^2-x^2 )
= (2/(pi r^2)) int _-r ^r [sqrt(r^2 -x^2)] xdx
= (2/(pi r^2)) int _0 ^0 [sqrt(u)] (-1/2)du
= 0 as int_a^b f(t) dt = F(a) - F(b) where F(x) is the antiderivative of f. So, int_0^0 f(t) dt = F(0)-F(0)=0
so, x_1 = 0
and now let us y_1 and so ,
y_1 is given as
y_1 = (1/(area)) int _a^b (1/2) [f^2(x)-g^2(x)] dx
where a= -r , b= r
= (1/((pi r^2)/2)) int _-r ^r (1/2)[(sqrt(r^2 -x^2))^2] dx
= (2/((pi r^2))) (1/2) int _-r ^r [(sqrt(r^2 -x^2))^2] dx
= (1/((pi r^2))) int _-r ^r [((r^2 -x^2))] dx
Because r^2-x^2 is an even function with respect to x, we can change the integral limits to
=(2/((pi r^2))) int _0 ^r [((r^2 -x^2))] dx
= (2/((pi r^2))) ((r^2)x -x^3 /3)_0 ^r
= (2/((pi r^2))) [ [(r^3 -r^3 /3)]-[((r^2)0 -0^3 /3)]]
= 2/(pi r^2) [(r^3 -r^3 /3)-0]
= (4r)/(3pi)

Therefore, the centroid coordinates are (0,(4r)/(3pi))