Calculus of a Single Variable, Chapter 7, 7.6, Section 7.6, Problem 49

Given curves are ,
y=(b/a)sqrt(a^2-x^2) , y=0

let f(x) =(b/a)sqrt(a^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)
=> (b/a)sqrt(a^2-x^2) =0
=> sqrt(a^2-x^2) =0
=> x^2 = a^2
=> x= +-a---------(1)
so the curves f(x)>=g(x) on [-a, a]
so the area = int _a ^b [f(x) -g(x)]dx where the lower bound is -a, and the upper bound is a
= int _-a ^a [(b/a)sqrt(a^2 - x^2) -0]dx
=int_-a^a[(b/a)sqrt(a^2 - x^2)]dx
The function which is being integrated is an even function so,
=2int_0^a[(b/a)sqrt(a^2 - x^2)]dx
=2(b/a) int_0^a[sqrt(a^2 - x^2)]dx


let x=a sin(theta) ------(2)
so , dx = a cos(theta) d(theta)
but from (1) and (2) we get
x=+-a , x= asin(theta) is
sin(theta) = +-1
=> theta = sin^(-1) (+-1)
so theta = +-(pi/2)
so ,now with the new integrals we get
area = 2(b/a) int_0^a[sqrt(a^2 - x^2)]dx
= 2(b/a) int_(0) ^(pi/2) sqrt(a^2 - a^2 sin^2 (theta)) a cos(theta) d(theta)
= 2(b/a) int_(0) ^(pi/2) sqrt(a^2(1 - sin^2 (theta))) a cos(theta) d(theta)
=2(b/a) int_(0) ^(pi/2) sqrt(a^2( cos^2 (theta))) a cos(theta) d(theta)
=2(b/a) int_(0) ^(pi/2) (a cos (theta)) a cos(theta) d(theta)
=2(b/a)int_(0) ^(pi/2) (acos (theta))^2 d(theta)
=2(b/a)(a^2)int_(0) ^(pi/2) (cos^2 (theta)) d(theta)
we can right the above integral as
=(2)(b/a) (a^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)(b/a) (a^2)int_0 ^(pi/2) (cos^2 (theta)) d(theta)
=(2)(b/a) (a^2) [(1/2)(x+(1/2)sin (2x))]_0 ^(pi/2)
= 2(b/a)(a^2)[((1/2)((pi/2)+(1/2)sin (2(pi/2))))-(1/2)((0)+(1/2)sin (2(0)))]
=2(b/a)(a^2)[((1/2)((pi/2)+(0)))-0]
=2(b/a) a^2 [pi/4]
= ((pi a^2)/2)*(b/a)
=(pi*a*b)/(2)
Now the centroid of the region bounded the curves is given as ,
let (x_1,y_1) be the co- ordinates of the centroid so ,
x_1 is given as
x_1 = (1/(area)) int _a^b x[f(x)-g(x)] dx
where the limits a= -a , b= a
so,
x_1 = (1/(area)) int _-a^a x[((b/a)sqrt(a^2-x^2))-(0)] dx
=(1/((pi*a*b)/2)) int _-a^a x[((b/a)sqrt(a^2-x^2))] dx
=(2/((pi*a*b))) int _-a^a x[((b/a)sqrt(a^2-x^2))] dx

let u = a^2 -x^2 => du = -2x dx
(-1/2)du = xdx
The bounds are then u = a^2 - (a^2) = 0 and u=a^2-(-a)^2 = 0
so,
= (2/((pi*a*b))) int _0^0 [((b/a)sqrt(a^2-x^2))] dx
= 0 since for int_a^b g(u) du = G(a) - G(b) it follows that int_0^0 g(u) du = G(0) - G(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 the a= -a , b= a
so ,
= (1/((pi*a*b)/(2))) int _-a ^a (1/2)[((b/a)sqrt(a^2 -x^2))^2 -[0]^2] dx
= (2/((pi*a*b))) int _-a ^a (1/2)[((b/a)sqrt(a^2 -x^2))^2] dx
= (2/((pi*a*b))) (1/2) int _-a ^a [((b/a)sqrt(a^2 -x^2))^2] dx
= (1/((pi*a*b))) int _-a ^a [((b/a)sqrt(a^2 -x^2))^2] dx
= (1/((pi*a*b)))(b/a)^2 int _-a ^a [(sqrt(a^2 -x^2))^2] dx
since the function which is being integrated is even function so ,we can write the above equation as
= (2/((pi*a*b)))(b/a)^2 int _0 ^a [(sqrt(a^2 -x^2))^2] dx
= (2/((pi*a*b)))(b/a)^2 int _0 ^a [(a^2 -x^2)] dx
= (2/((pi*a*b)))(b/a)^2 [((a^2)x -(x^3)/3)]_0 ^a
= (2/((pi*a*b)))(b/a)^2 [[((a^2)a -(a^3)/3)]-[((0^2)a -(0^3)/3)]]
= (2/((pi*a*b)))(b/a)^2 [[((a^2)a -(a^3)/3)]-[0]]
= (2/((pi*a*b)))(b/a)^2 [((a^3) -(a^3)/3)]
= (2/((pi*a*b)))(b/a)^2 [ (2*(a^3))/3)]
= (2/((pi*b)))(b)^2 [ (2)/3]
= (2/((pi)))(b) [ (2)/3]
=((4b)/(3pi))
so the centroid of the area bounded by the curves is
= (x_1,y_1)= (0,(4b)/(3pi))