# Find the centroid of the region bounded by the graphs of `y=b/asqrt(a^2-x^2)` and `y=0`

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))`

Approved by eNotes Editorial Team