Find the centroid of the region bounded by the graphs of `y=sqrt(r^2-x^2)` and `y=0`

Expert Answers
kseddy123 eNotes educator| Certified Educator

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)` 


`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[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))`