y = 9-x^2 , y = 0 Find b such that the line y = b divides the region bounded by the graphs of the equations into two regions of equal area.

Given ,
y = 9-x^2 , y = 0
first let us find the total area of the bounded by the curves.
so we shall proceed as follows
as given ,
y = 9-x^2 , y = 0
=> 9-x^2=0
=> x^2 -9 =0
=> (x-3)(x+3)=0
so x=+-3
 
the the area of the region is = int _-3 ^3 (9-x^2 -0) dx
=[9x-x^3/3] _-3 ^3
= [27-9]-[-27+9]
=18-(-18) = 36
So now we have  to find the horizonal line that splits the region into two regions with area 18
as when the line y=b intersects the curve y=9-x^2 then the ared bounded is 18,so
let us solve this as follows
first we shall find the intersecting points
as ,
9-x^2=b
x^2= 9-b
x=+-sqrt(9-b)
so the area bound by these curves y=b and y=9-x^2 is as follows
A= int _-sqrt(9-b) ^sqrt(9-b) (9-x^2-b)dx = 18
=> int _-sqrt(9-b) ^sqrt(9-b)(9-x^2-b)dx=18
=> [-bx +9x-x^3/3]_-sqrt(9-b) ^sqrt(9-b)
=>[x(9-b)-x^3/3]_-sqrt(9-b) ^sqrt(9-b)
=>[((sqrt(9-b))*(9-b))-[(sqrt(9-b))^(3)]/3 ]-[((-sqrt(9-b))*(9-b))-[(-sqrt(9-b))^(3)]/3]
=>[(9-b)^(3/2) - ((9-b)^(3/2))/3]-[-(9-b)^(3/2)-(-((9-b)^(3/2))/3)]
=>[(2/3)[9-b]^(3/2)]-[-(9-b)^(3/2)+((9-b)^(3/2))/3]
=>(2/3)[9-b]^(3/2) -[-(2/3)[9-b]^(3/2)]
=(4/3)[9-b]^(3/2)
but we know half the Area of the region between y=9-x^2,y=0 curves =18
so now ,
(4/3)[9-b]^(3/2)=18
 
let t= 9-b
=> t^(3/2)= 18*3/4
=> t=(27/2)^(2/3)
=> 9-b= 9/(root3 (4))
 
=> b= 9-9/(root3 (4))
 
=9(1-1/(root3 (4))) = 3.330
 
so b= 3.330