Calculus of a Single Variable, Chapter 7, 7.1, Section 7.1, Problem 72

Given ,
y^2=4-x , x = 0
=>x=4-y^2 , x=0
first let us find the total area of the bounded by the curves.
so we shall proceed as follows
x=4-y^2 ,x=0
=> 4-y^2=0
=> y^2 -4 =0
=> (y-2)(y+2)=0
so y=+-2

the the area of the region is = int _-2 ^2 ((4-y^2)-0 ) dy
=>int _-2 ^2 (4-y^2) dy
=[4y-y^3/3] _-2 ^2
=[ [8-8/3]-[-8 -(-8)/3]]
=[[8-8/3]+[8 +(-8)/3]] = (16-16/3)=(2*16)/3=32/3
So now we have to find the vertical line that splits the region into two regions with area 16/3 as it is half of area of region covered by two curves y^2=4-x and x=0.
as when the line x=a intersects the curve x=4-y^2 then the area bounded is 16/3 ,so
let us solve this as follows
first we shall find the intersecting points
as ,
4-y^2=a
4-a=y^2
y=+-sqrt(4-a)
so the area bound by these curves x=a and x=4-y^2 is as follows
A= int _-sqrt(4-a) ^sqrt(4-a) (4-y^2-a)dy = 16/3
=> int _-sqrt(4-a) ^sqrt(4-a)(4-y^2-a)dy=16/3
=> [(4-a)(y)-y^3/3]_-sqrt(4-a) ^sqrt(4-a)
=>[(4-a)(sqrt(4-a))-(sqrt(4-a))^3/3]-[(4-a)(-sqrt(4-a))-(-sqrt(4-a))^3/3]
let t= sqrt(4-a)
so,
=>[t^2*(t)-(t)^3/3]-[t^2*(-t)-(-t)^3/3]
=>[t^3-t^3/3 +t^3-t^3/3]
=>2(t^3-t^3/3]
=>4/3t^3

but we know half the area of the region between x=4-y^2, x=0 curves =16/3
so now ,
4/3 t^3=16/3
=>4t^3 = 16
=>t^3=4
Substituting t = sqrt(4-a) ,
(4-a)^(3/2)= 4
4-a=4^(2/3)
a=4-4^(2/3)
=1.4801

so a= 1.4801