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`y^2 = 4-x , x = 0` Find a such that the line x = a divides the region bounded by the graphs of the equations into two regions of equal area.

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`

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