`x = 4-y^2 , x = y-2` Find the area of the region by integrating with (a) respect to x and (b) respect to y

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`x=4-y^2 , x=y-2` 

(a) first let us find the area of the region with respect to x

so ,

the lines are


=> `y^2=4-x`

`y=sqrt(4-x)` -------------(1)



=> `y=x+2`---------(2)

let us find the curves where they intersect

so ,

`sqrt(4-x) =x+2`




=>` x(x+5)=0`

=>`x=0 or x=-5`

and the let let us get the points where they intersect with respect to y



=> `y^2+y-6=0`




so `y=2 or y= -3`


so the points of intersection of the curves are `(0,2)` & `(-5,-3)`

but the curve `x=4-y^2` is beyond x=0 and intersects the x-axis (setting y=0) at x=4.


So the area of the region with respect to x -axis needs to have two integrals.

Area=`int_-5^0[(x+2)-(-(sqrt(4-x)))] dx + ` 

`int_0^4[(sqrt(4-x)-(-(sqrt(4-x)))]` `dx` 


=`int_-5^0 [(x+2)+((sqrt(4-x)))] dx`
`+int_0^4 [(sqrt(4-x)+((sqrt(4-x)))] dx`

=`[x^2/2+2x-2/3(4-x)^(3/2)]_-5^0 +[-2/3(4-x)^(3/2)-2/3(4-x)^(3/2)]_0^4 `


=`-16/3+31/2 +0+32/3`

=`61/6+32/3` =`125/6`

 so the Area is `125/6 `


`(b)` area of the region with respect to y is

    Area = `int _-3 ^2 [(y-2)-(4-y^2)] dy`

   = `[y^2/2 -2y -4y+y^3/3]_-3 ^2`

  =` [4/2 -4 -8 +8/3]-[9/2 +6+12-27/3]`

  `=-22/3 -27/2`


`= -20.833`


But since the area cannot be negative then the area is `20.833` or `125/6`

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