Calculus of a Single Variable, Chapter 7, 7.1, Section 7.1, Problem 15

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
x=4-y^2
=> y^2=4-x
y=sqrt(4-x) -------------(1)
and
x=y-2
=> y=x+2---------(2)
let us find the curves where they intersect
so ,
sqrt(4-x) =x+2
4-x=(x+2)^2
4-x=x^2+4x+4
=>x^2+5x=0
=> x(x+5)=0
=>x=0 or x=-5
and the let let us get the points where they intersect with respect to y
4-y^2=y-2
6=y+y^2
=> y^2+y-6=0
=>y^2+3y-2y-6=0
=>y(y+3)-2(y+3)=0
=>(y-2)(y+3)=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
=[(-2/3)(4)^(3/2)]-[25/2-10-(2/3)(9)^(3/2)]+[0]-[(-4/3)(4)^(3/2)]
=-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
=-125/6
= -20.833

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