Question

How do you find the area between `x=4-y^2` and `x=y-2`?

Answer 1

`125/6 ~~ 20.833 " units"^2`

Explanation:

First you need to find the intersection point(s) between the two curves by setting the two functions equal since they share the same points at the intersections:
`x = 4 - y^2; " " x = y - 2`

`4 - y^2 = y - 2`

Rearrange: `y^2 + y -6 = 0`

Factor: `(y + 3)(y - 2) = 0; " so " y = -3, 2`

Intersection points: `(-5, -3), (0, 2)`

Sketch or graph the functions: `y = +- sqrt(4-x), y = x+2`

The sketch reveals that the function `x = 4 - y^2` is to the right.

`A = int_-3^2 [(4 - y^2)- (y - 2)] dy = int_-3^2 (6 - y^2 - y) dy`

`A = 6y - 1/3y^3 - 1/2y^2 |_-3^2`

`A = 12 - 8/3 - 2 - (18 + 9 - 9/2) = 10 - 8/3 + 9 + 9/2`

`A = 19 - 8/3 + 9/2 = 114/6 - 16/6 + 27/6`

`A = 125/6 ~~ 20.833 " units"^2`