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

Apr 7, 2017

$\frac{125}{6} \approx 20.833 {\text{ units}}^{2}$

#### 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: $\left(- 5 , - 3\right) , \left(0 , 2\right)$

Sketch or graph the functions: $y = \pm \sqrt{4 - x} , y = x + 2$

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

$A = {\int}_{-} {3}^{2} \left[\left(4 - {y}^{2}\right) - \left(y - 2\right)\right] \mathrm{dy} = {\int}_{-} {3}^{2} \left(6 - {y}^{2} - y\right) \mathrm{dy}$

$A = 6 y - \frac{1}{3} {y}^{3} - \frac{1}{2} {y}^{2} {|}_{-} {3}^{2}$

$A = 12 - \frac{8}{3} - 2 - \left(18 + 9 - \frac{9}{2}\right) = 10 - \frac{8}{3} + 9 + \frac{9}{2}$

$A = 19 - \frac{8}{3} + \frac{9}{2} = \frac{114}{6} - \frac{16}{6} + \frac{27}{6}$

$A = \frac{125}{6} \approx 20.833 {\text{ units}}^{2}$