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

Question

Aurora Alvarado · Accepted Answer

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

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#

Axel Alvarez · Answer

undefined To find the area between the curves $x = 4 - y^2$ and $x = y - 2$, you first need to find the points of intersection of the two curves. Set $4 - y^2 = y - 2$ and solve for $y$. This gives you $y^2 + y - 6 = 0$, which factors to $(y + 3)(y - 2) = 0$. So, $y = -3$ or $y = 2$.

The area can be found by integrating the difference of the two curves with respect to $y$ from $-3$ to $2$, because the curves intersect at these points and $4 - y^2$ is the right curve between these limits:

$$
\text{Area} = \int_{-3}^{2} [(y - 2) - (4 - y^2)] \, dy
$$

Isaiah Allen · Answer

undefined To find the area between the curves $ x = 4 - y^2 $ and $ x = y - 2 $, you need to first determine the points of intersection between the two curves by setting them equal to each other and solving for $ y $. Once you have the $ y $-values for the points of intersection, you can integrate the difference of the two functions with respect to $ y $ over the interval between these $ y $-values. The integral represents the area between the curves. Remember to consider absolute values if the curves intersect in a way that one function lies above the other for some part of the interval and vice versa.