# Decide whether to integrate with respect to x or y. Then find the area of the region. x=16−y2,x=y2−16?

Relevance
• Neither.

Intersections of x=16−y² and x=y²−16 are (0,4) and (0,-4), with a vertical separation of 4 - -4 = 8 units.

The distance between the two graphs on their intersection with y=0 is 16 - -16 = 32 units.

So the area is ⅔ (8) (32) = 512/3 square units. Source(s): The area of a parabola is two-thirds times the base times the perpendicular height. Area = | ⅔ (x₁-x₀) (f(x₂)-g(x₂)) | where the points of intersection of the two parabolas are (x₀,f(x₀)=g(x₀)) and (x₁, f(x₁)= g(x₁)) and x₂ = ½(x₀+x₁).
• Do you mean this?

x = 16 - y², x = y² - 16

And those equations define boundaries of a closed region, and you want the area of that region? Is that right? Communicate.

I would not integrate at all. The curves are parabolas. The region is symmetric on both coordinate axes. The x-axis is the central diameter of both parabolas, and the common chord is on the y-axis.

vertices: U(-16, 0), V(16, 0)

points of intersection: A(0, -4), B(0, 4)

area of rhombus AUBV = 1/2(UV)(AB) = 128

The closed region comprises two parabola segments. Its area is 4/3 the area of the rhombus.

4/3(128) = 512/3

I get that this was supposed to be a calculus problem, but with second-degree curves you can usually get along without the integration.

• Because the functions can have the same x value for different y values you must integrate with respect to y.

Set them equal to each other to find the limits of integration.

16−y^2 = y^2−16

32 = y^2

y = ± sqrt(32)

Now integrate the difference between y = -sqrt(32) to sqrt(32)

(y^2−16) - (16−y^2) dy

2y^2 - 32 dy

(2/3)y^3 - 32y + C