Find the area bounded by y=8-x^2 and y=x^2.

First, we'll have to identify the limits of integration.
For this rason, we'll determine the intercepting points of the
graphs.

Since y = x^2 and y = 8-x^2, we'll
get:

x^2 = 8-x^2

We'll move
all terms to one side:

x^2 - 8 + x^2 =
0

2x^2 - 8 = 0

We'll divide by
2:

x^2 - 4 = 0

x^2 =
4

x1 = 2 and x2 = -2

Over the
interval [-2 ; 2] the graph 8-x^2 >= x^2

The area of
the region is the definite integral of the function 8 - x^2 - x^2, for x = -2 to x =
2.

A = Int (8 - x^2 - x^2)dx

A
= Int (8 - 2x^2)dx

A = Int 8dx -  Int
2x^2dx

A = 8x - 2x^3/3, for x = -2 to x =
2

A = 8(2 + 2) - (2/3)(8 +
8)

A = 32 - 32/3

A =
(96-32)/3

A = 64/3 square
units.

The area is A =  64/3 square
units.