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

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### 2 Answers

We first determine the points where the curves y = 8 - x^2 and y = x^2, meet.

8 - x^2 = x^2

=> x^2 = 4

=> x = 2 , x = -2

Now we find the integral of 8 - x^2 - x^2 between the limits x = -2 and x = 2

Int [ 8 - 2x^2 ]

=> 8x - 2x^3/3

Between the limits x = -2 and x = 2

8x - 2x^3/3 - 8x + 2x^3/3

=> 8*2 - 2*8/3 + 8*2 - 2*8/3

=> 32 - 32/3

=> 64/3

**The area bounded by the curves is 64/3.**

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.**