Question

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

Accepted Answer

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.

Answer

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.

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

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