Find the area of the region between the curves y=8−x2 y=x2 x=−3 and x=3
To find the area of the region between the curves, you can use the concept of definite integration. Here's how you can approach the problem:
Step 1: Determine the limits of integration.
In this case, the curves intersect at x = -3 and x = 3, so those will be our limits of integration.
Step 2: Determine the upper and lower functions.
Since the curves are given by y = 8 - x^2, y = x^2, x = -3, and x = 3, we need to determine which curve is the upper function and which one is the lower function in the given interval.
To do this, we can compare the y-values of the curves at a few points within the interval, such as x = 0.
For x = 0, The value of y for the first curve y = 8 - x^2 is y = 8 - (0)^2 = 8.
The value of y for the second curve y = x^2 is y = (0)^2 = 0.
Since the first curve y = 8 - x^2 has a higher y-value at x = 0 than the second curve y = x^2, the first curve is the upper function and the second curve is the lower function within the given interval.
Step 3: Set up the integral.
Since the curves intersect at x = -3 and x = 3, the integral will be the difference between the integrals of the upper and lower functions.
The area can be calculated using the following integral:
A = ∫(upper function - lower function) dx
In this case, the upper function is y = 8 - x^2 and the lower function is y = x^2. Therefore, the integral becomes:
A = ∫(8 - x^2) - x^2 dx
Step 4: Evaluate the integral.
Integrate the expression (8 - x^2) - x^2 with respect to x between the limits of integration x = -3 and x = 3.
A = ∫[8 - 2x^2] dx
= [8x - (2/3)x^3] between x = -3 and x = 3
Evaluate the expression at the upper limit (x = 3) and subtract the value at the lower limit (x = -3):
A = [8(3) - (2/3)(3)^3] - [8(-3) - (2/3)(-3)^3]
= [24 - 18] - [-24 + 18]
= 6 - (-6)
= 6 + 6
= 12
Therefore, the area of the region between the curves y = 8 - x^2, y = x^2, x = -3, and x = 3 is 12 square units.
To find the area between the curves, we need to integrate the difference of the curves over the given interval.
First, let's find the points of intersection of the curves by setting them equal to each other:
8 - x^2 = x^2
Simplifying the equation, we have:
2x^2 = 8
Dividing both sides by 2, we get:
x^2 = 4
Taking the square root of both sides, we get two solutions:
x = 2 and x = -2
We have two different regions to calculate:
1. From x = -3 to x = -2.
2. From x = 2 to x = 3.
Let's start with the first region:
∫[x = -3 to -2] (8 - x^2 - x^2) dx
Simplifying the integral:
∫[x = -3 to -2] (8 - 2x^2) dx
Now integrate:
[8x - (2/3)x^3] | from -3 to -2
Plugging in the values:
[(8 * -2) - (2/3)*(-2)^3] - [(8 * -3) - (2/3)*(-3)^3]
Simplifying:
(-16 + (16/3)) - (-24 + (54/3))
(-16 + 16/3) - (-24 + 54/3)
(-48 + 16) / 3
-32/3
Now let's calculate the area for the second region:
∫[x = 2 to 3] (8 - x^2 - x^2) dx
Simplifying the integral:
∫[x = 2 to 3] (8 - 2x^2) dx
Now integrate:
[8x - (2/3)x^3] | from 2 to 3
Plugging in the values:
[(8 * 3) - (2/3)*(3)^3] - [(8 * 2) - (2/3)*(2)^3]
Simplifying:
(24 - (54/3)) - (16 - (16/3))
(24 - 18) - (16 - 16/3)
6 - 16/3
(18/3) - (16/3)
2/3
Finally, add the two areas together to get the total area between the curves:
(-32/3) + (2/3)
-30/3
-10
Therefore, the area between the curves y = 8 - x^2, y = x^2, x = -3, and x = 3 is -10 units squared.