Find c>0 such that the area of the region enclosed by the parabolas y=x^2−c^2 and y=c^2−x^2 is 90.
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To find the value of c, such that the area of the region enclosed by the two parabolas is 90, we need to set up the integral and solve it.
The region enclosed by the parabolas can be defined as the area between the two curves. We can find this area by integrating the difference between the upper and lower curves with respect to x.
The given parabolas are:
1) y = x^2 − c^2
2) y = c^2 − x^2
To find the points of intersection between the two parabolas, we can equate the y-values, since they represent the same region:
x^2 − c^2 = c^2 − x^2
Simplifying the equation leads to:
2x^2 = 2c^2
x^2 = c^2
Taking the square root of both sides gives:
x = ±c
The points of intersection occur at x = -c and x = c.
Now, let's set up the integral to find the area between the two curves. We need to determine the upper and lower functions for the integration. Looking at the given equations, we can see that y = c^2 - x^2 is the upper curve and y = x^2 - c^2 is the lower curve (based on their positions in the coordinate plane).
The integral to calculate the area is:
A = ∫[a,b] [upper curve - lower curve] dx
In this case, the limits of integration are -c and c, since those are the points of intersection.
A = ∫[-c,c] [(c^2 - x^2) - (x^2 - c^2)] dx
Simplifying, we have:
A = ∫[-c,c] (2c^2 - 2x^2) dx
Now we need to integrate:
A = [2c^2x - (2/3)x^3] | from -c to c
Evaluating the integral, we get:
A = 2c^3 - (2/3)c^3 - [-2c^3 - (2/3)c^3]
A = 2c^3 - (2/3)c^3 + 2c^3 + (2/3)c^3
A = 6c^3
According to the problem statement, the area of the region enclosed by the parabolas is 90, so we have:
6c^3 = 90
Dividing both sides by 6:
c^3 = 15
Taking the cube root of both sides gives:
c = ∛15
Therefore, the value of c that satisfies the given condition and makes the area of the region enclosed by the parabolas equal to 90 is c = ∛15 (approximately 2.466).