Find c>0 such that the area of the region enclosed by the parabolas y=x2−c2 and y=c2−x2 is 230.
the x-intercepts for both functions are
c and -c
If you look at your sketch there is symmetry about the y-axis as well, so we could just take our area from 0 to c
4∫(c^2 - x^2) dx = 230 from 0 to c
∫(c^2 - x^2) dx
= [c^2 x- (1/3)x^3] from 0 to c
= (c^3 - (1/3)c^3 - 0
= (2/3)c^3
4(2/3)c^3 = 230
c^3 = 86.25
c = 4.418...
check my arithmetic
To find the value of c such that the area of the region enclosed by the parabolas is 230, we need to use integration to find the area between the two curves.
First, let's find the x-coordinates of the points where the two parabolas intersect. We can set the two equations equal to each other:
x^2 - c^2 = c^2 - x^2
2x^2 = 2c^2
x^2 = c^2
Taking the square root of both sides, we get:
x = ± c
Now, we can set up the integral to find the area between the two parabolas. The formula for finding the area between two curves is:
Area = ∫[a, b] [f(x) - g(x)] dx
where a and b are the x-coordinates of the intersection points, and f(x) and g(x) are the equations of the curves.
In this case, the area is given to be 230, so we set up the equation:
230 = ∫[-c, c] [(x^2 - c^2) - (c^2 - x^2)] dx
Simplifying the equation, we get:
230 = ∫[-c, c] (2x^2 - 2c^2) dx
Using the properties of integrals, we integrate each term separately:
230 = [2/3 * x^3 - 2c^2 * x] |[-c, c]
Plugging in the limits of integration, we get:
230 = [2/3 * c^3 - 2c^3] - [2/3 * (-c)^3 - 2c^3]
Simplifying further:
230 = (2/3 * c^3 - 2c^3) - (2/3 * c^3 + 2c^3)
Combining like terms:
230 = -4c^3
Dividing both sides by -4, we get:
-57.5 = c^3
Taking the cube root of both sides:
c ≈ -3.786
However, this value of c is negative, and we are given that c > 0. Hence, there is no c>0 such that the area of the region enclosed by the parabolas is 230.
To find the value of c that satisfies the given condition, we need to determine the intersection points of the two parabolas and calculate the area enclosed between them.
Let's start by finding the intersection points of the parabolas y = x^2 - c^2 and y = c^2 - x^2.
Setting the equations equal to each other, we get:
x^2 - c^2 = c^2 - x^2
Simplifying the equation, we have:
2x^2 = 2c^2
Dividing both sides by 2, we obtain:
x^2 = c^2
Taking the square root of both sides, we get two cases:
Case 1: x = c
Case 2: x = -c
Now, let's calculate the y-coordinates of these intersection points by substituting the x-values into either of the original equations.
For Case 1 (x = c):
y = (c)^2 - c^2 = 0
For Case 2 (x = -c):
y = c^2 - (-c)^2 = c^2 - c^2 = 0
So, both intersection points have a y-coordinate of 0.
To find the area enclosed between the parabolas, we can integrate the difference between the upper parabola (y = x^2 - c^2) and the lower parabola (y = c^2 - x^2) with respect to x.
The integral expression is:
∫[a,b] (x^2 - c^2) - (c^2 - x^2) dx
To determine the limits of integration, a and b, we need to find the x-values where the two parabolas intersect.
From the previous calculations, we know that the intersection points are x = c and x = -c.
Therefore, the limits of integration are a = -c and b = c.
The integral expression becomes:
∫[-c,c] (x^2 - c^2) - (c^2 - x^2) dx
Simplifying, we have:
∫[-c,c] 2x^2 dx
Evaluating the integral, we get:
[(2/3)x^3] from -c to c
Substituting the limits, we obtain:
[(2/3)c^3] - [(2/3)(-c)^3]
Since (c)^3 = (-c)^3, the expression simplifies to:
[(2/3)c^3] - [(2/3)c^3]
The terms cancel out:
[(2/3)c^3] - [(2/3)c^3] = 0
Therefore, the area enclosed between the parabolas y = x^2 - c^2 and y = c^2 - x^2 is always 0, regardless of the value of c.