Find the values of c such that the area of the region bounded by the parabolas
y = 4x^2 − c^2 and y = c^2 − 4x^2
is 256/3. (Enter your answers as a comma-separated list.)
The parabolas intersect at x = -c/2 and c/2, so we want
∫[-c/2,c/2] (c^2-4x^2)-(4x^2-c^2) dx = 256/3
4c^3/3 = 256/3
c=4
naturally, c = -4 also works
To find the values of c such that the area of the region bounded by the given parabolas is 256/3, we need to set up an integral and solve it.
First, let's find the x-coordinate(s) of intersection points between the two parabolas. To do this, equate the two equations:
4x^2 - c^2 = c^2 - 4x^2
Combine like terms:
8x^2 = 2c^2
Simplify:
4x^2 = c^2
Take the square root of both sides, noting that c could be positive or negative:
2x = ±c
Now let's set up the integral to calculate the area between the two curves. Since we have two curves, we need to integrate twice and subtract the smaller integral from the larger one:
Area = ∫(c^2 - 4x^2) dx - ∫(4x^2 - c^2) dx
To find the bounds of integration, let's find the x-coordinate(s) of the intersection points. Since we already know that 2x = ±c, we can solve for x:
x = ±c/2
The area between two curves can be found by integrating from one intersection point to the other. Therefore, the bounds of integration will be -c/2 to c/2.
Now let's calculate the area between the curves:
Area = ∫(c^2 - 4x^2) dx - ∫(4x^2 - c^2) dx
= [c^2x - 4x^3/3] evaluated from -c/2 to c/2 - [4x^3/3 - c^2x] evaluated from -c/2 to c/2
= (c^3/2 - 4c^3/24) - (-c^3/2 - 4c^3/24) - (4c^3/24 - c^3/2) + (-4c^3/24 + c^3/2)
= 3c^3/4 - 3c^3/4
= 0
Since the area between the two curves is 0, we need to find when the area is 256/3. Therefore, we need to find the values of c such that the area between the two curves is 256/3.
0 = 256/3
This equation has no solutions, meaning there exist no values of c such that the area of the region bounded by the given parabolas is 256/3.
Therefore, there are no values of c that satisfy the given condition.