
What are all the values of c such that the area of the region bounded by the parabolae y=x2-c2 and y=c2-x2 is 4608


We are given the parabolas \( y = x^2 - c^2 \) and \( y = c^2 - x^2 \), and we need to find the values of \( c \) such that the area of the region bounded by these curves is 4608 square units.

### Step 1: Identify the Intersection Points

The intersection points of the curves occur where \( y = x^2 - c^2 \) and \( y = c^2 - x^2 \) are equal:

\[
x^2 - c^2 = c^2 - x^2
\]

Simplifying this equation:

\[
2x^2 = 2c^2
\]

\[
x^2 = c^2
\]

\[
x = \pm c
\]

Thus, the curves intersect at \( x = -c \) and \( x = c \).

### Step 2: Set Up the Integral for the Area

The area \( A \) between the curves can be found by integrating the difference of the functions from \( x = -c \) to \( x = c \):

\[
A = \int_{-c}^{c} \left[(c^2 - x^2) - (x^2 - c^2)\right] \, dx
\]

Simplifying the integrand:

\[
A = \int_{-c}^{c} \left[2c^2 - 2x^2\right] \, dx
\]

\[
A = 2 \int_{-c}^{c} \left(c^2 - x^2\right) \, dx
\]

### Step 3: Compute the Integral

Now, compute the integral:

\[
A = 2 \left[ c^2x - \frac{x^3}{3} \right]_{-c}^{c}
\]

Evaluating at the limits:

\[
A = 2 \left[ \left(c^3 - \frac{c^3}{3}\right) - \left(-c^3 + \frac{(-c)^3}{3}\right)\right]
\]

Since \( (-c)^3 = -c^3 \):

\[
A = 2 \left[ \frac{2c^3}{3} - \left(-\frac{2c^3}{3}\right) \right]
\]

\[
A = 2 \times \frac{4c^3}{3}
\]

\[
A = \frac{8c^3}{3}
\]

### Step 4: Set the Area Equal to 4608

We are given that the area is 4608 square units, so:

\[
\frac{8c^3}{3} = 4608
\]

Multiply both sides by 3:

\[
8c^3 = 13824
\]

Divide by 8:

\[
c^3 = 1728
\]

Taking the cube root of both sides:

\[
c = \sqrt[3]{1728} = 12
\]

Thus, \( c = 12 \) is the value that satisfies the given area condition.

### Final Answer

The value of \( c \) that satisfies the condition is \( \boxed{12} \).

Would you like more details on any part of this solution, or do you have any other questions?

Here are 5 related questions:
1. How do you solve for the intersection points of more complex curves?
2. What methods can be used to approximate areas under curves when the integral is difficult to evaluate?
3. How would the area change if the parabolas had different coefficients?
4. What if the area was required between the curves for only a part of their intersection range?
5. Can this method be generalized to parabolas with higher powers or different coefficients?

**Tip:** Remember that when integrating symmetric functions over symmetric intervals, you can simplify calculations by considering just one half and doubling the result.

Learn how to find values of c such that the area of the region bounded by the parabolas y=x^2-c^2 and y=c^2-x^2 is 4608 square units. This problem involves calculus and integral calculus concepts, including finding intersection points and evaluating integrals. Suitable for advanced high school students and college calculus courses.

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