Question

Find the value(s) of c such that the area of the region bounded by the parabolae y = x^2-c^2 and y = c^2 - x^2 is 4608

Answer (separate by commas): c= ??

Answer 1

Either
`c = 12` or
`c = (-12)`.

Explanation:

Find the
`x`-coordinate of the intersection of the two curves by equating their
`y`-values and solving for
`x`:

`x^(2) - c^(2) = c^(2) - x^(2)`.

`x^(2) = c^(2)`.

`x = c` or
`x = (-c)`.

In other words, the two curves intersect both when
`x = c` and when
`x = (-c)`.

Assume that
`c \ge 0`. The area enclosed would be over the range
`(-c) < x < c`. Note that over this range,
`x^(2) \le c^(2)`, so
`x^(2) - c^(2) \le 0 \le c^(2) - x^(2)`. The curve
`y = c^(2) - x^(2)` would be above
`y = x^(2) - c^(2)`.

Integrate the upper curve minus the lower curve
`(c^(2) - x^(2)) - (x^(2) - c^(2))` over this interval to find the area enclosed:

`\begin{aligned}& \int\limits_(-c)^(c) \left((c^(2) - x^(2)) - (x^(2) - c^(2))\right)\, dx \\ =\; & \int\limits_(-c)^(c) \left(2\, c^(2) - 2\, x^(2)\right)\, dx \\ =\; & {\left[2\, c^(2)\, x - (2)/(3)\, x^(3)\right]}_(-c)^(c) && (\text{power rule}) \\ =\; & \left(2\, c^(2)\, c -(2)/(3)\, c^(3)\right) - \left(2\, c^(2)\, (-c) - (2)/(3)\, (-c)^(3)\right) \\ =\; & (8)/(3)\, c^(3) \end{aligned}`.

In other words, assuming that
`c \ge 0`, area enclosed between these two curves would be
`(8/3)\, c^(3)`. Since this area is given, the value of
`c` can be found as:

`\begin{aligned}c &= \sqrt[3]{(3)/(8)\, (\text{Area})} = 12\end{aligned}`.

However, note that if
`c < 0`, the order of limits of the integral would need to be reversed:

`\begin{aligned}& \int\limits_(c)^(-c) \left((c^(2) - x^(2)) - (x^(2) - c^(2))\right)\, dx \\ =\; & \cdots \\ =\; & {\left[2\, c^(2)\, x - (2)/(3)\, x^(3)\right]}_(c)^(-c) \\ =\; & \left(2\, c^(2)\, (-c) - (2)/(3)\, (-c)^(3)\right) - \left(2\, c^(2)\, c -(2)/(3)\, c^(3)\right) \\ =\; & \left(-(8)/(3)\, c^(3)\right) \end{aligned}`.

Hence,
`c = (-12)` would also be a solution to this question.