# Find the values of c such that the area...?

## Find the 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 576.

Sep 19, 2017

$c = 6$

#### Explanation:

The two curves are:

${y}_{1} \left(x\right) = {x}^{2} - {c}^{2}$

and

${y}_{2} \left(x\right) = {c}^{2} - {x}^{2}$

We can note that for every $x$ we have: ${y}_{1} \left(x\right) = - {y}_{2} \left(x\right)$ so the two parabolas are symmetric with respect to the $x$ axis.The two curves thus intercept when ${y}_{1} \left(x\right) = {y}_{2} \left(x\right) = 0$, that is for $x = \pm c$

Given the symmetry, the area bounded by the two parabolas is twice the area bounded by either parabola and the $x$ axis.

If we choose ${y}_{2} \left(x\right) = {c}^{2} - {x}^{2}$, which is positive in the interval, we thus have:

$A = 2 {\int}_{- c}^{c} \left({c}^{2} - {x}^{2}\right) \mathrm{dx} = 2 {\left[{c}^{2} x - {x}^{3} / 3\right]}_{- c}^{c} = \frac{8}{3} {c}^{3}$

and posing

$\frac{8}{3} {c}^{3} = 576$

we get:

$c = \sqrt[3]{\frac{3 \times 576}{8}} = 6$