#### Answer

$c=\dfrac{\sqrt[3] 9}{4}$

#### Work Step by Step

The area of under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$Area, A= \int_{-\sqrt c}^{\sqrt c} (c-x^2-x^2+c) \ dx \\ = [2cx -\dfrac{2x^3}{3}]_{-\sqrt c}^{\sqrt c} \\=(2c^{3/2} -\dfrac{2 c^{3/2}}{3})-(-2c^{3/2} +\dfrac{2 c^{3/2}}{3}) \\=\dfrac{8 c^{3/2}}{3}$
Since, $Area= 1$
$\dfrac{8 c^{3/2}}{3} =1 \implies c=\dfrac{\sqrt[3] 9}{4}$