## Calculus (3rd Edition)

$c=\dfrac{\sqrt[3] 9}{4}$
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, $Area= 1$ $\dfrac{8 c^{3/2}}{3} =1 \implies c=\dfrac{\sqrt[3] 9}{4}$