# How do you find the volume of the solid with base region bounded by the curves y=1-x^2 and y=x^2-9 if cross sections perpendicular to the x-axis are squares?

Nov 3, 2014

Since the length of a square cross-section can be expressed by

$1 - {x}^{2} - \left({x}^{2} - 9\right) = 10 - 2 {x}^{2} = 2 \left(5 - {x}^{2}\right)$,

the area of the cross-section can be found by

$A \left(x\right) = {\left[2 \left(5 - {x}^{2}\right)\right]}^{2} = 4 \left(25 - 10 {x}^{2} + {x}^{4}\right)$.

Let us find the $x$-coordinates of the intersections of the two parabola.

$2 \left(5 - {x}^{2}\right) = 0 \implies {x}^{2} = 5 \implies x = \pm \sqrt{5}$,

Since the base of the solid spans from $- \sqrt{5}$ to $\sqrt{5}$, the volume $V$ of the solid can be found by

$V = 4 {\int}_{- \sqrt{5}}^{\sqrt{5}} \left(25 - 10 {x}^{2} + {x}^{4}\right) \mathrm{dx}$

by using the symmetry (even function),

$= 8 {\int}_{0}^{\sqrt{5}} \left(25 - 10 {x}^{2} + {x}^{4}\right) \mathrm{dx}$

$= 8 {\left[25 x - \frac{10}{3} {x}^{3} + \frac{1}{5} {x}^{5}\right]}_{0}^{\sqrt{5}}$

$= 8 \left(25 \sqrt{5} - \frac{50}{3} \sqrt{5} + 5 \sqrt{5}\right) = \frac{320}{3} \sqrt{5}$.

I hope that this was helpful.