# How do you find the volume of the solid with base region bounded by the curve y=1-x^2 and the x-axis if cross sections perpendicular to the x-axis are isosceles triangles with height equal base?

Since its triangular cross-sectional area $A \left(x\right)$ can be found by
$A \left(x\right) = \frac{1}{2}$(Base)(Height)$= \frac{1}{2} {\left(1 - {x}^{2}\right)}^{2}$,
the volume $V$ of the solid can be found by
$V = \frac{1}{2} {\int}_{- 1}^{1} {\left(1 - {x}^{2}\right)}^{2} \mathrm{dx}$
$= {\int}_{0}^{1} \left(1 - 2 {x}^{2} + {x}^{4}\right) \mathrm{dx}$
$= {\left[x - \frac{2}{3} {x}^{3} + \frac{1}{5} {x}^{5}\right]}_{0}^{1} = 1 - \frac{2}{3} + \frac{1}{5} = \frac{8}{15}$