Volume of a solid with equilateral-triangle cross-sections

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A solid has as its base the region between $y=1-x^2$ and the $x$-axis on $[-1,1]$. The cross-sections perpendicular to the $x$-axis are equilateral triangles whose side is the height of the parabola at that point. (a) Write the cross-section area $A(x)$ (the area of an equilateral triangle of side $L$ is $\tfrac{\sqrt3}{4}L^2$); (b) compute the volume.

Solution

(a) The side of the cross-section at height $x$ is $L(x)=1-x^2$, so $A(x)=\frac{\sqrt3}{4}\,L(x)^2=\frac{\sqrt3}{4}(1-x^2)^2.$

(b) The integrand is even, so $\int_{-1}^1=2\int_0^1$: $V=\int_{-1}^1\frac{\sqrt3}{4}(1-x^2)^2\,dx=\frac{\sqrt3}{2}\int_0^1(1-2x^2+x^4)\,dx=\frac{\sqrt3}{2}\left[x-\frac{2x^3}{3}+\frac{x^5}{5}\right]_0^1=\frac{\sqrt3}{2}\cdot\frac{8}{15}=\frac{4\sqrt3}{15}\approx0.4619.$

$\boxed{V=\frac{4\sqrt3}{15}}$