Question

Find the volume of the solid whose base is the region in the first quadrant bounded by y=x^3, y=1, and the y axis whose cross sections perpendicular to the y axis are equilateral triangles.

Answer 1

The region in question is the set

$R = \left\{ (x, y) : 0 \le x \le 1 \text{ and } x^3 \le y \le 1 \right\}$

or equivalently,

$R = \left\{ (x, y) : 0 \le y \le 1 \text{ and } 0 \le x \le \sqrt[3]{y} \right\}$

Cross sections are taken perpendicular to the y-axis, which means each section has a base length equal to the horizontal distance between the curve y = x³ and the line x = 0 (the y-axis). This horizontal distance is given by

y = x³   ⇒   x = ∛y

so that each triangular cross section has side length ∛y.

The area of an equilateral triangle with side length s is √3/4 s², so each cross section contributes an infinitesimal area of √3/4 ∛(y²).

Then the volume of this solid is

$\displaystyle \frac{\sqrt3}4 \int_0^1 \sqrt[3]{y^2} \, dy = \frac{\sqrt3}4 \int_0^1 y^{2/3} \, dy = \frac{\sqrt3}4\cdot\frac35 y^{5/3} \bigg|_0^1 = \boxed{\frac{3\sqrt3}{20}}$

I've attached some sketches of the solid with 16 and 64 such cross sections to give an idea of what this solid looks like.