# What is the volume of the solid given the base of a solid is the region in the first quadrant bounded by the graph of #y=-x^2+5x-4^ and the x-axis and the cross-sections of the solid perpendicular to the x-axis are equilateral triangles?

Mar 22, 2015

The volume is $\setminus \frac{81 \setminus \sqrt{3}}{40} \setminus \approx 3.5074$.

The curve $y = - {x}^{2} + 5 x - 4 = - \left(x - 1\right) \left(x - 4\right)$ has $x$-intercepts at $x = 1$ and $x = 4$ (and is above the $x$-axis for $1 < x < 4$).

For $1 < x < 4$, let $b \left(x\right) = - {x}^{2} + 5 x - 4$. This will be the base of the equilateral triangle cross-section at $x$.

The solid itself looks something like this: Now draw the cross-section equilateral triangle, label the sides $b \left(x\right)$, and draw a vertical line for the height $h \left(x\right)$ from the top vertex down perpendicular to the base.
By the Pythagorean Theorem, ${\left(b \left(x\right)\right)}^{2} = {\left(\setminus \frac{b \left(x\right)}{2}\right)}^{2} + {\left(h \left(x\right)\right)}^{2}$. so that $h \left(x\right) = \setminus \sqrt{{\left(b \left(x\right)\right)}^{2} - {\left(\setminus \frac{b \left(x\right)}{2}\right)}^{2}} = \setminus \sqrt{\setminus \frac{3}{4} {\left(b \left(x\right)\right)}^{2}} = \setminus \frac{\setminus \sqrt{3}}{2} b \left(x\right)$.

The cross-sectional area is $A \left(x\right) = \setminus \frac{1}{2} b \left(x\right) h \left(x\right) = \setminus \frac{\setminus \sqrt{3}}{4} {\left(b \left(x\right)\right)}^{2} = \setminus \frac{\setminus \sqrt{3}}{4} \left({x}^{4} - 10 {x}^{3} + 33 {x}^{2} - 40 x + 16\right)$, which means the volume of the solid is $V = \setminus {\int}_{1}^{4} A \left(x\right) \mathrm{dx} = \setminus \frac{\setminus \sqrt{3}}{4} \setminus {\int}_{1}^{4} \left({x}^{4} - 10 {x}^{3} + 33 {x}^{2} - 40 x + 16\right) \mathrm{dx} = \setminus \frac{\setminus \sqrt{3}}{4} \left(\setminus \frac{{x}^{5}}{5} - \setminus \frac{5 {x}^{4}}{2} + 11 {x}^{3} - 20 {x}^{2} + 16 x\right) {|}_{x = 1}^{x = 4}$.

Continuing to simplify gives $V = \setminus \frac{\setminus \sqrt{3}}{4} \left(\left(\setminus \frac{1024}{5} - \setminus \frac{1280}{2} + 704 - 320 + 64\right) - \left(\setminus \frac{1}{5} - \setminus \frac{5}{2} + 11 - 20 + 16\right)\right) = \setminus \frac{\setminus \sqrt{3}}{4} \left(\setminus \frac{64}{5} - \setminus \frac{47}{10}\right) = \setminus \frac{81 \setminus \sqrt{3}}{40}$.