Calculate polygon using some integrals

Question

I have a polygon below. The length of it is $a$, base is $b$, and height is $h$. I want to have an integral to calculate volume of it. This is so different from what I know before, and it's not graph volume with disks. But, I have a big hint that the volume is $$ \frac{\sqrt{3} h}{12} \bigl( a^2 + ab + b^2 \bigr). $$ I want to ask what to make to set up the calculation. And how to show that the volume is the answer here. I have ideas, like using the method of calculating volume of disk, $\pi r_{\rm out} - \pi r_{\rm in}$. But where do I find radius? Thank you for helping..

Answer 1

Assuming that the base of this frustum of a pyramid has an equilateral triangle cross section, the first thing you have to work out is the area of the triangle in terms of its side length: $$ A = \frac{\sqrt{3}}{4} s^2 $$

Next you imagine slicing the solid region perpendicular to the height, so that each slice is an equilateral triangle. These triangles vary in size, with their side lengths varying linearly from $s = b$ to $s = a$ over the course of a height change of $h$. Let's parametrize this side length in terms of this vertical coordinate $z$: $$ s = b + \frac{a-b}{h} z. $$ Notice that this parametrization gives $s=b$ when $z=0$ and $s=a$ when $z=h$, as it must. Substituting this expression for $s$ into the triangle area formula yields: $$ \begin{align*} A(z) &= \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \biggl( b + \frac{a-b}{h} z \biggr)^{\!2} = \frac{\sqrt{3}}{4} \cdot \frac{1}{h^2} \Bigl( bh + (a-b) z \Bigr)^2 \\ &= \frac{\sqrt{3}}{4h^2} \bigl( b^2h^2 + 2bh(a-b)z + (a-b)^2 z^2 \bigr) \end{align*} $$

Now, we assemble all these cross-sectional areas in an integral to find the volume: $$ \begin{align*} V = \int_0^h A(z) \, dz &= \frac{\sqrt{3}}{4h^2} \int_0^h \bigl( b^2h^2 + 2bh(a-b)z + (a-b)^2 z^2 \bigr) \, dz \\ &= \frac{\sqrt{3}}{4h^2} \biggl. \Bigl( b^2h^2z + bh(a-b)z^2 + (a-b)^2 \cdot \frac{1}{3} z^3 \Bigr) \biggr\rvert_0^h \\ &= \frac{\sqrt{3}}{4h^2} \Bigl( b^2h^2(h) + bh(a-b)(h)^2 + (a-b)^2 \cdot \frac{1}{3}(h)^3 \Bigr) \\ &= \frac{\sqrt{3}}{4h^2} \cdot \frac{h^3}{3} \Bigl( 3b^2 + 3b(a-b) + (a-b)^2 \Bigr) \\ &= \frac{\sqrt{3}h}{12} \Bigl( a^2 + ab + b^2 \Bigr) \end{align*} $$

Answer 2

The side length of the equilateral triangle at the base is $a$ and at height $h$ is $b$. So at height $x$, the side length will be $a - \frac{(a-b)x}{h}, 0 \leq x \leq h$.

The area of equilateral triangle at height $x$, $A(x) = \frac{\sqrt3}{4}\big(a - \frac{(a-b)x}{h}\big)^2$.

Now at height $x$, if we consider the volume of a very thin wedge of this solid, of thickness $dx$,

$dV = A(x) \cdot dx$

So $V = \displaystyle \int_0^h \frac{\sqrt3}{4h^2}\big(a(h-x) + bx\big)^2 \ dx$

$ = \frac{\sqrt3}{4h^2}\displaystyle \int_0^h \big(a^2(h-x)^2 + b^2x^2 + 2 abx(h-x)\big) \ dx$

$ = \displaystyle \frac{\sqrt3}{12} (a^2 + ab + h^2) h$