Sam G's question at Yahoo! Answers regarding the volume of a truncated square pyramid

MarkFL

Staff member
Here is the question:

Truncated Pyramid Question?

If the angle of repose is given as 32 degrees, the height is 18cm and the top square is 20*20cm, can I calculate the volume?
I have posted a link there to this topic so the OP can see my work.

MarkFL

Staff member
Hello Sam G,

One method we can use to calculate the volume of the pyramid is to use a technique from integral calculus called volumes by slicing. We will slice the pyramid into horizontal square slices and then add the slices by integrating.

I would choose to orient the cross-section of the pyramid as follows: We can now see that the volume of an arbitrary slice, a square slice of side length $s$ and thickness $dy$ is:

$$\displaystyle dV=s^2\,dy$$

where:

$$\displaystyle s=w+2x=w+2\cot(\theta)y$$

and so we have:

$$\displaystyle dV=\left(w+2\cot(\theta)y \right)^2\,dy= \left(w^2+4w\cot(\theta)y+4\cot^2(\theta)y^2 \right)\,dy$$

Now, summing the slices by integration, we find:

$$\displaystyle V=\int_0^h w^2+4w\cot(\theta)y+4\cot^2(\theta)y^2\,dy$$

Applying the FTOC, we obtain:

$$\displaystyle V=\left[w^2y+2w\cot(\theta)y^2+\frac{4}{3}\cot^2(\theta)y^3 \right]_0^h=w^2h+2w\cot(\theta)h^2+\frac{4}{3}\cot^2( \theta)h^3$$

$$\displaystyle V=\frac{h}{3}\left(4\cot^2( \theta)h^2+6w\cot(\theta)h+3w^2 \right)$$

Using the given data:

$$\displaystyle h=18\text{ cm},\,w=20\text{ cm},\,\theta=32^{\circ}$$

we find:

$$\displaystyle V\approx47855.220519615943\text{ cm}^3$$