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:
View attachment 1132
We can now see that the volume of an arbitrary slice, a square slice of side length $s$ and thickness $dy$ is:
$$dV=s^2\,dy$$
where:
$$s=w+2x=w+2\cot(\theta)y$$
and so we have:
$$dV=\left(w+2\cot(\theta)y \right)^2\,dy= \left(w^2+4w\cot(\theta)y+4\cot^2(\theta)y^2 \right)\,dy$$
Now, summing the slices by integration, we find:
$$V=\int_0^h w^2+4w\cot(\theta)y+4\cot^2(\theta)y^2\,dy$$
Applying the FTOC, we obtain:
$$V=\left[w^2y+2w\cot(\theta)y^2+\frac{4}{3}\cot^2(\theta)y^3 \right]_0^h=w^2h+2w\cot(\theta)h^2+\frac{4}{3}\cot^2( \theta)h^3$$
$$V=\frac{h}{3}\left(4\cot^2( \theta)h^2+6w\cot(\theta)h+3w^2 \right)$$
Using the given data:
$$h=18\text{ cm},\,w=20\text{ cm},\,\theta=32^{\circ}$$
we find:
$$V\approx47855.220519615943\text{ cm}^3$$
