How to split pyramid volume into slices for Riemann sum.

Question

I'm working on a problem: A pyramid of height H has a square base a by a. Find the area A at height h. If the rock used to build the pyramid weighs approximately w=100lb/ft3, how much work did it take to lift all the rock? source

My plan is:

• split the pyramid into infinitely thin layers, obtain formula for volume of each individual slice;
• multiply volume by density given to get weight of slice;
• multiply weight by earth gravity g to get force needed to lift slice;
• multiply force by distance the slice needs to be lifted to get work;
• use resulting work formula in a Riemann sum and convert it to a definite integral;

I need a check on volume formula. I took x axis starting at the top of the pyramid and pointing down.

Side of slice: $$ a_i=a\cdot \frac{x_i}{H}=\frac{a}{H}x_i $$

Area of slice:

$$ A_i=\left(\frac{a}{H}\right)^2x^2_i $$

Using pyramid frustum volume formula:

$$ V_{frustrum}=\frac{1}{3}\cdot h_{frustrum}\cdot \left(A_{top}+A_{bottom}+\sqrt{A_{top}\cdot A_{bottom}}\right) $$

I get volume of a slice:

$$ V_i=\frac{1}{3}\cdot \left(x_i-x_{i-1}\right)\cdot \left(\left(\frac{a}{H}\right)^2x^2_{i-1}+\left(\frac{a}{H}\right)^2x^2_i+\sqrt{\left(\frac{a}{H}\right)^2x^2_{i-1}\cdot \left(\frac{a}{H}\right)^2x^2_{i}}\right) $$

$$ V_i=\frac{1}{3}\cdot Δx\cdot \left(\frac{a}{H}\right)^2\left(x^2_{i-1}+x^2_i+x_{i-1}\cdot x_i\right) $$

Now, my question, can I assume $$ x_{i-1}=\:x_i $$

and simplify the above to

$$ V_i=\frac{1}{3}\cdot \:Δx\cdot \:\left(\frac{a}{H}\right)^2\cdot 3x^2_i=\:Δx\cdot \:\:\left(\frac{a}{H}\right)^2\cdot \:x^2_i $$

?

And in general, when can I assume $x_{i-1}=\:x_i$ ?

Answer 1

The idea of doing a Riemann sum is that you use square blocks, not frustums! You know that the volume of a frustum lies between $$\frac{a^2}{H^2}x_{i-1}^2(x_i-x_{i-1})\le V_i\le\frac{a^2}{H^2}x_i^2(x_i-x_{i-1})$$ And so the biggest error you can make by choosing either $x_i$ or $x_{i-1}$ is no more than $$\begin{align}\Delta V_i&\le\frac{a^2}{H^2}(x_i^2-x_{i-1}^2)(x_i-x_{i-1})\\ &=\frac{a^2}{H^2}(x_i+x_{i-1})(x_i-x_{i-1})^2\le2\frac{a^2}{H}(x_i-x_{i-1})\max(x_i-x_{i-1})\end{align}$$ So the biggest error you could make in volume is $$\Delta V\le2\frac{a^2}{H}\max(x_i-x_{i-1})\sum_{i=1}^N(x_i-x_{i-1})=2a^2\max(x_i-x_{i-1})$$ Thus you can make the error as small as you want by making the biggest step in $x$ small enough. Then if we let $x_i=\frac{iH}N$ so $x_i-x_{i-1}=\frac HN$ we get $$\begin{align}V&=\lim_{N\rightarrow\infty}\sum_{i=1}^N\frac{a^2}{H^2}x_i^2(x_i-x_{i-1})=\lim_{N\rightarrow\infty}\sum_{i=1}^N\frac{a^2Hi^2}{N^3}\\ &=\lim_{N\rightarrow\infty} \frac{a^2HN(N+1)(2N+1)}{6N^3}=\frac{a^2H}{3}\end{align}$$ Just a hint of course because you seem to be on the right track otherwise.