Calculate the volume of mount Fuji using the disc method.

Question

Problem

I am trying to approximate the volume of Mount Fuji using the disc method. Now, I know that the precise result depends on which sedimentary materials the mountain is made from (as these have different weights). However, I wanted to see if I could get in the right ballpark using simple calculus.

Attempt

According to some sources the base of Mount Fuji is $50$km, while the height of $3776$ meters, it looks like this

Here is a summary I found

Using some very simple Calculus I thought I could approximate the mountain using the following cosine function

$$ z(r) = \frac{h}{2} \cos \frac{\pi r}{w} + \frac{h}{2} $$

Where $h = 3776$m is the height, and $w=50\cdot 10^3$m is the radius of the base. Overlaying this function over Mount Fuji it looks roughly like this

Now I have the height as a function of the radius, but in order to use the cylinder method I need to express the radius in terms of the height. Thus

$$ r(z) = \frac{2w}{\pi} \arccos \sqrt{\frac{z}{h}} $$

The next step would be to sum up thin slizes of size $\pi r(z)^2$ where $h$ get smaller and smaller. E.g doing the integration similar to the following picture

Giving

$$ V = \int_0^h \pi z(r)^2 \mathrm{d}z $$

Using the formula above and my values of $w = 50\cdot 10^3$m and $h=3776$ I obtain $$ V = 4.72\cdot 10^{12} \Bigl(\pi-\frac{4}{\pi} \Bigr) \text{m} \approx 8.8186\cdot 10^{12}\text{m} $$ after some troublesome calculus. Converting my answer back to km$^3$ I finally get

$$ V \approx 8819 \text{km}^3 $$

However, this answer is way to high. It is not even the correct magnitude! My question is where does my mistake lie?

Questions

• Is there a better way to roughly calculate the volume of mount Fuji using the shell method?
• Is the problem with my choice of function? Is there a better function?
• Is there something wrong with my procedure of calculating the volume?
• Does one need to study the types of rocks contained in the mountain just to obtain an answer with the right order of magnitude?

Answer 1

The formula for cylinder volume is wrong. You have cylindrical shells, with the axis vertical, of height $z$, and radius $r$, with thickness $dr$. You need to integrate as a function of $r$. The volume of the cylindrical shell is $$dV=2\pi\ r\ dr\ z$$ So you need to write $z$ as a function of $r$. For simplicity, start with a simple linear function. $$z(0)=h\\z(R)=0$$ You should recover the volume of a cone with radius $R=25km$ and height $h=3.776km$.

Note that the volume of the mountain does not depend on the type of rock it is made out of. It is the mass the quantity that will depend on the rock density.