# 3 dimensional ball and story of my school life about study

Hello, guys! It’s me! Warotan!

Japanese version

This time, let’s calculate the volume of a 3 dimensional ball toward that of an $$n$$ dimensional ball.
Also, my misunderstanding about study in high school is given.

I drunk too much today, so there might be some mistakes.
The year-end party was so much fun!!

## The volume of a 3 dimensional ball

We know that the volume of a 3 dimensional ball is written as

$$\begin{eqnarray}V_{3}=\frac{4}{3}\pi r^{3}.\end{eqnarray}$$

The suffix 3 shows that it is a THREE dimensional one.

How can we derive this formula?

Let us first consider a semicircle as a preparation for the 3 dimensional volume. A semicircle of radius $$r$$

If we rotate this around the $$x$$ axis, we can make a three dimensional ball. Rotating the semicircle around the axis, we get a 3 dimensional ball. Rotating the blue line around the $$x$$ axis, we get a circle of radius $$\xi \equiv \sqrt{r^2-X^2}$$.

The area of a circle created by turning the blue line at $$X$$ around the $$x$$ axis is given by $$\pi \xi^2$$, $$\xi \equiv \sqrt{r^2-X^2}$$.
If we consider a disk of thickness $${\rm d}X$$, the volume is $$\pi \xi^2{\rm d}X$$.
Summing up this volume in $$X\in\left[-r,r\right]$$ gives us the desired volume;

$$\begin{eqnarray} V_{3} &=& \int_{-r}^{r}\pi\xi^2 {\rm d}X \\ &=& \pi\int_{-r}^{r}\left(r^2-X^2\right){\rm d}X \\ &=& \pi \, 2\int_{0}^{r}\left(r^2-X^2\right){\rm d}X \\ &=& 2\pi\left[r^2X-\frac{1}{3}X^3\right]^{r}_{0} \\ &=& 2\pi\left(r^3-\frac{1}{3}r^3\right) \\ &=& 2\pi\frac{2}{3}r^3 \\ &=& \frac{4}{3}\pi r^3 \end{eqnarray}$$

It is a kind of shell integration.

## Story of my school life~Integration is a method of calculating the area of a graph!?~

In addition to the main story, I tell you my mistake about study.

When I was a high school student, I was not (and am not so) good at studying.
One of my big misinterpretation was about integration.

In those days, I thought that integration is a method of calculating an area of a graph.
Exercises about integration given by the school was all to calculate an area such as
“calculate the area between $$y=-x^2+3$$ and the $$x$$ axis” or
“calculate the area between $$y=\sin\left(x\right)\,\left(0\leq x\leq \pi\right)$$ and the $$x$$ axis”
and so on…
I came to think that
Ah… ok. Integration gives us an area!

BUT!!! That was a mistake.
After I passed the entrance exam of a university and, in the lecture,
a professor wrote some equations like

$$\begin{eqnarray} M &=& \int{\rm d}m \\ &=& \int\rho{\rm d}V. \end{eqnarray}$$

Here $$M$$ is mass and $$\rho$$ is density.

It confused me and I got into a panic.
“What does it mean??? Integration gives me an area but he says he calculates the mass of something???”

Of course it’s impossible for one who thinks integration is a method of calculating an area to understand the equation.

Now that makes me laugh but I was serious and asked him what the equation means.
He was surprised but taught me so kindly.

## Summary

Now, that’s all what I wanted to write.

This time, we did a warming up for the next step, volume of an $$n$$ dimensional ball and I gave you a story of my school study.
Dividing a large problem into small ones is a usual practice (integration).