Hello, guys! It's me! Warotan!
This time, let's calculate the volume of a 3 dimensional ball toward that of an [latex]n[/latex] 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!!
I would like to tell you the construction of this page and please select where you start to read to your interests.
The volume of a 3 dimensional ball
We know that the volume of a 3 dimensional ball is written as
[latex]\begin{eqnarray}V_{3}=\frac{4}{3}\pi r^{3}.\end{eqnarray}[/latex]
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.
If we rotate this around the [latex]x[/latex] axis, we can make a three dimensional ball.
The area of a circle created by turning the blue line at [latex]X[/latex] around the [latex]x[/latex] axis is given by [latex]\pi \xi^2[/latex], [latex]\xi \equiv \sqrt{r^2-X^2}[/latex].
If we consider a disk of thickness [latex]{\rm d}X[/latex], the volume is [latex]\pi \xi^2{\rm d}X[/latex].
Summing up this volume in [latex]X\in\left[-r,r\right][/latex] gives us the desired volume;
[latex]\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}[/latex]
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 [latex]y=-x^2+3[/latex] and the [latex]x[/latex] axis" or
"calculate the area between [latex]y=\sin\left(x\right)\,\left(0\leq x\leq \pi\right)[/latex] and the [latex]x[/latex] 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
[latex]\begin{eqnarray} M &=& \int{\rm d}m \\ &=& \int\rho{\rm d}V. \end{eqnarray}[/latex]
Here [latex]M[/latex] is mass and [latex]\rho[/latex] 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 [latex]n[/latex] dimensional ball and I gave you a story of my school study.
Dividing a large problem into small ones is a usual practice (integration).
Thank you for reading and please spread this blog if you like.
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tsunetthi(at)gmail.com
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