数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 体積 - 回転体の体積(楕円、放物線、三角関数(正弦、余弦)、x軸、y軸)

数学読本〈5〉

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
• 参考書籍
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

数学読本〈5〉微分法の応用/積分法/積分法の応用/行列と行列式(松坂 和夫(著)、岩波書店)の第20章(面積、体積、長さ - 積分法の応用)、20.2(体積)、回転体の体積、問19.を取り組んでみる。

19. 1. 
       $\begin{array}{l}{y}^2=b^2\left(1-\frac{x^2}{a^2}\right)\\ \\ \frac{x^2}{a^2}+\frac{0}{b^2}=1\\ x^2=a^2\\ \\ \pi {\displaystyle {\int }_{-a}^a{y}^{2}dx}\\ =2\pi {\displaystyle {\int }_0^a{b}^2\left(1-\frac{x^2}{a^2}\right)dx}\\ =2b^2\pi {\left[x-\frac{1}{3a^2}{x}^3\right]}_0^a\\ =2b^2\pi \left(a-\frac{1}{3}a\right)\\ =2b^2\pi ·\frac{2}{3}a\\ =\frac{4}{3}\pi a{b}^2\end{array}$
    2. 
       $\begin{array}{l}{x}^2=a^2\left(1-\frac{y^2}{b^2}\right)\\ \\ \frac{0}{a^2}+\frac{y^2}{b^2}=1\\ y^2=b^2\\ \\ 2\pi {\displaystyle {\int }_0^b{x}^{2}dy}\\ =2\pi {\displaystyle {\int }_0^b{a}^2\left(1-\frac{y^2}{b^2}\right)dy}\\ =2\pi a^2{\left[y-\frac{y^3}{3b^2}\right]}_0^b\\ =2\pi a^2\left(b-\frac{1}{3}b\right)\\ =2\pi a^2·\frac{2b}{3}\\ =\frac{4}{3}\pi a^{2}b\end{array}$
    3. 
       $\begin{array}{l}0=-x^2+h^2\\ x^2=h^2\\ \\ \pi {\displaystyle {\int }_{-h}^h{y}^{2}dx}\\ =\pi {\displaystyle {\int }_{-h}^h\left(x^4-2h^2{x}^2+h^4\right)dx}\\ =2\pi {\displaystyle {\int }_0^h\left(x^4-2h^2{x}^2+h^4\right)dx}\\ =2\pi {\left[\frac{1}{5}{x}^5-\frac{2}{3}{h}^2{x}^3+h^{4}x\right]}_0^h\\ =2\pi \left(\frac{1}{5}{h}^5-\frac{2}{3}{h}^5+h^5\right)\\ =2\pi ·\frac{3-10+15}{15}{h}^5\\ =\frac{16}{15}\pi h^5\end{array}$
    4. 
       $\begin{array}{l}y=-0+h^2\\ y=h^2\\ \pi {\displaystyle {\int }_0^{h^2}{x}^{2}dy}\\ =\pi {\displaystyle {\int }_0^{h^2}\left(h^2-y\right)dy}\\ =\pi {\left[h^{2}y-\frac{y^2}{2}\right]}_0^{h^2}\\ =\pi \left(h^4-\frac{h^4}{2}\right)\\ =\frac{1}{2}\pi h^4\end{array}$
    5. 
       

       平行移動。

       $\begin{array}{l}g=\left(\left(x+\frac{a+b}{2}\right)-a\right)\left(\left(x+\frac{a+b}{2}\right)-b\right)\\ =\left(x-\frac{a-b}{2}\right)\left(x+\frac{a-b}{2}\right)\end{array}$

       体積を求める。

       $\begin{array}{l}\pi {\displaystyle {\int }_a^b{y}^{2}dx}\\ =\pi {\displaystyle {\int }_{-\frac{b-a}{2}}^{\frac{b-a}{2}}{g}^{2}dx}\\ =\pi {\displaystyle {\int }_{-\frac{b-a}{2}}^{\frac{b-a}{2}}{\left(x^2-{\left(\frac{a-b}{2}\right)}^2\right)}^{2}dx}\\ =\pi {\displaystyle {\int }_{-\frac{b-a}{2}}^{\frac{b-a}{2}}\left(x^4-2{\left(\frac{a-b}{2}\right)}^2{x}^2+{\left(\frac{a-b}{2}\right)}^4\right)dx}\\ =2\pi {\displaystyle {\int }_0^{\frac{b-a}{2}}\left(x^4-2{\left(\frac{a-b}{2}\right)}^2{x}^2+{\left(\frac{a-b}{2}\right)}^4\right)dx}\\ =2\pi {\left[\frac{1}{5}{x}^5-\frac{2}{3}{\left(\frac{a-b}{2}\right)}^2{x}^3+{\left(\frac{a-b}{2}\right)}^{4}x\right]}_0^{\frac{b-a}{2}}\\ =2\pi \left(\frac{1}{5}-\frac{2}{3}+1\right){\left(\frac{b-a}{2}\right)}^5\\ =\pi ·\frac{3-10+15}{15}·\frac{{\left(b-a\right)}^5}{2^4}\\ =\pi \frac{8}{15}·\frac{{\left(b-a\right)}^5}{2^4}\\ =\frac{\pi }{30}{\left(b-a\right)}^5\end{array}$
    6. 
       $\begin{array}{l}\pi {\displaystyle {\int }_0^{\pi }{\sin }^{2}xdx}\\ =\pi {\displaystyle {\int }_0^{\pi }\frac{\cos \left(x-x\right)-\cos \left(x+x\right)}{2}dx}\\ =\pi {\displaystyle {\int }_0^{\pi }\frac{\cos 0-\cos 2x}{2}dx}\\ =\frac{\pi }{2}{\displaystyle {\int }_0^{\pi }\left(1-\cos 2x\right)dx}\\ =\frac{\pi }{2}{\left[x-\frac{1}{2}\sin 2x\right]}_0^{\pi }\\ =\frac{\pi }{2}\left(\left(\pi -\frac{1}{2}\sin 2\pi \right)-\left(0-\frac{1}{2}\sin 0\right)\right)\\ =\frac{\pi }{2}\pi \\ =\frac{{\pi }^2}{2}\end{array}$
    7. 
       $\begin{array}{l}\cos x=\sin x\\ \frac{\sin x}{\sin x}=1\\ \tan x=1\\ x=\frac{\pi }{4}\\ \\ \pi {\displaystyle {\int }_0^{\frac{\pi }{4}}{\cos }^{2}xdx}-\pi {\displaystyle {\int }_0^{\frac{\pi }{4}}{\sin }^{2}xdx}\\ =\pi {\displaystyle {\int }_0^{\frac{\pi }{4}}\left({\cos }^{2}x-{\sin }^{2}x\right)dx}\\ =\pi {\displaystyle {\int }_0^{\frac{\pi }{4}}\left(2{\cos }^{2}x-1\right)dx}\\ =\pi \left(2{\displaystyle {\int }_0^{\frac{\pi }{4}}{\cos }^{2}xdx}-{\displaystyle {\int }_0^{\frac{\pi }{4}}1dx}\right)\\ =\pi \left(2{\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{1}{2}\left(\cos \left(x+x\right)+\cos \left(x-x\right)\right)dx}-{\left[x\right]}_0^{\frac{\pi }{4}}\right)\\ =\pi \left({\displaystyle {\int }_0^{\frac{\pi }{4}}\left(\cos 2x+\cos 0\right)dx}-\frac{\pi }{4}\right)\\ =\pi \left({\displaystyle {\int }_0^{\frac{\pi }{4}}\left(\cos 2x+1\right)dx}-\frac{\pi }{4}\right)\\ =\pi \left({\left[\frac{1}{2}\sin 2x+x\right]}_0^{\frac{\pi }{4}}-\frac{\pi }{4}\right)\\ =\pi \left(\frac{1}{2}\sin \frac{\pi }{2}+\frac{\pi }{4}-\frac{1}{2}\sin 0-0-\frac{\pi }{4}\right)\\ =\frac{\pi }{2}\end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import pprint, symbols, Integral, pi, sin, cos

x, a, b, h = symbols('x a b h')

fs = [(b ** 2 * (1 - x ** 2 / a ** 2), (-a, a)),
      (a ** 2 * (1 - x ** 2 / b ** 2), (-b, b)),
      ((-x ** 2 + h ** 2) ** 2, (-h, h)),
      (h ** 2 - x, (0, h ** 2)),
      (((x - a) * (x - b)) ** 2, (a, b)),
      (cos(x) ** 2 - sin(x) ** 2, (0, pi / 4))]

for i, (f, (x1, x2)) in enumerate(fs, 1):
    print(f'({i})')
    I = Integral(f * pi, (x, x1, x2))
    for t in [I, I.doit().factor()]:
        pprint(t)
        print()
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample19.py
(1)
a                  
⌠                  
⎮       ⎛     2⎞   
⎮     2 ⎜    x ⎟   
⎮  π⋅b ⋅⎜1 - ──⎟ dx
⎮       ⎜     2⎟   
⎮       ⎝    a ⎠   
⌡                  
-a                 

       2
4⋅π⋅a⋅b 
────────
   3    


(2)
b                  
⌠                  
⎮       ⎛     2⎞   
⎮     2 ⎜    x ⎟   
⎮  π⋅a ⋅⎜1 - ──⎟ dx
⎮       ⎜     2⎟   
⎮       ⎝    b ⎠   
⌡                  
-b                 

     2  
4⋅π⋅a ⋅b
────────
   3    


(3)
h                 
⌠                 
⎮             2   
⎮    ⎛ 2    2⎞    
⎮  π⋅⎝h  - x ⎠  dx
⌡                 
-h                

      5
16⋅π⋅h 
───────
   15  


(4)
 2              
h               
⌠               
⎮    ⎛ 2    ⎞   
⎮  π⋅⎝h  - x⎠ dx
⌡               
0               

   4
π⋅h 
────
 2  


(5)
b                         
⌠                         
⎮           2         2   
⎮ π⋅(-a + x) ⋅(-b + x)  dx
⌡                         
a                         

          5 
-π⋅(a - b)  
────────────
     30     


(6)
π                           
─                           
4                           
⌠                           
⎮   ⎛     2         2   ⎞   
⎮ π⋅⎝- sin (x) + cos (x)⎠ dx
⌡                           
0                           

π
─
2


$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="0.5">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.0001" value="0.001">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-10">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="10">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-10">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="10">
<br>
<label for="a0">a = </label>
<input id="a0" type="number" value="-2">
<label for="b0">b = </label>
<input id="b0" type="number" value="3">

<button id="draw0">draw</button>
<button id="clear0">clear</button>

<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/d3/4.2.6/d3.min.js" integrity="sha256-5idA201uSwHAROtCops7codXJ0vja+6wbBrZdQ6ETQc=" crossorigin="anonymous"></script>

<script src="sample19.js"></script>

JavaScript

let div0 = document.querySelector('#graph0'),
    pre0 = document.querySelector('#output0'),
    width = 600,
    height = 600,
    padding = 50,
    btn0 = document.querySelector('#draw0'),
    btn1 = document.querySelector('#clear0'),
    input_r = document.querySelector('#r0'),
    input_dx = document.querySelector('#dx'),
    input_x1 = document.querySelector('#x1'),
    input_x2 = document.querySelector('#x2'),
    input_y1 = document.querySelector('#y1'),
    input_y2 = document.querySelector('#y2'),
    input_a0 = document.querySelector('#a0'),
    input_b0 = document.querySelector('#b0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_a0, input_a0],
    p = (x) => pre0.textContent += x + '\n',
    range = (start, end, step=1) => {
        let res = [];
        for (let i = start; i < end; i += step) {
            res.push(i);
        }
        return res;
    };

let draw = () => {
    pre0.textContent = '';

    let r = parseFloat(input_r.value),
        dx = parseFloat(input_dx.value),
        x1 = parseFloat(input_x1.value),
        x2 = parseFloat(input_x2.value),
        y1 = parseFloat(input_y1.value),
        y2 = parseFloat(input_y2.value),
        a0 = parseFloat(input_a0.value),
        b0 = parseFloat(input_b0.value);

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }
    
    let points = [],
        lines = [[a0, y1, a0, y2, 'red'],
                 [b0, y1, b0, y2, 'red'],
                 [-(a0 - b0) / 2, y1, -(a0 - b0) / 2, y2, 'green'],
                 [(a0 - b0) / 2, y1, (a0 - b0) / 2, y2, 'green']],
        f = (x) => (x - a0) * (x - b0),
        g = (x) => (x - (a0 - b0) / 2) * (x + (a0 - b0) / 2),
        fns = [[f, 'blue'],
               [g, 'orange']],
        fns1 = [],
        fns2 = [];

    fns.forEach((o) => {
        let [fn, color] = o;
        for (let x = x1; x <= x2; x += dx) {
            let y = fn(x);

            if (Math.abs(y) < Infinity) {
                points.push([x, y, color]);
            }
        }
    });
    fns1.forEach((o) => {
        let [fn, color] = o;
        
        lines.push([x1, fn(x1), x2, fn(x2), color]);
    });
    fns2.forEach((o) => {
        let [fn, color] = o;

        for (let x = x1; x <= x2; x += dx0) {
            let g = fn(x);
            
            lines.push([x1, g(x1), x2, g(x2), color]);
        }        
    });
    let xscale = d3.scaleLinear()
        .domain([x1, x2])
        .range([padding, width - padding]);
    let yscale = d3.scaleLinear()
        .domain([y1, y2])
        .range([height - padding, padding]);

    let xaxis = d3.axisBottom().scale(xscale);
    let yaxis = d3.axisLeft().scale(yscale);
    div0.innerHTML = '';
    let svg = d3.select('#graph0')
        .append('svg')
        .attr('width', width)
        .attr('height', height);

    svg.selectAll('line')
        .data([[x1, 0, x2, 0], [0, y1, 0, y2]].concat(lines))
        .enter()
        .append('line')
        .attr('x1', (d) => xscale(d[0]))
        .attr('y1', (d) => yscale(d[1]))
        .attr('x2', (d) => xscale(d[2]))
        .attr('y2', (d) => yscale(d[3]))
        .attr('stroke', (d) => d[4] || 'black');
    
    svg.selectAll('circle')
        .data(points)
        .enter()
        .append('circle')
        .attr('cx', (d) => xscale(d[0]))
        .attr('cy', (d) => yscale(d[1]))
        .attr('r', r)
        .attr('fill', (d) => d[2] || 'green');
    
    svg.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

    svg.append('g')
        .attr('transform', `translate(${padding}, 0)`)
        .call(yaxis);

    [fns, fns1, fns2].forEach((fs) => p(fs.join('\n')));
};

inputs.forEach((input) => input.onchange = draw);
btn0.onclick = draw;
btn1.onclick = () => pre0.textContent = '';
draw();

(x) => (x - a0) * (x - b0),blue
(x) => (x - (a0 - b0) / 2) * (x + (a0 - b0) / 2),orange

r = dx =
x1 = x2 =
y1 = y2 =
a = b =