2019年2月9日土曜日

数学 - Python - 解析学 - 積分 - 積分の応用 - 面積(三角関数(正弦)、倍角、置換積分法、極座標表示)

1. $\begin{array}{}\int \frac{\pi {r}^{2}}{2\pi }d\theta \\ =\frac{1}{2}\int {\left(1+\mathrm{sin}\left(2\theta \right)\right)}^{2}d\theta \\ =\frac{1}{2}\int \left(1+2\mathrm{sin}\left(2\theta \right)+{\mathrm{sin}}^{2}\left(2\theta \right)\right)d\theta \\ =\frac{1}{2}\left(\theta -\mathrm{cos}\left(2\theta \right)\right)+\frac{1}{2}\int {\mathrm{sin}}^{2}\left(2\theta \right)d\theta \\ t=2\theta \\ \frac{\mathrm{dt}}{d\theta }=2\\ \theta =0,t=0\\ \theta =2\pi ,t=4\pi \\ \int {\mathrm{sin}}^{2}\left(2\theta \right)d\theta \\ =\frac{1}{2}\int {\mathrm{sin}}^{2}t\mathrm{dt}\\ =\frac{1}{2}\left(-\frac{1}{2}\mathrm{sin}t\mathrm{cos}t+\frac{1}{2}\int 1\mathrm{dt}\right)\\ =\frac{1}{4}\left(-\mathrm{sin}t\mathrm{cos}t+t\right)\end{array}$

よって求める面積は、

$\begin{array}{}\frac{1}{2}\left({\left[\theta -\mathrm{cos}\left(2\theta \right)\right]}_{0}^{2\pi }+\frac{1}{4}{\left[-\mathrm{sin}t\mathrm{cos}t+t\right]}_{0}^{4\pi }\right)\\ =\frac{1}{2}\left(2\pi +\frac{1}{4}·4\pi \right)\\ =\frac{3}{2}\pi \end{array}$

コード

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, Integral, cos, sin, pi, exp, sqrt
from sympy.plotting import plot_parametric

theta = symbols('θ')
r = 1 + sin(2 * theta)
x = r * cos(theta)
y = r * sin(theta)

I = Integral(r ** 2 / 2, (theta, 0, 2 * pi))
for o in [I, I.doit()]:
pprint(o.simplify())
print()

p = plot_parametric((x, y, (theta, 0, pi / 2)),
(x, y, (theta, pi / 2, pi)),
(x, y, (theta, pi, 3 * pi / 2)),
(x, y, (theta, 3 * pi / 2, 2 * pi)),
show=False)

colors = ['red', 'green', 'blue', 'brown', 'orange', 'pink']
for i, s in enumerate(p):
s.line_color = colors[i]
p.save('sample8.png')


C:\Users\...> py -3 sample8.py
2⋅π
⌠
⎮                2
⎮  (sin(2⋅θ) + 1)
⎮  ─────────────── dθ
⎮         2
⌡
0

3⋅π
───
2

C:\Users\...>