積分法定積分の計算、曲線、平方根、座標軸、図形の面積

学習環境

微分積分学 (ちくま学芸文庫) (吉田洋一(著)、筑摩書房)のⅣ.(積分法)、演習問題18.の解答を求めてみる。

\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1

\sqrt{\frac{y}{b}} = 1 - \sqrt{\frac{x}{a}}

\frac{y}{b} = 1 - 2 \sqrt{\frac{x}{a}} + \frac{x}{a}

y = b (1 - 2 \sqrt{\frac{x}{a}} + \frac{x}{a})

\sqrt{\frac{x}{a}} = 1

\frac{x}{a} = 1

x = a

よって求める面積は、

\int_{0}^{a} b (1 - 2 \sqrt{\frac{x}{a}} + \frac{x}{a}) dx

= b {[x - \frac{4}{3} \cdot \frac{x^{\frac{3}{2}}}{\sqrt{a}} + \frac{x^{2}}{2 a}]}_{0}^{a}

= b (a - \frac{4 a^{\frac{3}{2}}}{3 \sqrt{a}} + \frac{a^{2}}{2 a})

= b (a - \frac{4}{3} a + \frac{1}{2} a)

= \frac{6 - 8 + 3}{6} a b

= \frac{1}{6} a b

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import Integral, symbols, sqrt, solve, pprint, plot from sympy.abc import x, y print('18.') a, b = symbols('a, b', positive=True) eq = sqrt(x / a) + sqrt(y / b) - 1 ys = solve(eq, y) pprint(ys) y0 = ys[0] xs = solve(y0, x) pprint(xs) x0 = xs[0] class Test(TestCase): def test(self): self.assertEqual( Integral(y0, (x, 0, x0)).doit(), a * b / 6 ) a0 = 3 b0 = 2 d = {a: a0, b: b0} p = plot(y0.subs(d), (x, 0, a0), ylim=(0, a0), legend=True) p.save('sample18.png') if __name__ == "__main__": main()

% ./sample18.py -v 18. ⎡ 2⎤ ⎢b⋅(√a - √x) ⎥ ⎢────────────⎥ ⎣ a ⎦ [a] test (__main__.Test) ... ok ---------------------------------------------------------------------- Ran 1 test in 0.145s OK %