広島大学 先進理工系科学研究科 情報科学プログラム 2021年1月実施 専門科目I 問題2

Author

samparker, 祭音Myyura

Description

以下の問いに答えよ。

(1) 不定積分 \(\int \frac{d\theta}{\sin\theta}\) を求めよ。

(2) 領域 \(\left\{(x,y) \in \mathbb{R}^2 \mid |xy| \geq 1, x^2 + y^2 \leq R^2 \right\}, R \geq 0\) の面積 \(S(R)\) を求めよ。

(3) \(\lim_{R \to \infty} \frac{S(R)}{\pi R^2}\) を求めよ。

---

Answer the following questions:

(1) Calculate the integral: \(\int \frac{d\theta}{\sin\theta}.\)

(2) Find the area \(S(R), R \geq 0,\) of the region \(\left\{(x,y) \in \mathbb{R}^2 \mid |xy| \geq 1, x^2 + y^2 \leq R^2 \right\}.\)

(3) Find the limit: \(\lim_{R \to \infty} \frac{S(R)}{\pi R^2}.\)

Kai

(1)

\[ \int \frac{1}{\sin \theta} d \theta = \int \frac{\sin \theta}{\sin^2 \theta} d \theta = \int \frac{\sin \theta}{1 - \cos^2 \theta} d\theta \]

substitute \(\cos \theta\) by \(t\), we have

\[ \begin{aligned} \int \frac{\sin \theta}{1 - \cos^2 \theta} d\theta &= \int \frac{-dt}{1-t^2} = \int \frac{-dt}{(1+t)(1-t)} \\ &= -\frac{1}{2} \int \left( \frac{1}{1-t} + \frac{1}{1+t} \right) dt \\ &= -\frac{1}{2} (-\log |1-t| + \log |1+t|) + C \\ &= -\frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + C \\ &= \frac{1}{2} \log \left| \frac{1-t}{1+t} \right| + C \\ &= \frac{1}{2} \log \left( \frac{1 - \cos \theta}{1 + \cos \theta} \right) + C \end{aligned} \]

(2)

Since \((x^2 + y^2) / 2 \geq \sqrt{x^2 y^2}\), we have

\[ x^2 + y^2 \geq 2 |xy| \geq 2 \]

Hence the region can be rewritten as \(\{(x,y) \in \mathbb{R}^2 \mid 2 \leq x^2 + y^2 \leq R^2\}\), and the area of which is

\[ S(R) = \pi R^2 - \pi \cdot2 = \pi (R^2 - 2) \]

(3)

\[ \lim_{R \to \infty} \frac{S(R)}{\pi R^2} = \lim_{R \to \infty} \frac{\pi (R^2 - 2)}{\pi R^2} = \lim_{R \to \infty} \left(1 - \frac{2}{R^2} \right) = 1 \]