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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

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Author​

samparker, 祭音Myyura

Description​

以下の問いに答えよ。

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

(2) 領域 {(x,y)∈R2∣∣xy∣≥1,x2+y2≤R2},R≥0\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)S(R) を求めよ。

(3) lim⁡R→∞S(R)πR2\lim_{R \to \infty} \frac{S(R)}{\pi R^2} を求めよ。

Answer the following questions:

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

(2) Find the area S(R),R≥0,S(R), R \geq 0, of the region {(x,y)∈R2∣∣xy∣≥1,x2+y2≤R2}.\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→∞S(R)πR2.\lim_{R \to \infty} \frac{S(R)}{\pi R^2}.

题目描述​

回答以下各问：

  1. 求不定积分

    ∫dθsin⁡θ.\int\frac{d\theta}{\sin\theta}.

  2. 对 R≥0R\ge0，求区域

    {(x,y)∈R2: ∣xy∣≥1, x2+y2≤R2}\{(x,y)\in\mathbb R^2:\ |xy|\ge1,\ x^2+y^2\le R^2\}

    的面积 S(R)S(R)。

  3. 求极限

    lim⁡R→∞S(R)πR2.\lim_{R\to\infty}\frac{S(R)}{\pi R^2}.

Kai​

(1)​

∫1sin⁡θdθ=∫sin⁡θsin⁡2θdθ=∫sin⁡θ1−cos⁡2θdθ\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⁡θ\cos \theta by tt, we have

∫sin⁡θ1−cos⁡2θdθ=∫−dt1−t2=∫−dt(1+t)(1−t)=−12∫(11−t+11+t)dt=−12(−log⁡∣1−t∣+log⁡∣1+t∣)+C=−12log⁡∣1+t1−t∣+C=12log⁡∣1−t1+t∣+C=12log⁡(1−cos⁡θ1+cos⁡θ)+C\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)​

In polar coordinates,

∣xy∣=r22∣sin⁡2θ∣.|xy|=\frac{r^2}{2}|\sin2\theta|.

Hence S(R)=0S(R)=0 for 0≤R≤20\leq R\leq\sqrt2. For R>2R>\sqrt2, put

θ0=12arcsin⁡2R2.\theta_0=\frac12\arcsin\frac{2}{R^2}.

Using symmetry in the four quadrants,

S(R)=4∫θ0π2−θ0∫2/sin⁡2θRr dr dθ=2R2(π2−2θ0)−4ln⁡(cot⁡θ0)=πR2−2R2arcsin⁡2R2−4ln⁡ ⁣(cot⁡ ⁣(12arcsin⁡2R2)).\begin{aligned} S(R) &=4\int_{\theta_0}^{\frac{\pi}{2}-\theta_0} \int_{\sqrt{2/\sin2\theta}}^R r\,dr\,d\theta\\ &=2R^2\left(\frac{\pi}{2}-2\theta_0\right) -4\ln(\cot\theta_0)\\ &=\pi R^2-2R^2\arcsin\frac{2}{R^2} -4\ln\!\left(\cot\!\left(\frac12\arcsin\frac{2}{R^2}\right)\right). \end{aligned}

(3)​

cot⁡θ0=R22(1+1−4R4),\cot\theta_0 =\frac{R^2}{2}\left(1+\sqrt{1-\frac4{R^4}}\right),

so ln⁡(cot⁡θ0)=O(ln⁡R)\ln(\cot\theta_0)=O(\ln R). Therefore

lim⁡R→∞S(R)πR2=1.\lim_{R\to\infty}\frac{S(R)}{\pi R^2}=1.
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