國立臺灣大學 106 學年度轉學生招生考試試題：微積分(A)詳解

※禁止使用計算機
※注意：請於答案卷內之「非選擇題作答區」標明題號依序作答。

1. ($20\%$) Determine the convergence or divergence of the following series:

   $\displaystyle\sum_{n=1}^\infty\left(\frac{\cos(\pi n)}{\tanh(\pi n)}\frac{1-\cosh\left(\frac1n\right)}{\sin\left(\frac1n\right)}\right)$.

訣竅
運用交錯級數審歛法與 Dirichlet 審斂法即可。
解法
由於 $\cos(\pi n)=(-1)^n$，因此本題之級數可寫為 $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{\tanh\left(\pi n\right)}\frac{\cosh(1/n)-1}{\sin(1/n)}$。
考慮函數 $\displaystyle f(x)=\frac1{\tanh(x)}-1=\coth(x)-1=\frac{e^x+e^{-x}}{e^x-e^{-x}}-1=\frac2{e^{2x}-1}$，那麼求導可知
$\displaystyle f'(x)=-\frac{4e^{2x}}{(e^{2x}-1)^2}<0\quad\text{for}~x>0$.
因此 $f$ 嚴格遞減。如此根據交錯級數審斂法可知級數 $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\left(\frac1{\tanh(n\pi)}-1\right)$ 收斂。
因此可知存在正數 $M$ 使得對任何 $N\in\mathbb N$ 均有
$\displaystyle\left|\sum_{n=1}^N\frac{(-1)^{n+1}}{\tanh(n\pi)}\right|-\left|\sum_{n=1}^N(-1)^{n+1}\right|\leq\left|\sum_{n=1}^N(-1)^{n+1}\left(\frac1{\tanh(n\pi)}-1\right)\right|\leq M$.
這表示 $\displaystyle\left|\sum_{n=1}^N\frac{(-1)^{n+1}}{\tanh(n\pi)}\right|\leq M+1$。
另一方面，設 $g:[0,1]\to\mathbb R$ 定義為 $\displaystyle g(x)=\frac{\cosh(x)-1}{\sin x}$，$0<x\leq1$，而 $g(0)=1$。容易注意到 $g$ 在 $[0,1]$ 上連續，那麼求導有
$\displaystyle g'(x)=\frac{\sinh(x)\sin(x)-(\cosh(x)-1)\cos(x)}{\sin^2(x)}\quad\text{for}~x\in(0,1)$.
進一步考慮 $h(x)=\sinh(x)\sin(x)-(\cosh(x)-1)\cos(x)$，$0\leq x\leq1$。求導可知
$h'(x)=(2\cosh(x)-1)\sin(x)>0\quad\text{for}~x\in(0,1)$.
這表明 $h$ 在 $[0,1]$ 上嚴格遞增且當 $x\in(0,1]$ 時 $h(x)>h(0)=0$。由此可知當 $x\in(0,1)$ 時有 $g'(x)>0$，即 $g$ 在 $(0,1)$ 上嚴格遞增。如此一來便可發現數列 $\displaystyle\left\{\frac{\cosh(1/n)-1}{\sin(1/n)}\right\}_{n=1}^\infty=\{g(1/n)\}_{n=1}^\infty$ 嚴格遞減且趨於零。因此根據 Dirichlet 審斂法知原先給定的級數收斂。


2. ($20\%$) For a second continuously differentiable function, $f\left(x,y,z\right)$, defined on the space $\mathbb R^3$, its Laplacian is defined to be

   $\displaystyle\bigtriangleup f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}$.

   Derive the formula of $\bigtriangleup f$ in the spherical coordinate system.
訣竅
熟悉球極座標變換並按照多變數連鎖律仔細計算之即可。
解法

由球極座標變換，設 $x=\rho\sin\phi\cos\theta$、$y=\rho\sin\phi\sin\theta$、$z=\rho\cos\phi$，即有

$\rho=\sqrt{x^2+y^2+z^2},\quad\phi=\arccos(z/\rho),\quad\theta=\arctan(y/x).$

首先分別表達 $f$ 對 $x$、$y$ 與 $z$ 的一階偏導數如下

$\displaystyle\begin{aligned}&\frac{\partial f}{\partial x}=\frac{\partial f}{\partial\rho}\frac{\partial\rho}{\partial x}+\frac{\partial f}{\partial\phi}\frac{\partial\phi}{\partial x}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial x},\\&\frac{\partial f}{\partial y}=\frac{\partial f}{\partial\rho}\frac{\partial\rho}{\partial y}+\frac{\partial f}{\partial\phi}\frac{\partial\phi}{\partial y}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial y},\\&\frac{\partial f}{\partial z}=\frac{\partial f}{\partial\rho}\frac{\partial\rho}{\partial z}+\frac{\partial f}{\partial\phi}\frac{\partial\phi}{\partial z}+\frac{\partial f}{\partial\theta}\frac{\partial\theta}{\partial z}.\end{aligned}$

因此我們計算下列各項偏導有

$\begin{array}{lll}\displaystyle\frac{\partial\rho}{\partial x}=\sin\phi\cos\theta,\quad&\displaystyle\frac{\partial\phi}{\partial x}=\frac{\cos\theta\cos\phi}\rho,\quad&\displaystyle\frac{\partial\theta}{\partial x}=-\frac{\sin\theta}{\rho\sin\phi},\\\displaystyle\frac{\partial\rho}{\partial y}=\sin\phi\sin\theta,\quad&\displaystyle\frac{\partial\phi}{\partial y}=\frac{\sin\theta\cos\phi}\rho,\quad&\displaystyle\frac{\partial\theta}{\partial y}=\frac{\cos\theta}{\rho\sin\phi},\\\displaystyle\frac{\partial\rho}{\partial z}=\cos\phi,\quad&\displaystyle\frac{\partial\phi}{\partial z}=-\frac{\sin\phi}\rho,\quad&\displaystyle\frac{\partial\theta}{\partial z}=0.\end{array}$

由此可得

$\begin{array}{l}\displaystyle\frac{\partial f}{\partial x}=\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta},\\\displaystyle\frac{\partial f}{\partial y}=\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta},\\\displaystyle\frac{\partial f}{\partial z}=\cos\phi\frac{\partial f}{\partial\rho}-\frac{\sin\phi}\rho\frac{\partial f}{\partial\phi}.\end{array}$

據此計算二階偏導數。首先是 $\displaystyle\frac{\partial^2f}{\partial x^2}$：

$\displaystyle\begin{aligned}\frac{\partial^2f}{\partial x^2}&=\sin\phi\cos\theta\frac\partial{\partial\rho}\frac{\partial f}{\partial x}+\frac{\cos\theta\cos\phi}\rho\frac\partial{\partial\phi}\frac{\partial f}{\partial x}-\frac{\sin\theta}{\rho\sin\phi}\frac\partial{\partial\theta}\frac{\partial f}{\partial x}\\&=\sin\phi\cos\theta\frac\partial{\partial\rho}\left(\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&\quad+\frac{\cos\theta\cos\phi}\rho\frac\partial{\partial\phi}\left(\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&\quad-\frac{\sin\theta}{\rho\sin\phi}\frac\partial{\partial\theta}\left(\sin\phi\cos\theta\frac{\partial f}{\partial\rho}+\frac{\cos\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}-\frac{\sin\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&=\sin^2\phi\cos^2\theta\frac{\partial^2f}{\partial\rho^2}+\frac{\cos^2\theta\cos^2\phi}{\rho^2}\frac{\partial^2f}{\partial\phi^2}+\frac{\sin^2\theta}{\rho^2\sin^2\phi}\frac{\partial^2f}{\partial\theta^2}\\&\quad+\frac{2\cos^2\theta\sin\phi\cos\phi}\rho\frac{\partial^2f}{\partial\rho\partial\phi}-\frac{2\sin\theta\cos\theta}\rho\frac{\partial^2f}{\partial\rho\partial\theta}-\frac{2\sin\theta\cos\theta\cos\phi}{\rho^2\sin\phi}\frac{\partial^2f}{\partial\phi\partial\theta}\\&\quad+\frac{\cos^2\theta\cos^2\phi+\sin^2\theta}\rho\frac{\partial f}{\partial\rho}+\frac{\sin^2\theta\cos\phi-2\sin\phi^2\cos\phi\cos^2\theta}{\rho^2\sin\phi}\frac{\partial f}{\partial\phi}\\&\quad+\frac{2\sin\theta\cos\theta}{\rho^2\sin^2\phi}\frac{\partial f}{\partial\theta}.\end{aligned}$

接著是 $\displaystyle\frac{\partial^2f}{\partial y^2}$：

$\displaystyle\begin{aligned}\frac{\partial^2f}{\partial y^2}&=\sin\phi\sin\theta\frac\partial{\partial\rho}\frac{\partial f}{\partial y}+\frac{\sin\theta\cos\phi}\rho\frac\partial{\partial\phi}\frac{\partial f}{\partial y}+\frac{\cos\theta}{\rho\sin\phi}\frac\partial{\partial\theta}\frac{\partial f}{\partial y}\\&=\sin\phi\sin\theta\frac\partial{\partial\rho}\left(\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&\quad+\frac{\sin\theta\cos\phi}\rho\frac\partial{\partial\phi}\left(\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&\quad+\frac{\cos\theta}{\rho\sin\phi}\frac\partial{\partial\theta}\left(\sin\phi\sin\theta\frac{\partial f}{\partial\rho}+\frac{\sin\theta\cos\phi}\rho\frac{\partial f}{\partial\phi}+\frac{\cos\theta}{\rho\sin\phi}\frac{\partial f}{\partial\theta}\right)\\&=\sin^2\phi\sin^2\theta\frac{\partial^2f}{\partial\rho^2}+\frac{\sin^2\theta\cos^2\phi}{\rho^2}\frac{\partial^2f}{\partial\phi^2}+\frac{\cos^2\theta}{\rho^2\sin^2\phi}\frac{\partial^2f}{\partial\theta^2}\\&\quad+\frac{2\sin^2\theta\sin\phi\cos\phi}\rho\frac{\partial^2f}{\partial\rho\partial\phi}+\frac{2\sin\theta\cos\theta}\rho\frac{\partial^2f}{\partial\rho\partial\theta}+\frac{2\sin\theta\cos\theta\cos\phi}{\rho^2\sin\phi}\frac{\partial^2f}{\partial\phi\partial\theta}\\&\quad+\frac{\sin^2\theta\cos^2\phi+\cos^2\theta}{\rho^2}\frac{\partial f}{\partial\rho}+\frac{\cos^2\theta\cos\phi-2\sin^2\theta\sin^2\phi\cos\theta}{\rho^2\sin\phi}\frac{\partial f}{\partial\phi}\\&\quad-\frac{2\sin\theta\cos\theta}{\rho^2\sin^2\phi}\frac{\partial f}{\partial\theta}.\end{aligned}$

最後是 $\displaystyle\frac{\partial^2f}{\partial z^2}$：

$\displaystyle\begin{aligned}\frac{\partial^2f}{\partial z^2}&=\cos\phi\frac\partial{\partial\rho}\frac{\partial f}{\partial z}-\frac{\sin\phi}\rho\frac\partial{\partial\phi}\frac{\partial f}{\partial z}\\&=\cos\phi\frac\partial{\partial\rho}\left(\cos\phi\frac{\partial f}{\partial\rho}-\frac{\sin\phi}\rho\frac{\partial f}{\partial\phi}\right)-\frac{\sin\phi}\rho\frac\partial{\partial\phi}\left(\cos\phi\frac{\partial f}{\partial\rho}-\frac{\sin\phi}\rho\frac{\partial f}{\partial\phi}\right)\\&=\cos^2\phi\frac{\partial^2f}{\partial\rho^2}+\frac{\sin^2\phi}{\rho^2}\frac{\partial^2f}{\partial\phi^2}-\frac{2\sin\phi\cos\phi}\rho\frac{\partial^2f}{\partial\rho\partial\phi}+\frac{\sin^2\phi}\rho\frac{\partial f}{\partial\rho}+\frac{2\sin\phi\cos\phi}{\rho^2}\frac{\partial f}{\partial\phi}.\end{aligned}$

藉由以上的計算將三者加總可得

$\displaystyle\bigtriangleup f=\frac{\partial^2f}{\partial\rho^2}+\frac{1}{\rho^2}\frac{\partial^2f}{\partial\phi^2}+\frac{1}{\rho^2\sin^2\phi}\frac{\partial^2f}{\partial\theta^2}+\frac2\rho\frac{\partial f}{\partial\rho}+\frac{\cos\phi}{\rho^2\sin\phi}\frac{\partial f}{\partial\phi}.$



3. ($20\%$) Suppose that $f(x)>0$ is a continuously differentiable function defined on $x\geq1$. Let $S$ be the surface of revolution of the graph $y=f(x)$ about the $x$-axis. Let $E\subset\mathbb R^3$ be the solid enclosed by $S$ and $x=1$. More precisely,

   $E=\{(x,y,z)\in\mathbb R^3\,|\,x\geq1\text{ and }\sqrt{y^2+z^2}\leq f(x)\text{ for each }x\geq1\}$.

   If the surface area of $S$ is finite, prove that the volume of $E$ must also be finite.
訣竅
此定理之逆敘述為經典之悖論：存在一旋轉體其體積有限但表面積無限。在估算體積時應充分留意表面積有限可先得到函數值有界，並由此估量體積。
解法
由於表面積 $A$ 有限，因此我們可以注意到
$\begin{aligned}\displaystyle\lim_{x\to\infty}[f^2(x)-f^2(1)]&=\lim_{x\to\infty}\int_1^x\!(f^2(s))'\,\mathrm ds\leq\int_1^\infty\!|(f^2(s))'|\,\mathrm ds\\&=\int_1^\infty\!2f(s)|f'(s)|\,\mathrm ds\leq\int_1^\infty\!2f(s)\sqrt{1+(f'(s))^2}\,\mathrm ds=\frac A\pi<\infty.\end{aligned}$
由於連續函數 $\displaystyle\lim_{x\to\infty}|f(x)|$ 有限，故 $f$ 在 $[1,\infty)$ 上有界（記為 $M$），如此 $E$ 的體積可被估算如下：
$\displaystyle V=\int_1^\infty\!\pi f^2(x)\,\mathrm dx\leq\int_1^\infty\!\frac M2\cdot2\pi f(x)\,\mathrm dx\leq\frac M2\int_1^\infty\!2\pi f(x)\sqrt{1+(f'(x))^2}\,\mathrm dx=\frac{MA}2$.
如此由 $A$ 的有限性即可推得 $V$ 的有限性，證明完畢。


4. Let $C$ be the curve defined by $\{(x,y)\in\mathbb R^2\,|\,x^4+2x^2y^2+y^4-2x(x^2-y^2)=0\}$.
   1. ($10\%$) Sketch the curve $C$.
   2. ($10\%$) Calculate the area of the region enclosed by $C$.
訣竅
運用極座標變換後描點作圖並且運用極座標下的面積公式計算其面積。
解法

1. 運用極座標變換，令 $x=r\cos\theta$、$y=r\sin\theta$，如此方程可改寫為

   $r^4\cos^4\theta+2r^4\cos^2\theta\sin^2\theta+r^4\sin^4\theta-2r^3\cos\theta\cos2\theta=0$

   即有 $r=2\cos\theta\cos2\theta$。描點繪圖如下：
   
2. 運用面積公式可得

   $\displaystyle\begin{aligned}A&=\frac12\int_0^\pi\!r^2\,\mathrm d\theta\\&=2\int_0^\pi\!\cos^2\theta\cos^22\theta\,\mathrm d\theta\\&=\frac18\int_0^\pi\!(1+\cos2\theta)(1+\cos4\theta)\,\mathrm d\theta\\&=\frac12\int_0^\pi\!(1+\cos2\theta+\cos4\theta+\cos2\theta\cos4\theta)\,\mathrm d\theta\\&=\frac\pi2+\frac12\int_0^\pi\!\frac{\cos6\theta+\cos2\theta}2\,\mathrm d\theta\\&=\frac\pi2.\end{aligned}$



5. ($20\%$) Let $a$, $b$, $c$ be three positive numbers with $a>b>c$. Let $\Sigma$ be the surface defined by

   $\displaystyle\frac{\left(x-b\right)^2}{a^2}+\frac{\left(y-c\right)^2}{b^2}+\frac{z^2}{c^2}=4$.

   Note that $\Sigma$ encloses a bounded solid. Set $\vec n$ to be the unit outer normal with respect to that solid. Consider the vector field

   $\displaystyle\vec{F}=\left(\frac{x}{\left(x^2+y^2+z^2\right)^{\frac32}}+\cos\left(yz\right),\frac{y}{\left(x^2+y^2+z^2\right)^{\frac32}}+y,\frac{z}{\left(x^2+y^2+z^2\right)^{\frac32}}+e^{z^2}\right)$.

   Calculate the flux integral $\displaystyle\iint_\Sigma\!\vec F\cdot\vec n\,\mathrm dS$.
訣竅
運用 Gauss 散度定理並且對奇異點圈出另外處理。
解法

首先，記 $\vec G$ 與 $\vec H$ 如下

$\begin{aligned}&\vec G=\left(\frac x{(x^2+y^2+z^2)^{3/2}},\,\frac y{(x^2+y^2+z^2)^{3/2}},\,\frac z{(x^2+y^2+z^2)^{3/2}}\right),\qquad\vec H=\left(\cos(yz),\,y,\,e^{z^2}\right).\end{aligned}$

並且設 $\Omega$ 為 $\Sigma$ 所包圍之區域，那麼可以發現 $\vec H$ 在 $\Omega$ 上有一階地連續偏導函數，因此直接使用 Gauss 散度定理可得

$\begin{aligned}\iint_\Sigma\!\vec H\cdot\vec n\,\mathrm dS&=\iiint_\Omega\!\nabla\cdot\vec H\,\mathrm dV=\iiint_\Omega\!(1+2ze^{z^2})\,\mathrm dV=\iiint_\Omega\!1\,\mathrm dV+\iiint_\Omega\!2ze^{z^2}\,\mathrm dV\\&=\frac43\pi\cdot2a\cdot2b\cdot2c+\iiint_\Omega\!2ze^{z^2}\,\mathrm dV=\frac{32}3\pi abc,\end{aligned}$

其中最後一個等號使用了積分區域的對 $xy$ 平面的對稱性。另一方面，因為 $(0,0,0)\in\Omega$，因此 $\vec G$ 在 $\Omega$ 上沒有連續地一階偏導數。可以取充分小的正數 $\varepsilon$ 使得 $B_\varepsilon(0)\subset\Omega$，從而 $\vec G$ 在 $\Omega-B_\varepsilon(0)$ 上有連續地一階偏導函數，那麼在其上使用 Gauss 散度定理可以直接得到

$\displaystyle\iint_{\Sigma\cup\partial B_\varepsilon(0)}\!\vec G\cdot\vec n\,\mathrm dS=\iiint_{\Omega-B_\varepsilon(0)}\!\nabla\cdot\vec G\,\mathrm dV=0$.

因此

$\displaystyle\iint_\Sigma\!\vec G\cdot\vec n\,\mathrm dS=\iint_{\partial B_\varepsilon(0)}\vec G\cdot\vec n\,\mathrm dS$,

其中 $\vec n$ 均表示所積分之區域指向外的單位法向量（在不同的積分中可能所指之方向會隨積分範圍而有所不同）。藉由直接計算可知

$\displaystyle\iint_{\partial B_\varepsilon(0)}\!\vec G\cdot\vec n\,\mathrm dS=\iint_{\partial B_\varepsilon(0)}\!\frac{(x,y,z)}{\varepsilon^3}\cdot\frac{(x,y,z)}\varepsilon\,\mathrm dS=4\pi\varepsilon^2\cdot\frac1{\varepsilon^2}=4\pi$.

因此本題之所求為

$\displaystyle\iint_\Sigma\!\vec F\cdot\vec n\,\mathrm dS=\iint_\Sigma\!\vec G\cdot\vec n\,\mathrm dS+\iint_\Sigma\!\vec H\cdot\vec n\,\mathrm dS=4\pi+\frac{32}3\pi abc$.