- Find the integral $\displaystyle\int_0^2\frac{12x^2}{\sqrt{16-x^2}}dx$. ($15\%$)
- 當 $x=0$ 則有 $\theta=0$;
- 當 $x=2$ 則有 $\displaystyle\theta=\frac\pi6$;
- 求導便有 $dx=4\cos\theta\,d\theta$。
- Let $\displaystyle f\left(x\right)=\sum_{n=1}^{+\infty}\frac{n}{4^n}x^n$ for any $\left|x\right|<4$.
Find the function $f\left(x\right)$ and the value $f\left(2\right)$. ($15\%$) - Let $D=\left\{\left(x,y,z\right)|x^2+y^2+z^2=50\right\}$ and the temperature at the point $\left(x,y,z\right)$ on the $D$ is $T\left(x,y,z\right)=12xz+16yz$. Find the highest and lowest temperatures. ($20\%$)
- 若 $\lambda=0$,則 $z=0$,因此應解 $\left\{\begin{aligned} &3x+4y=0,\\&x^2+y^2=50.\end{aligned}\right.$,易得 $\left(x,y\right)=\left(4\sqrt2,-3\sqrt2\right)$ 或 $\left(x,y\right)=\left(-4\sqrt2,3\sqrt2\right)$。
- 若 $\lambda\neq0$,則 $x^2+y^2=z^2$,配合第四式有 $z^2=25$,即 $z=\pm5$,代回第一式與第二式分別有 $\pm30+\lambda x=0$、$\pm40+\lambda y=0$,且知 $x^2+y^2=25$,因此有 $\displaystyle\frac{900}{\lambda^2}+\frac{1600}{\lambda^2}=25$,故 $\lambda=\pm10$,若 $\lambda=10$,則有 $x=-3$、$y=-4$;若 $\lambda=-10$,則有 $x=3$、$y=4$。
- Let $f\left(x\right)$ be a function such that
- $f\left(x\right)$ is continuous on $\left[0,1\right]$, $f\left(0\right)=0$ and $f\left(1\right)=7$.
- $f'\left(x\right)$ and $f''\left(x\right)$ exist for any $0< x<1$.
- $f''\left(x\right)>0$ for any $0< x<1$.
[Hint: there is $0<\alpha<1$ such that $f'\left(\alpha\right)=7$.] - Let $D$ be the region bounded by the curve $C:\left\{\begin{aligned}&x=t^3+t^2\\&y=t^2+t\end{aligned}\right.$ for $0\leq t\leq1$ and the line $y=x$. Find the area of $D$. ($15\%$)
- Let $\vec{F}\left(x,y,z\right)=xy^2\vec{i}+yz^2\vec{j}+zx^2\vec{k}$ be a vector field and $S=\left\{\left(x,y,z\right)|x^2+y^2+z^2=25\right\}$ be a surface with outward orientation. Find the flux $\displaystyle\iint_S\vec{F}\cdot\vec{N}\,dS$ of $\vec{F}\left(x,y,z\right)$. ($20\%$)
訣竅
利用三角代換處理。解法
令 $x=4\sin\theta$,則$\begin{aligned}\displaystyle\int_0^2\frac{12x^2}{\sqrt{16-x^2}}dx&=\int_0^{\frac\pi6}\frac{12\times16\sin^2\theta}{4\cos\theta}\times4\cos\theta\,d\theta\\&=96\int_0^{\frac\pi6}\left(1-\cos2\theta\right)d\theta\\&=96\theta-48\sin2\theta\Big|_0^{\frac\pi6}=16\pi-24\sqrt3.\end{aligned}$
訣竅
利用冪級數的微分或積分的特性求解即可。解法
容易注意到 $f$ 可以作如下的改寫與計算:$\displaystyle f\left(x\right)=\frac{x}4\sum_{n=1}^{\infty}n\left(\frac{x}4\right)^{n-1}=x\left[\sum_{n=0}^{\infty}\left(\frac{x}4\right)^n\right]'=x\left(\frac1{\displaystyle1-\frac{x}4}\right)'=\frac{4x}{\left(4-x\right)^2}$.
故 $f\left(2\right)=2$。訣竅
可以運用拉格朗日乘子法或初等不等式的方法求解。解法一
考慮拉格朗日乘子函數如下:$F\left(x,y,z,\lambda\right)=12xz+16yz+\lambda\left(x^2+y^2+z^2-50\right)$
如此解下列聯立方程組:$\displaystyle\left\{\begin{aligned}&F_x\left(x,y,z,\lambda\right)=12z+2\lambda x=0,\\&F_y\left(x,y,z,\lambda\right)=16z+2\lambda y=0,\\&F_z\left(x,y,z,\lambda\right)=12x+16y+2\lambda z=0,\\F_\lambda\left(x,y,z,\lambda\right)=x^2+y^2+z^2-50=0.\end{aligned}\right.$
我們將第一式乘上 $x$、第二式乘上 $y$、第三式乘上 $z$,並且將前兩者相加後減去第三個,如此可得 $2\lambda\left(x^2+y^2-z^2\right)=0$。解法二
可以留意到如下的不等式:$\displaystyle500=10x^2+10y^2+10z^2=\left(10x^2+\frac{18}{5}z^2\right)+\left(10y^2+\frac{32}{5}z^2\right)\geq12xy+16yz$
以及$\displaystyle\begin{aligned}-500&=-10x^2-10y^2-10z^2=-\left(10x^2+\frac{18}{5}z^2\right)-\left(10y^2+\frac{32}{5}z^2\right)\\&\leq12xy+16yz\end{aligned}$
因此最大與最小值分別為 $500$ 與 $-500$,其中等號成立條件分別為 $\left(3,4,5\right)$、$\left(-3,-4,-5\right)$ 與 $\left(3,4,-5\right)$、$\left(-3,-4,5\right)$。訣竅
利用均值定理與單調性來證明問題。解法
設 $g\left(x\right)=f\left(x\right)-7x$。由條件 $1$ 可知 $g\left(0\right)=g\left(1\right)=0$,因此透過均值定理可以知道在 $\left(0,1\right)$ 中存在一個實數 $\alpha$使得 $g'\left(\alpha\right)=f\left(\alpha\right)-7=0$。
又由於 $g''\left(x\right)=f''\left(x\right)>0$,因此 $g'$ 為嚴格遞增函數,故 $g'$ 在 $\left(0,\alpha\right)$ 為負而在 $\left(\alpha,1\right)$ 上為正。從而表明 $g$ 在 $\left(0,\alpha\right)$ 上遞減且在 $\left(\alpha,1\right)$ 上遞增,亦即 $g$ 在 $x=\alpha$ 處有最小值。
從而在 $\left(0,\alpha\right)$ 上遞減可知 $0=g\left(0\right)>g\left(x\right)$,而在 $\left(\alpha,1\right)$ 遞增可知 $0=g\left(1\right)>g\left(x\right)$,總結而言,在 $\left(0,1\right)$ 上有 $g\left(x\right)=f\left(x\right)-7x<0$,從而證明本命題。訣竅
直接列式計算即可。解法一
容易注意到 $y\geq x$,由此可以直接列式計算如下:$\displaystyle\begin{aligned}D&=\int_0^1\left[y\left(t\right)-x\left(t\right)\right]x'\left(t\right)dt\\&=\int_0^1\left(t-t^3\right)\left(3t^2+2t\right)dt\\&=\int_0^1\left(2t^2+3t^3-2t^4-3t^5\right)dt\\&=\left.\frac{2t^3}{3}+\frac{3t^4}{4}-\frac{2t^5}{5}-\frac{t^6}{2}\right|_0^1\\&=\frac{31}{60}.\end{aligned}$
解法二
記 $\left(2,2\right)$ 為 $A$ 點,接著作過 $A$ 並垂直於 $x$ 軸的直線交於 $B\left(2,0\right)$。如此可以注意到三角形 $OAB$ 的面積為 $2$。故所求的面積可以列式如下:$\displaystyle\begin{aligned}D&=\int_{t=0}^{t=1}y\left(t\right)x'\left(t\right)dt-2\\&=\int_0^1\left(t^2+t\right)\left(3t^2+2t\right)dt-2\\&=\int_0^1\left(3t^4+5t^3+2t^2\right)dt-2\\&=\left.\frac{3t^5}{5}+\frac{5t^4}{4}+\frac{2t^3}{3}\right|_0^1-2\\&=\frac{31}{60}.\end{aligned}$
訣竅
利用高斯散度定理或運用曲面積分的定義計算之。解法一
設 $K=\left\{\left(x,y,z\right)\in\mathbb{R}^3|\,x^2+y^2+z^2\leq25\right\}$。運用高斯散度定理,我們將曲面積分改為三重積分並計算如下:$\displaystyle\iint_S\vec{F}\cdot\vec{N}\,dS=\iiint_K\nabla\cdot\vec{F}dV=\iiint_K\left(x^2+y^2+z^2\right)dV$
運用球座標,令 $\left\{\begin{aligned}&x=\rho\cos\theta\sin\phi\\&y=\rho\sin\theta\sin\phi\\&z=\rho\cos\phi\end{aligned}\right.$,如此原曲面積分可以改寫並計算如下:$\begin{aligned}\displaystyle\iint_S\vec{F}\cdot\vec{N}\,dS&=\int_0^\pi\int_0^{2\pi}\int_0^5\rho^2\times\rho^2\sin\phi\,d\rho\,d\theta\,d\phi\\&=\left(\int_0^\pi\sin\phi\,d\phi\right)\left(\int_0^{2\pi}d\theta\right)\left(\int_0^5\rho^4d\rho\right)\\&=2\times2\pi\times\frac{5^5}{5}\\&=2500\pi.\end{aligned}$
解法二
運用球面座標參數化 $S: {\bf r}\left(s,t\right)=\left(5\cos s\sin t,5\sin s\sin t,5\cos t\right)$,如此有$\begin{aligned}\vec{N}&=\left(\cos s\sin t,\sin s\sin t,\cos t\right),\\dS&=\left|\frac{\partial{\bf r}}{\partial s}\times\frac{\partial{\bf r}}{\partial t}\right|dsdt\\&=\left|\left(-25\cos s\sin^2t,-25\sin s\sin^2t,-25\sin t\cos t\right)\right|dsdt\\&=25\sin tdsdt\end{aligned}$
因此所求的曲面積分可以計算如下:$\begin{aligned}\displaystyle&\int_0^\pi\int_0^{2\pi}\left(125\cos s\sin^2s\sin^3t,125\sin s\sin t\cos^2t,125\cos^2s\cos t\sin^2t\right)\cdot\left(\cos s\sin t,\sin s\sin t,\cos t\right)\times25\sin tdsdt\\=&3125\int_0^\pi\int_0^{2\pi}\left(\cos^2s\sin^2s\sin^4t+\sin^2s\sin^2t\cos^2t+\cos^2s\cos^2t\sin^2t\right)\sin tdsdt\\=&3125\int_0^\pi\int_0^{2\pi}\cos^2s\sin^2s\sin^5tdsdt+3125\int_0^\pi\int_0^{2\pi}\cos^2t\sin^3tdsdt\\=&3125\left(\int_0^\pi\sin^5tdt\right)\left(\int_0^{2\pi}\frac{\sin^22s}{4}ds\right)+6250\pi\int_0^\pi\cos^2t\left(\cos^2t-1\right)d\cos t\\=&-\frac{3125}{4}\left(\int_0^\pi\left(1-\cos^2t\right)^2d\cos t\right)\left(\int_0^{2\pi}\frac{1-\cos4s}{2}ds\right)+6250\pi\left.\left(\frac{\cos^5s}{5}-\frac{\cos^3s}{3}\right)\right|_0^\pi\\=&-\frac{3125}{8}\left(\int_0^\pi\left(1-2\cos^2t+\cos^4t\right)d\cos t\right)\times\left.\left(s-\frac{\sin4s}{4}\right)\right|_0^{2\pi}+\frac{5000\pi}{3}\\=&-\frac{3125\pi}{4}\times\left.\left(\cos t-\frac{2\cos^3t}{3}+\frac{\cos^5t}{5}\right)\right|_0^\pi+\frac{5000\pi}{3}\\=&\frac{2500\pi}{3}+\frac{5000\pi}{3}\\=&2500\pi.\end{aligned}$
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