高等数学基础 – 第 38 页

问题

以下关于向量的模的描述中，正确的是哪个？

选项

[A]. 向量的模就是向量的长度

[B]. 向量的模就是向量的别称

[C]. 向量的模就是向量的一种模拟

[D]. 向量的模就是向量的倾斜角度

答案

向量的模就是向量的长度：

向量 $\overrightarrow{AB}$ 的长度就是向量 $\overrightarrow{AB}$ 的模，记作：

$\overrightarrow{|AB|}$.

有时候，我们也称向量 $\overrightarrow{AB}$ 的模为向量 $\overrightarrow{AB}$ 的大小.

问题

已知有与 $X$ 轴、$Y$ 轴和 $Z$ 轴方向分别相同的单位向量 $i$, $j$, $k$, 此外，有且只有一组合适的实数 $x$, $y$, $z$, 则一下对向量 $a$ 的表示方式中，正确的是哪个？

选项

[A]. $a$ $=$ $xi$ $+$ $yj$ $+$ $zk$

[B]. $a$ $=$ $\frac{i}{x}$ $+$ $\frac{j}{y}$ $+$ $\frac{k}{z}$

[C]. $a$ $=$ $x+i$ $\times$ $y+j$ $\times$ $z+k$

[D]. $a$ $=$ $xi$ $\times$ $yj$ $\times$ $zk$

答案

$\textcolor{red}{a}$ $\textcolor{green}{=}$ $\textcolor{cyan}{x} \textcolor{orange}{i}$ $\textcolor{green}{+}$ $\textcolor{cyan}{y} \textcolor{orange}{j}$ $\textcolor{green}{+}$ $\textcolor{cyan}{z} \textcolor{orange}{k}$

空间区域的形心公式（B007）

问题

若空间区域 $\Omega$ 的体密度函数 $\rho(x, y, z)$ 为常数 $C$, 则该空间区域的 [形心] 坐标 $($ $\textcolor{orange}{\bar{x}}, \textcolor{orange}{\bar{y}}, \textcolor{orange}{\bar{z}}$ $)$ $=$ $?$

选项

[A]. $\begin{cases} & \bar{x} = \frac{C \iiint_{\Omega} x \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{C \iiint_{\Omega} y \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{C \iiint_{\Omega} z \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

[B]. $\begin{cases} & \bar{x} = \frac{\iiint_{\Omega} x^{2} \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{\iiint_{\Omega} y^{2} \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{\iiint_{\Omega} z^{2} \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

[C]. $\begin{cases} & \bar{x} = \frac{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} x \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} y \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} z \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

[D]. $\begin{cases} & \bar{x} = \frac{\iiint_{\Omega} x \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{\iiint_{\Omega} y \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{\iiint_{\Omega} z \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

答案

$\begin{cases} & \textcolor{orange}{\bar{x}} = \frac{\iiint_{\Omega} \textcolor{red}{x} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}}{\iiint_{\Omega} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}} \\ & \textcolor{orange}{\bar{y}} = \frac{\iiint_{\Omega} \textcolor{red}{y} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}}{\iiint_{\Omega} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}} \\ & \textcolor{orange}{\bar{z}} = \frac{\iiint_{\Omega} \textcolor{red}{z} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}}{\iiint_{\Omega} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}} \end{cases}$

平面图形的形心公式（B007）

问题

若平面图形 $D$ 的线密度函数 $\rho(x, y)$ 为常数 $C$, 则该平面图形的 [形心] 横坐标 $\textcolor{orange}{\bar{x}}$ 和纵坐标 $\textcolor{orange}{\bar{y}}$ 分别是多少？

选项

[A]. $\begin{cases} & \bar{x} = \frac{\iint_{D} \mathrm{d} x \mathrm{d} y}{\iint_{D} x \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{\iint_{D} \mathrm{d} x \mathrm{d} y}{\iint_{D} y \mathrm{d} x \mathrm{d} y} \end{cases}$

[B]. $\begin{cases} & \bar{x} = \frac{\iint_{D} x \mathrm{d} x \mathrm{d} y}{\iint_{D} \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{\iint_{D} y \mathrm{d} x \mathrm{d} y}{\iint_{D} \mathrm{d} x \mathrm{d} y} \end{cases}$

[C]. $\begin{cases} & \bar{x} = \frac{C \iint_{D} x \mathrm{d} x}{\iint_{D} \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{C \iint_{D} y \mathrm{d} y}{\iint_{D} \mathrm{d} x \mathrm{d} y} \end{cases}$

[D]. $\begin{cases} & \bar{x} = \frac{\iint_{D} x^{2} \mathrm{d} x \mathrm{d} y}{\iint_{D} \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{\iint_{D} y^{2} \mathrm{d} x \mathrm{d} y}{\iint_{D} \mathrm{d} x \mathrm{d} y} \end{cases}$

答案

$\begin{cases} & \textcolor{orange}{\bar{x}} = \frac{\iint_{D} \textcolor{red}{x} \textcolor{green}{\cdot} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}}{\iint_{D} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}} \\ & \textcolor{orange}{\bar{y}} = \frac{\iint_{D} \textcolor{red}{y} \textcolor{green}{\cdot} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}}{\iint_{D} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}} \end{cases}$

平面曲线的形心公式（B007）

问题

若平面曲线 $L$ 的线密度函数为 $\rho(x)$ 为常数 $C$, 则该平面曲线形心坐标中的横坐标 $\textcolor{orange}{\bar{x}}$ 和纵坐标 $\textcolor{orange}{\bar{y}}$ 分别是多少？

选项

[A]. $\begin{cases}& \bar{x} = \frac{C \int_{L} x \mathrm{d} s}{\int_{L} \mathrm{d} s} \\ & \bar{y} = \frac{C \int_{L} y \mathrm{d} s}{\int_{L} \mathrm{d} s} \end{cases}$

[B]. $\begin{cases}& \bar{x} = \frac{\int_{L} x \mathrm{d} s}{\int_{L} \mathrm{d} s} \\ & \bar{y} = \frac{\int_{L} y \mathrm{d} s}{\int_{L} \mathrm{d} s} \end{cases}$

[C]. $\begin{cases}& \bar{x} = \frac{\int_{L} x^{2} \mathrm{d} s}{\int_{L} \mathrm{d} s} \\ & \bar{y} = \frac{\int_{L} y^{2} \mathrm{d} s}{\int_{L} \mathrm{d} s} \end{cases}$

[D]. $\begin{cases}& \bar{x} = \frac{\int_{L} \mathrm{d} s}{\int_{L} x \mathrm{d} s} \\ & \bar{y} = \frac{\int_{L} \mathrm{d} s}{\int_{L} y \mathrm{d} s} \end{cases}$

答案

$\begin{cases}& \textcolor{orange}{\bar{x}} = \frac{\int_{L} \textcolor{red}{x} \textcolor{green}{\cdot} \mathrm{d} s}{\int_{L} \mathrm{d} s} \\ & \textcolor{orange}{\bar{y}} = \frac{\int_{L} \textcolor{red}{y} \textcolor{green}{\cdot} \mathrm{d} s}{\int_{L} \mathrm{d} s} \end{cases}$

空间区域的质心公式（B007）

问题

若空间区域 $\Omega$ 的体密度为 $\rho$, 则该空间区域质心坐标 $($ $\textcolor{orange}{\bar{x}}, \textcolor{orange}{\bar{y}}, \textcolor{orange}{\bar{z}}$ $)$ $=$ $?$

选项

[A]. $\begin{cases} & \bar{x} = \frac{\iiint_{\Omega} x \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{\iiint_{\Omega} y \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{\iiint_{\Omega} z \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

[B]. $\begin{cases} & \bar{x} = \frac{\iiint_{\Omega} x^{2} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{\iiint_{\Omega} y^{2} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{\iiint_{\Omega} z^{2} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

[C]. $\begin{cases} & \bar{x} = \frac{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} x \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} y \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} z \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

[D]. $\begin{cases} & \bar{x} = \frac{\iiint_{\Omega} x \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{y} = \frac{\iiint_{\Omega} y \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \\ & \bar{z} = \frac{\iiint_{\Omega} z \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z}{\iiint_{\Omega} \rho \mathrm{d} x \mathrm{d} y \mathrm{d} z} \end{cases}$

答案

$\begin{cases} & \textcolor{orange}{\bar{x}} = \frac{\iiint_{\Omega} \textcolor{red}{x} \textcolor{green}{\cdot} \textcolor{red}{\rho} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}}{\iiint_{\Omega} \textcolor{red}{\rho} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}} \\ & \textcolor{orange}{\bar{y}} = \frac{\iiint_{\Omega} \textcolor{red}{y} \textcolor{green}{\cdot} \textcolor{red}{\rho} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}}{\iiint_{\Omega} \textcolor{red}{\rho} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}} \\ & \textcolor{orange}{\bar{z}} = \frac{\iiint_{\Omega} \textcolor{red}{z} \textcolor{green}{\cdot} \textcolor{red}{\rho} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}}{\iiint_{\Omega} \textcolor{red}{\rho} \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y} \mathrm{d} \textcolor{cyan}{z}} \end{cases}$

平面图形的质心公式（B007）

问题

若平面图形 $D$ 的线密度为 $\rho(x, y)$, 则该平面图形质心坐标中的横坐标 $\textcolor{orange}{\bar{x}}$ 和纵坐标 $\textcolor{orange}{\bar{y}}$ 分别是多少？

选项

[A]. $\begin{cases} & \bar{x} = \frac{\iint_{D} x^{2} \rho(x, y) \mathrm{d} x \mathrm{d} y}{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{\iint_{D} y^{2} \rho(x, y) \mathrm{d} x \mathrm{d} y}{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y} \end{cases}$

[B]. $\begin{cases} & \bar{x} = \frac{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y}{\iint_{D} x \rho(x, y) \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y}{\iint_{D} y \rho(x, y) \mathrm{d} x \mathrm{d} y} \end{cases}$

[C]. $\begin{cases} & \bar{x} = \frac{\iint_{D} x \rho(x, y) \mathrm{d} x \mathrm{d} y}{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{\iint_{D} y \rho(x, y) \mathrm{d} x \mathrm{d} y}{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y} \end{cases}$

[D]. $\begin{cases} & \bar{x} = \frac{\iint_{D} x \mathrm{d} x}{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y} \\ & \bar{y} = \frac{\iint_{D} y \mathrm{d} y}{\iint_{D} \rho(x, y) \mathrm{d} x \mathrm{d} y} \end{cases}$

答案

$\begin{cases} & \textcolor{orange}{\bar{x}} = \frac{\iint_{D} \textcolor{red}{x} \textcolor{green}{\cdot} \textcolor{red}{\rho}(x, y) \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}}{\iint_{D} \textcolor{red}{\rho}(x, y) \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}} \\ & \textcolor{orange}{\bar{y}} = \frac{\iint_{D} \textcolor{red}{y} \textcolor{green}{\cdot} \textcolor{red}{\rho}(x, y) \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}}{\iint_{D} \textcolor{red}{\rho}(x, y) \mathrm{d} \textcolor{cyan}{x} \mathrm{d} \textcolor{cyan}{y}} \end{cases}$

平面曲线的质心公式（B007）

问题

若平面曲线 $L$ 的线密度为 $\rho(x)$, 则该平面曲线质心坐标中的横坐标 $\textcolor{orange}{\bar{x}}$ 和纵坐标 $\textcolor{orange}{\bar{y}}$ 分别是多少？

选项

[A]. $\begin{cases} & \bar{x} = \frac{\int_{L} x^{2} \rho \mathrm{d} s}{\int_{L} \rho \mathrm{d} s} \\ & \bar{y} = \frac{\int_{L} y^{2} \rho \mathrm{d} s}{\int_{L} \rho \mathrm{d} s} \end{cases}$

[B]. $\begin{cases} & \bar{x} = \frac{\int_{L} \rho \mathrm{d} s}{\int_{L} x \rho \mathrm{d} s} \\ & \bar{y} = \frac{\int_{L} \rho \mathrm{d} s}{\int_{L} y \rho \mathrm{d} s} \end{cases}$

[C]. $\begin{cases} & \bar{x} = \frac{\int_{L} x \rho \mathrm{d} s}{\int_{L} \rho \mathrm{d} s} \\ & \bar{y} = \frac{\int_{L} y \rho \mathrm{d} s}{\int_{L} \rho \mathrm{d} s} \end{cases}$

[D]. $\begin{cases} & \bar{x} = \frac{\int_{L} x \mathrm{d} s}{\int_{L} \rho \mathrm{d} s} \\ & \bar{y} = \frac{\int_{L} y \mathrm{d} s}{\int_{L} \rho \mathrm{d} s} \end{cases}$

答案

$\begin{cases} & \textcolor{orange}{\bar{x}} = \frac{\int_{L} \textcolor{red}{x} \textcolor{green}{\cdot} \textcolor{red}{\rho} \mathrm{d} s}{\int_{L} \textcolor{red}{\rho} \mathrm{d} s} \\ & \textcolor{orange}{\bar{y}} = \frac{\int_{L} \textcolor{red}{y} \textcolor{green}{\cdot} \textcolor{red}{\rho} \mathrm{d} s}{\int_{L} \textcolor{red}{\rho} \mathrm{d} s} \end{cases}$

空间区域的质量公式（B007）

问题

若空间区域 $\Omega$ 的体密度函数为 $\mu(x, y, z)$, 则空间区域 $\Omega$ 的质量 $m$ $=$ $?$

选项

[A]. $m$ $=$ $\iiint_{\Omega}$ $\mu(x, y, z)$ $\mathrm{d} x$ $\mathrm{d} y$ $\mathrm{d} z$

[B]. $m$ $=$ $\iiint_{\Omega}$ $\mu(x, y, z)$ $\mathrm{d} x$

[C]. $m$ $=$ $\iiint_{\Omega}$ $|\mu(x, y, z)|$ $\mathrm{d} x$ $\mathrm{d} y$ $\mathrm{d} z$

[D]. $m$ $=$ $\iint_{\Omega}$ $\mu(x, y, z)$ $\mathrm{d} x$ $\mathrm{d} y$

答案

$m$ $=$ $\iiint_{\textcolor{orange}{\Omega}}$ $\textcolor{red}{\mu}(x, y, z)$ $\mathrm{d} \textcolor{cyan}{x}$ $\mathrm{d} \textcolor{cyan}{y}$ $\mathrm{d} \textcolor{cyan}{z}$

平面图形的质量公式（B007）

问题

若平面图形 $D$ 的面密度为 $\rho(x, y)$, 则平面图形 $D$ 的质量 $m$ $=$ $?$

选项

[A]. $m$ $=$ $\iint_{D}$ $| \rho(x, y) |$ $\mathrm{d} x$ $\mathrm{d} y$

[B]. $m$ $=$ $\int_{D}$ $\rho(x, y)$ $\mathrm{d} x$ $\mathrm{d} y$

[C]. $m$ $=$ $\iint_{D}$ $\rho(x, y)$ $\mathrm{d} x$

[D]. $m$ $=$ $\iint_{D}$ $\rho(x, y)$ $\mathrm{d} x$ $\mathrm{d} y$

答案

$m$ $=$ $\iint_{\textcolor{orange}{D}}$ $\textcolor{red}{\rho}(x, y)$ $\mathrm{d} \textcolor{cyan}{x}$ $\mathrm{d} \textcolor{cyan}{y}$

平面曲线的质量公式（B007）

问题

若曲线 $L$ 的线密度为 $\rho(x)$, 则曲线 $L$ 的质量 $m$ $=$ $?$

选项

[A]. $m$ $=$ $\int_{L}$ $\rho(x)$ $\mathrm{d} y$

[B]. $m$ $=$ $\int_{L}$ $\rho(x)$ $\mathrm{d} x$

[C]. $m$ $=$ $\int$ $\rho(x)$ $\mathrm{d} s$

[D]. $m$ $=$ $\int_{L}$ $\rho(x)$ $\mathrm{d} s$

答案

若以曲线积分的方式描述，则线密度为 $\rho(x)$ 的曲线 $L$ 的质量为：

$m$ $=$ $\int_{\textcolor{cyan}{L}}$ $\textcolor{red}{\rho}(x)$ $\mathrm{d} \textcolor{orange}{s}$

计算旋转体的侧面积（B007）

问题

若平面图形由曲线 $y$ $=$ $y(x)$ 与直线 $x$ $=$ $a$, $x$ $=$ $b$ 和 $x$ 轴围成，则该图形绕 $x$ 轴旋转一周所形成的旋转体的侧面积 $S$ $=$ $?$

选项

[A]. $S$ $=$ $\pi$ $\int_{a}^{b}$ $|f(x)| \sqrt{1+f^{\prime 2}(x)}$ $\mathrm{d} x$

[B]. $S$ $=$ $2 \pi$ $\int_{a}^{b}$ $|f(x)| \sqrt{1+f^{\prime 2}(x)}$ $\mathrm{d} x$

[C]. $S$ $=$ $2 \pi$ $\int_{a}^{b}$ $|f(x)| \sqrt{1+f^{2}(x)}$ $\mathrm{d} x$

[D]. $S$ $=$ $2 \pi$ $\int_{a}^{b}$ $f(x) \sqrt{1+f^{\prime 2}(x)}$ $\mathrm{d} x$

答案

$S$ $=$ $\textcolor{cyan}{2} \textcolor{green}{\cdot} \textcolor{cyan}{\pi}$ $\int_{\textcolor{orange}{a}}^{\textcolor{orange}{b}}$ $\big[$ $\textcolor{Red}{|f(x)|} \textcolor{Red}{\sqrt{1+f^{\prime 2}(x)}}$ $\big]$ $\mathrm{d} x$

计算平行截面面积已知的立体体积（B007）

问题

如下图所示，若已知 $S(x)$ 为某立体垂直于 $x$ 轴的截面面积函数，则，如何使用定积分表示该立体在 $x$ $=$ $a$ 和 $x$ $=$ $b$ ($a$ $<$ $b$) 两个截面之间的体积 $V$ ?

选项

[A]. $V$ $=$ $\int_{a}^{b}$ $S(x)$ $\mathrm{d} x$

[B]. $V$ $=$ $\int_{a}^{b}$ $S^{2}(x)$ $\mathrm{d} x$

[C]. $V$ $=$ $\int_{a}^{b}$ $|S(x)|$ $\mathrm{d} x$

[D]. $V$ $=$ $\int_{b}^{a}$ $S(x)$ $\mathrm{d} x$

答案

$V$ $=$ $\int_{\textcolor{orange}{a}}^{\textcolor{orange}{b}}$ $\textcolor{red}{S(x)}$ $\mathrm{d} x$

基于极坐标系计算平面曲线的弧长（B007）

问题

若有极坐标系下的方程 $\rho$ $=$ $\rho(\theta)$, 且 $\alpha$ $\leqslant$ $\theta$ $\leqslant$ $\beta$, 则该极坐标方程在角度 $\theta$ 的取值范围 $[\alpha, \beta]$ 内的弧长 $L$ $=$ $?$

选项

[A]. $L$ $=$ $\int_{\alpha}^{\beta}$ $\sqrt{\rho^{2}(\theta) – \rho^{\prime 2}(\theta)}$ $\mathrm{d} \theta$

[B]. $L$ $=$ $\int_{\alpha}^{\beta}$ $\sqrt{\rho^{\prime 2}(\theta) + \rho^{2}(\theta)}$ $\mathrm{d} \theta$

[C]. $L$ $=$ $\int_{\alpha}^{\beta}$ $\sqrt{\rho^{2}(\theta) + \rho^{\prime 2}(\theta)}$ $\mathrm{d} \theta$

[D]. $L$ $=$ $\int_{\alpha}^{\beta}$ $\sqrt{\rho^{\prime 2}(\theta) – \rho^{\prime 2}(\theta)}$ $\mathrm{d} \theta$

答案

$L$ $=$ $\int_{\textcolor{orange}{\alpha}}^{\textcolor{orange}{\beta}}$ $\textcolor{red}{\sqrt{\rho^{2}(\theta) + \rho^{\prime 2}(\theta)}}$ $\mathrm{d} \theta$

基于参数方程计算平面曲线的弧长（B007）

问题

若有参数方程 $\begin{cases} & x = x(t) \\ & y = y(t) \end{cases}$, 且参数 $a$ $\leqslant$ $t$ $\leqslant$ $b$, 则该参数方程在区间 $[a, b]$ 上的弧长 $L$ $=$ $?$

选项

[A]. $L$ $=$ $\int_{a}^{b}$ $\sqrt{x^{2}(t) + y^{2}(t)} \mathrm{d} t$

[B]. $L$ $=$ $\int_{a}^{b}$ $\sqrt{x^{\prime}(t) + y^{\prime}(t)} \mathrm{d} t$

[C]. $L$ $=$ $\int_{a}^{b}$ $\sqrt{x^{\prime 2}(t) – y^{\prime 2}(t)} \mathrm{d} t$

[D]. $L$ $=$ $\int_{a}^{b}$ $\sqrt{x^{\prime 2}(t) + y^{\prime 2}(t)} \mathrm{d} t$

答案

$L$ $=$ $\int_{\textcolor{orange}{a}}^{\textcolor{orange}{b}}$ $\textcolor{red}{\sqrt{x^{\prime 2}(t) + y^{\prime 2}(t)}} \mathrm{d} t$