荒原之梦 - 第102页共241页 - 专注于考研数学|高等数学|线性代数|概率统计

向量 $\vec{a}$ 相对于 $y$ 轴的方向余弦：$\cos \beta$（B008）

问题

若向量 $\vec{a}$ 的坐标为 $(x, y, z)$ 且向量 $\vec{a}$ 与 $y$ 轴之间的夹角为 $\beta$, 则向量 $\vec{a}$ 相对于 $y$ 轴的方向余弦 $\cos \beta$ $=$ $?$

选项

[A]. $\cos \beta$ $=$ $\frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}}$

[B]. $\cos \beta$ $=$ $\frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}}$

[C]. $\cos \beta$ $=$ $\frac{y}{\sqrt{x + y + z}}$

[D]. $\cos \beta$ $=$ $\frac{y^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}}$

答案

$\textcolor{orange}{\cos} \textcolor{cyan}{\beta}$ $\textcolor{green}{=}$ $\frac{\textcolor{red}{y}}{\textcolor{red}{\sqrt{x^{2} + y^{2} + z^{2}}}}$

向量 $\vec{a}$ 相对于 $x$ 轴的方向余弦：$\cos \alpha$（B008）

问题

若向量 $\vec{a}$ 的坐标为 $(x, y, z)$ 且向量 $\vec{a}$ 与 $x$ 轴之间的夹角为 $\alpha$, 则向量 $\vec{a}$ 相对于 $x$ 轴的方向余弦 $\cos \alpha$ $=$ $?$

选项

[A]. $\cos \alpha$ $=$ $\frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}}$

[B]. $\cos \alpha$ $=$ $\frac{x}{\sqrt{x + y + z}}$

[C]. $\cos \alpha$ $=$ $\frac{|x|}{\sqrt{x^{2} + y^{2} + z^{2}}}$

[D]. $\cos \alpha$ $=$ $\frac{x^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}}$

答案

$\textcolor{orange}{\cos} \textcolor{cyan}{\alpha}$ $\textcolor{green}{=}$ $\frac{\textcolor{red}{x}}{\textcolor{red}{\sqrt{x^{2} + y^{2} + z^{2}}}}$

向量的单位化（B008）

问题

若有向量 $\vec{a}$ 和其模 $\vec{|a|}$, 且向量 $\vec{a}$ 的坐标为 $(x, y, z)$, 则如何计算向量 $\vec{a}$ 的单位化向量的坐标？

选项

[A]. $\frac{\vec{a}}{\vec{|a|}}$ $=$ $($ $\frac{\sqrt{x}}{\sqrt{x^{2} + y^{2} + z^{2}}}$, $\frac{\sqrt{y}}{\sqrt{x^{2} + y^{2} + z^{2}}}$, $\frac{\sqrt{z}}{\sqrt{x^{2} + y^{2} + z^{2}}}$ $)$

[B]. $\frac{\vec{a}}{\vec{|a|}}$ $=$ $($ $\frac{x^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}}$, $\frac{y^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}}$, $\frac{z^{2}}{\sqrt{x^{2} + y^{2} + z^{2}}}$ $)$

[C]. $\frac{\vec{a}}{\vec{|a|}}$ $=$ $($ $\frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}}$, $\frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}}$, $\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}$ $)$

[D]. $\frac{\vec{a}}{\vec{|a|}}$ $=$ $($ $\frac{x}{\sqrt{x^{2} \cdot y^{2} \cdot z^{2}}}$, $\frac{y}{\sqrt{x^{2} \cdot y^{2} \cdot z^{2}}}$, $\frac{z}{\sqrt{x^{2} \cdot y^{2} \cdot z^{2}}}$ $)$

答案

向量 $\vec{a}$ 单位向量的坐标为：

$\frac{\textcolor{orange}{\vec{a}}}{\textcolor{orange}{\vec{|a|}}}$ $\textcolor{green}{=}$ $\big($ $\frac{\textcolor{red}{x}}{\textcolor{cyan}{\sqrt{x^{2} + y^{2} + z^{2}}}}$, $\frac{\textcolor{red}{y}}{\textcolor{cyan}{\sqrt{x^{2} + y^{2} + z^{2}}}}$, $\frac{\textcolor{red}{z}}{\textcolor{cyan}{\sqrt{x^{2} + y^{2} + z^{2}}}}$ $\big)$

如何计算向量的模（B008）

问题

若向量 $\vec{a}$ 的坐标为 $(x, y, z)$, 则向量 $\vec{a}$ 的模 $\vec{|a|}$ $=$ $?$

选项

[A]. $\vec{|a|}$ $=$ $\sqrt{x^{3} + y^{3} + z^{3}}$

[B]. $\vec{|a|}$ $=$ $\sqrt{x + y + z}$

[C]. $\vec{|a|}$ $=$ $\sqrt{x^{2} + y^{2} + z^{2}}$

[D]. $\vec{|a|}$ $=$ $\sqrt{|x|^{2} + |y|^{2} + |z|^{2}}$

答案

$\textcolor{orange}{\vec{|a|}}$ $=$ $\sqrt{x^{\textcolor{red}{2}} + y^{\textcolor{red}{2}} + z^{\textcolor{red}{2}}}$

什么是向量的模（B008）

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别称

[B]. 向量的模就是向量的一种模拟

[C]. 向量的模就是向量的倾斜角度

[D]. 向量的模就是向量的长度

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.

向量 $a$ 的坐标表示（B008）

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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：

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