考研数学二 – 第 116 页

向量的加法运算法则（B008）

问题

若向量 $\vec{a}$ $=$ $(x_{1}, y_{1}, z_{1})$, 向量 $\vec{b}$ $=$ $(x_{2}, y_{2}, z_{2})$, 则 $\vec{a}$ $+$ $\vec{b}$ $=$ $?$

选项

[A]. $\vec{a}$ $+$ $\vec{b}$ $=$ $($ $x_{1} – x_{2}$, $y_{1} – y_{2}$, $z_{1} – z_{2}$ $)$

[B]. $\vec{a}$ $+$ $\vec{b}$ $=$ $($ $x_{1} + x_{2}$, $y_{1} + y_{2}$, $z_{1} + z_{2}$ $)$

[C]. $\vec{a}$ $+$ $\vec{b}$ $=$ $($ $\frac{x_{1}}{x_{2}}$, $\frac{y_{1}}{y_{2}}$, $\frac{z_{1}}{z_{2}}$ $)$

[D]. $\vec{a}$ $+$ $\vec{b}$ $=$ $($ $x_{2} – x_{1}$, $y_{2} – y_{1}$, $z_{2} – z_{1}$ $)$

答案

$\textcolor{red}{\vec{a}}$ $\textcolor{green}{+}$ $\textcolor{red}{\vec{b}}$ $=$ $($ $\textcolor{orange}{x}_{\textcolor{cyan}{1}} \textcolor{green}{+} \textcolor{orange}{x}_{\textcolor{cyan}{2}}$, $\textcolor{orange}{y}_{\textcolor{cyan}{1}} \textcolor{green}{+} \textcolor{orange}{y}_{\textcolor{cyan}{2}}$, $\textcolor{orange}{z}_{\textcolor{cyan}{1}} \textcolor{green}{+} \textcolor{orange}{z}_{\textcolor{cyan}{2}}$ $)$

两点间有向线段的坐标表示（B008）

问题

若点 $M_{1}$ 的坐标为 $(x_{1}, y_{1}, z_{1})$, 点 $M_{2}$ 的坐标为 $(x_{2}, y_{2}, z_{2})$, 则有向线段 $\overrightarrow{M_{1} M_{2}}$ 的坐标为多少？

选项

[A]. $\overrightarrow{M_{1} M_{2}}$ $=$ $($ $\frac{x_{2}}{x_{1}}$, $\frac{y_{2}}{y_{1}}$, $\frac{z_{2}}{z_{1}}$ $)$

[B]. $\overrightarrow{M_{1} M_{2}}$ $=$ $($ $x_{2} \times x_{1}$, $y_{2} \times y_{1}$, $z_{2} \times z_{1}$ $)$

[C]. $\overrightarrow{M_{1} M_{2}}$ $=$ $($ $x_{2} + x_{1}$, $y_{2} + y_{1}$, $z_{2} + z_{1}$ $)$

[D]. $\overrightarrow{M_{1} M_{2}}$ $=$ $($ $x_{2} – x_{1}$, $y_{2} – y_{1}$, $z_{2} – z_{1}$ $)$

答案

$\textcolor{red}{\overrightarrow{M_{1} M_{2}}}$ $=$ $($ $\textcolor{cyan}{x_{2}} \textcolor{orange}{-} \textcolor{cyan}{x_{1}}$, $\textcolor{cyan}{y_{2}} \textcolor{orange}{-} \textcolor{cyan}{y_{1}}$, $\textcolor{cyan}{z_{2}} \textcolor{orange}{-} \textcolor{cyan}{z_{1}}$ $)$

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

问题

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

选项

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

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

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

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

答案

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

答案

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