定理1
设\(\mathbb{R}^2\)上的曲线\(\Gamma\)由极坐标方程
\[ r = r(\theta) \quad (\boldsymbol{\alpha} \le \theta \le \boldsymbol{\beta}) \]
表示，从而由此曲线和射线
\[ \theta = \alpha, \quad \theta = \beta \]
围城的区域的面积\(S\)为
\[ S = \frac{1}{2} \int_{\alpha}^{\beta} r^2 (\theta) \mathrm{d}(\theta) \]
证：对\(\theta\)的变化范围\([\alpha, \beta]\)作一分割
\[
\alpha = \theta_1 < \theta_2 < \cdots < \theta_n = \beta
\]
任取\(\xi_i \in [\theta_{i-1}, \theta_{i}]
(i=1,2,\cdots,n)\)，从而夹在\(\theta =
\theta_{i-1}, \theta = \theta_i\)与\(r
= r(\theta)\)之间的区域的面积，在\(\Delta \theta_i = \theta_{i} -
\theta_{i-1}\)很小时可表示为
\[
\Delta S_i \approx \frac{1}{2} r^2(\theta_{i}) \Delta \theta_i
\]
令\(\max \limits_{1 \le i \le n} \Delta
\theta_i \to 0\)，得到
\[
S = \frac{1}{2} \int_{\alpha}^{\beta} r^2 (\theta)
\mathrm{d}(\theta)
\]
Q.E.D.
引理1
设\(x(t), y(t), z(t)\)在\([\alpha, \beta]\)上有连续的导函数，取\(t\)的一个分割
\[ \pi: \alpha = t_0 < t_1 < \cdots < t_n = \beta \]
那么有
\[ \lim \limits_{\Vert \pi \Vert \to 0} \sum_{i=1}^n \sqrt{(x^\prime(\xi_i))^2 + (y^\prime(\eta_i))^2 + (z^\prime(\zeta_i))^2} \Delta t_i= \int_{\alpha}^{\beta} \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2} \mathrm{d}t \]
其中\(\xi_i, \eta_i, \zeta_i\)分别为\(x, y, z\)函数的值点，即\(\xi_i,\eta_i,\zeta_i \in [t_{i-1}, t_i]\)。
证： 由三角不等式
\[
| \Vert \boldsymbol{a} \Vert - \Vert \boldsymbol{b}
\Vert| \le \Vert \boldsymbol{a} - \boldsymbol{b} \Vert
\]
得
\[
\begin{aligned}
& \left| \sqrt {(x^\prime(\xi_i))^2 + (y^\prime(\xi_i))^2 +
(z^\prime(\xi_i))^2} - \sqrt {(x^\prime(\xi_i))^2 + (y^\prime(\eta_i))^2
+ (z^\prime(\zeta_i))^2} \right| \\
& \le \sqrt {(y^\prime(\xi_i) - y^\prime(\eta_i))^2 +
(z^\prime(\xi_i) - z^\prime(\zeta_i))^2} \\
& \le |y^\prime(\xi_i) - y^\prime(\eta_i)| + |z^\prime(\xi_i) -
z^\prime(\zeta_i)|
\end{aligned}
\]
由于\(y^\prime(t)\)与\(z^\prime(t)\)在\([\alpha,
\beta]\)上连续，从而一致连续，因此对任意的\(\varepsilon > 0\)，存在\(\delta_1 > 0\)，当\(\Vert \pi \Vert < \delta_1\)时，
\[
\left\{
\begin{aligned}
|y^\prime(\xi_i) - y^\prime(\eta_i)| <
\frac{\varepsilon}{4(\beta - \alpha)} \\
|z^\prime(\xi_i) - z^\prime(\zeta_i)| <
\frac{\varepsilon}{4(\beta - \alpha)} \\
\end{aligned}
\right.
\]
对\(i=1,2,\cdots,n\)都成立，令
\[
\begin{aligned}
I = \int_{\alpha}^{\beta} \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 +
(z^\prime(t))^2} \mathrm{d}t \\
S = \sum_{i=1}^n \sqrt {(x^\prime(\xi_i))^2 + (y^\prime(\xi_i))^2 +
(z^\prime(\xi_i))^2} \Delta t_i
\end{aligned}
\]
由积分的定义可知，对上述\(\varepsilon\)，存在\(\delta_2 > 0\)，当\(\Vert \pi \Vert \le \delta_2\)时，有
\[
|S - I| < \frac{\varepsilon}{2}
\]
而不论\(\xi_i \in [t_{i-1}, t_i]
(i=1,2,\cdots,n)\)如何选取，当\(\Vert
\pi \Vert < \min(\delta_1, \delta_2)\)时，有
\[
\begin{aligned}
& |\sum_{i=1}^n \sqrt {(x^\prime(\xi_i))^2 +
(y^\prime(\eta_i))^2 + (z^\prime(\zeta_i))^2} \Delta t_i- I| \\
& \le |\sum_{i=1}^n \sqrt {(x^\prime(\xi_i))^2 +
(y^\prime(\eta_i))^2 + (z^\prime(\zeta_i))^2} \Delta t_i - S| + |S - I|
\\
& < \sum_{i=1}^n (|y^\prime(\xi_i) - y^\prime(\eta_i)| +
|z^\prime(\xi_i) - z^\prime(\zeta_i)|) \Delta t_i +
\frac{\varepsilon}{2} \\
& < \left( \frac{\varepsilon}{4(\beta - \alpha)} +
\frac{\varepsilon}{4(\beta - \alpha)} \right) \sum_{i=1}^n \Delta t_i +
\frac{\varepsilon}{2} \\
& = \varepsilon
\end{aligned}
\]
Q.E.D.
定理2
设\(\mathbb{R}^3\)的曲线\(\Gamma\)的参数方程为
\[ \left\{ \begin{aligned} x = x(t), \\ y = y(t), \\ z = z(t), \end{aligned} \right. \quad (\alpha \le t \le \beta) \]
或用向量形式表示为
\[ \boldsymbol{r} = \boldsymbol{r}(t) \quad (\alpha \le t \le \beta) \]
其中\(x(t), y(t), z(t)\)在\([\alpha, \beta]\)上有连续的导数，点\(A = \boldsymbol{r}(\alpha)\)与\(B = \boldsymbol{r}(\beta)\)分别是\(\Gamma\)的起点与终点。则该曲线的弧长公式为
\[ s(\Gamma) = \int_{\alpha}^{\beta} \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2} \mathrm{d}t \]
证：沿\(A\)到\(B\)的方向在\(\Gamma\)上取\(n+1\)个点：
\[
A = A_0, A_1, A_2, \cdots, A_n = B
\]
并把这\(n\)条线段之和\(\sum_{i=1}^n |A_{i-1}A_i|\)作为\(\Gamma\)弧长的一个近似值，可知当分割点越来越细时，近似值的极限就是弧长。设点\(A_i\)对应着参数值\(t_i (i=0,1,2,\cdots,n)\)，则
\[
\alpha = t_0 < t_1 < t_2 < \cdots < t_n = \beta
\]
记为分割\(\pi\)，从而
\[
\begin{aligned}
|A_{i-1}A_i| &= \Vert \boldsymbol{r}(t_i) -
\boldsymbol{r}(t_{i-1}) \Vert \\
&= \sqrt {((x(t_i) - x(t_{i-1}))^2 + (y(t_i) - y(t_{i-1}))^2 +
(z(t_i) - z(t_{i-1}))^2)} \\
&= \sqrt {(x^\prime(\xi_i))^2 + (y^\prime(\eta_i))^2 +
(z^\prime(\zeta_i))^2} \Delta t_i
\end{aligned}
\]
其中\(\xi_i,\eta_i, \zeta_i \in (t_{i-1},
t_i)\)，又由于\(x^\prime, y^\prime,
z^\prime\)都是连续函数，所以存在一个常数\(K\)，使得
\[
|A_{i-1}A_i| \le K \Delta t_i \le K \Vert \pi \Vert
\]
所以
\[
\max \limits_{1 \le i \le n} | A_{i-1}A_i | \le K \Vert \pi \Vert
\]
这表明，将分割\(\pi\)无限加细，对应的曲线\(\Gamma\)上的分割也会无限加细，从而\(\Gamma\)的弧长为
\[
\lim \limits_{\Vert \pi \Vert \to 0} \sum_{i=1}^n
\sqrt{(x^\prime(\xi_i))^2 + (y^\prime(\eta_i))^2 +
(z^\prime(\zeta_i))^2} \Delta t_i
\]
再由引理1可知，弧长
\[
S(\Gamma) = \int_{\alpha}^{\beta} \sqrt {(x^\prime(t))^2 +
(y^\prime(t))^2 + (z^\prime(t))^2} \mathrm{d} t
\]
Q.E.D.
定理3
（1）设\(y = f(x) \ge 0\)是区间\([a,b]\)上的一条连续曲线，让这条曲线绕\(x\)轴旋转一周，得到的旋转体的体积公式为
\[ V = \pi \int_a^b f^2(x) \mathrm{d} x \]
（2）设曲线\(\Gamma\)
\[ x= x(t), y = y(t) \quad (\alpha \le t \le \beta) \]
是一条在上半平面不自交的\(C^1\)类曲线，让这条曲线绕\(Ox\)轴旋转一周，生成的旋转曲面的面积为
\[ S = 2\pi \int_{\alpha}^{\beta} y(t) \sqrt {(x^\prime(t))^2 + (y^\prime(t))^2} \mathrm{d} t \] b
证：（1）作区间\([a,b]\)的一个分割
\[
\pi: a = x_0 < x_1 < \cdots < x_n = b
\]
将旋转体介于平面\(x = x_{k-1}, x=
x_{k}\)之间的体积记为\(V_k\)，任取\(\xi_k \in [x_{k-1}, x_k]\)，则
\[
V_k \approx \pi f^2(\xi_k) \Delta x_k
\]
从而
\[
V = \lim \limits_{\Vert \pi \Vert \to 0} \sum_{k=1}^n \pi f^2(\xi_k)
\Delta x_k = \pi \int_a^b f^2(x) \mathrm{d} x
\]
（2）在\(\Gamma\)上取\(n+1\)个点
\[
A_0,A_1,\cdots,A_n
\]
从而对应的参数区间\([\alpha,
\beta]\)也有分点
\[
\alpha = t_0 < t_1 < \cdots < t_n = \beta
\]
这时曲线上的第\(i\)段\(A_{i-1}A_i\)的弧长可近似为
\[
\sqrt {(x^\prime(\xi_i))^2 + (y^\prime(\eta_i))^2} \Delta t_i
\]
其中\(\xi_i, \eta_i \in [t_{i-1},
t_i]\)，由这段曲线弧旋转而成的曲面面积可表示为
\[
\Delta S_i \approx 2 \pi y(\zeta_i) \sqrt {(x^\prime(\xi_i))^2 +
(y^\prime(\eta_i))^2} \Delta t_i
\]
其中\(\zeta_i \in [t_{i-1},
t_i]\)，所以旋转曲面的总面积可表示为
\[
S \approx \sum_{i=1}^n 2\pi y(\zeta_i) \sqrt {(x^\prime(\xi_i))^2 +
(y^\prime(\eta_i))^2} \Delta t_i
\]
当\(\Vert \pi \Vert \to
0\)时，利用类似引理1的证法，可得
\[
S = \lim \limits_{\Vert \pi \Vert \to 0} \sum_{i=1}^n 2\pi
y(\zeta_i) \sqrt {(x^\prime(\xi_i))^2 + (y^\prime(\eta_i))^2} \Delta t_i
= 2\pi \int_{\alpha}^{\beta} y(t) \sqrt {(x^\prime(t))^2 +
(y^\prime(t))^2} \mathrm{d} t
\]
Q.E.D.