6.7 定积分的应用

1 平面图形的面积

1.1 直角坐标的情形

从定义看出, Riemann 积分实际上是函数有向面积的代数和.

$x = a, x = b, y = f (x), y = g (x)$ 围成的图形面积为 $S = \int_{a}^{b} (f (x) - g (x)) d x$ .
$y = c, y = d, x = φ (y), x = ψ (y)$ 围成的图形面积为 $S = \int_{c}^{d} (φ (y) - ψ (y)) d y$ .

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266872c1029f979dc7c01b022517d8e903a8da32: $φ (x)$

5d7a1b4d6925da32f944dce7f8d6fdd405effbca: $ψ (x)$

例子

计算 $y^{2} = 2 x, y = x - 4$ 围成图形的面积.

按照 $x$ 方向, $S_{1} = \int_{0}^{2} (\sqrt{2 x} - (- \sqrt{2 x})) d x = \frac{16}{3}, S_{2} = \int_{2}^{8} (\sqrt{2 x} - (x - 4)) d x = \frac{156}{3} - 6$ , 从而 $S = 18$ .
按照 $y$ 方向, $S = \int_{- 2}^{4} (y + 4 - \frac{y^{2}}{2}) d y = 18$ .

1.2 参数方程的情形

考虑参数方程 $C : {\begin{aligned} x = x (t), \\ y = y (t), \end{aligned}$ $x^{' 2} (t) + y^{' 2} (t) \neq 0$ , $x^{'} (t), y^{'} (t)$ 连续. 不妨设 $x^{'} (t) \neq 0, x^{'} (t) > 0$ , 则 $x^{'} (t)$ 有反函数, 从而 $y = y (x^{- 1} (x))$ , $S = \int_{a}^{b} y (x^{- 1} (x)) d x$ . 令 $t = x^{- 1} (x)$ , 则 $\begin{matrix} (1.1) & S = \int_{t_{0}}^{t_{1}} y (t) x^{'} (t) d t . \end{matrix}$
一般地, $S = \int_{t_{0}}^{t_{1}} | y (t) x^{'} (t) | d t, t \in [t_{0}, t_{1}]$ .

例子

考虑摆线 $C : {\begin{aligned} x = a (t - \sin t), \\ y = a (1 - \cos t), \end{aligned}$ $0 \leq t \leq 2 π$ . 令 $x = a (1 - \sin t)$ , 则 $S = \int_{0}^{2 π a} y d x = \int_{0}^{2 π} a (1 - \cos t) d a (t - \sin t) = \int_{0}^{2 π} a^{2} (1 - \cos t)^{2} d t .$

1.3 极坐标情形

考虑 $C : r = r (θ)$ , $α \leq θ \leq β$ . 取分割 $π : α = θ_{0} < θ_{1} < \dots < θ_{n} = β$ , 回顾 Darbox和和 Riemann可积准则, $\begin{aligned} \underset{―}{S} (\frac{1}{2} r^{2} (θ), π) = \sum_{i = 1}^{n} \frac{1}{2} m_{i}^{2} Δ θ_{i} \leq \sum_{i = 1}^{n} \frac{1}{2} r_{i}^{2} (ξ_{i}) Δ θ_{i} \\ \leq & \sum_{i = 1}^{n} \frac{1}{2} M_{i}^{2} Δ θ_{i} = \overset{―}{S} (\frac{1}{2} r^{2} (θ), π) . \end{aligned}$ 令 $| | π | | \to 0$ , $\underset{―}{S}, \overset{―}{S} \to \int_{α}^{β} \frac{1}{2} r^{2} (θ) d θ$ . 得 $\begin{matrix} (1.2) & S = \frac{1}{2} \int_{α}^{β} r^{2} (θ) d θ . \end{matrix}$

例子

计算 $r = 1 + \cos θ$ 与 $r = 3 \cos θ$ 所围图形的面积.
注意到两条曲线交于 $θ = \frac{π}{3}$ . 因此划分出 $S_{1} = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{2} \cdot (3 \cos θ)^{2} d θ, S_{2} = \int_{0}^{\frac{π}{3}} \frac{1}{2} (1 + \cos θ)^{2} d θ .$

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2 旋转体的体积

3 曲线的弧长

考虑 $C : {\begin{aligned} x = x (t), \\ y = y (t), \end{aligned}$ $α \leq t \leq β$ . $t = α, β$ 时分别得到 $a, b$ . 考虑 $π : a = x_{0} < \dots < x_{n} = b$ , $x_{i} = x (t_{i}), y_{i} = y (t_{i})$ . 记 $P_{i} (x_{i}, y_{i})$ . 当 $| | π | | \to 0$ , 如果 $\sum_{π} | \overset{―}{P_{i - 1} P_{i}} | = \sum_{π} \sqrt{(x_{i} - x_{i - 1})^{2} + (y_{i} - y_{i - 1})^{2}}$ 有极限 $l$ , 则称曲线 $C$ 是可求长的.

推导

注意到 $\begin{aligned} \sum_{π} \sqrt{(x_{i} - x_{i - 1})^{2} + (y_{i} - y_{i - 1})^{2}} \\ = & \sum_{π} \sqrt{[x (t_{i}) - x (t_{i - 1})]^{2} + [y (t_{i}) - y (t_{i - 1})]^{2}} \\ = & \sum_{π} \sqrt{x^{' 2} (η_{i}) (Δ t_{i})^{2} + y^{' 2} (τ_{i}) (Δ t_{i})^{2}} \\ = & \sum_{π} \sqrt{x^{' 2} (η_{i}) + y^{' 2} (τ_{i})} Δ t_{i}, τ_{i}, η_{i} \in (x_{i - 1}, x_{i}) . \end{aligned}$ (回顾 Lagrange中值定理). 则 $\forall ξ_{i} \in [x_{i - 1}, x_{i}]$ : $\begin{aligned} | \sum_{π} \sqrt{x^{' 2} (η_{i}) + y^{' 2} (τ_{i})} Δ t_{i} - \sum_{π} \sqrt{x^{' 2} (ξ_{i}) + y^{' 2} (ξ_{i})} Δ t_{i} | \\ \leq & \sum_{π} | \sqrt{x^{' 2} (η_{i}) + y^{' 2} (τ_{i})} - \sqrt{x^{' 2} (ξ_{i}) + y^{' 2} (ξ_{i})} | Δ t_{i} \\ \leq & \sum_{π} \sqrt{[x^{'} (η_{i}) - y^{'} (τ_{i})]^{2} + [x^{'} (ξ_{i}) - y^{'} (ξ_{i})]^{2}} Δ t_{i} \\ \leq & \sum_{π} \sqrt{ω_{i}^{2} (x^{'} (t)) + ω_{i}^{2} (y^{'} (t))} Δ t_{i} . \end{aligned}$
这里 $ω_{i}$ 是振幅. 从而 $| | π | | \to 0$ 时, 如果 $x^{'} (t), y^{'} (t) \in R [α, β]$ , 有 $l = lim_{| | π | | \to 0} \sum_{π} \sqrt{x^{' 2} (η_{i}) + y^{' 2} (τ_{i})} Δ t_{i} = \int_{α}^{β} \sqrt{x^{' 2} (t) + y^{' 2} (t)} d t .$

从而 $\begin{matrix} (3.1) & l = \int_{α}^{β} \sqrt{x^{' 2} (t) + y^{' 2} (t)} d t . \end{matrix}$
特别地,

$C : y = y (x)$ , $l = \int_{a}^{b} \sqrt{1 + y^{' 2} (x)} d x$ .
$C : r = r (θ)$ , $l = \int_{α}^{β} \sqrt{[(r (θ) \cos θ)^{'}]^{2} + [(r (θ) \sin θ)^{'}]^{2}} d θ = \int_{α}^{β} \sqrt{r^{2} (θ) + r^{' 2} (θ)} d θ .$

3.1 微元法

直角坐标所围图形的面积 $S = \int_{a}^{b} [f (x) - g (x)] d x .$
参数方程所围图形的面积 $S = \frac{1}{2} \int_{α}^{β} r^{2} (θ) d θ .$ 平行截面面积已知的立体体积 $V = \int_{a}^{b} A (x) d x .$
绕 $x$ 轴旋转一周所得立体体积 $V = \int_{a}^{b} π f^{2} (x) d x .$
绕 $y$ 轴旋转一周所得立体体积 $V = \int_{c}^{d} π f^{2} (y) d y .$
直角坐标弧长 $l = \int_{a}^{b} \sqrt{1 + (y^{'} (x))^{2}} d x .$