📝 题目
例 9 设 $f\left( x\right) \in {C}^{1}\left\lbrack {a,b}\right\rbrack$ ,且 $f\left( x\right) \geq 0\left( {a \leq x \leq b}\right) ,a > 0$ .
( 1 )求由曲线 $y = f\left( x\right) \left( {a \leq x \leq b}\right)$ 绕 $y$ 轴旋转所成曲面的侧面积;
(2)求由平面图形 $\{ \left( {x,y}\right) \mid a \leq x \leq b,0 \leq y \leq f\left( x\right) \}$ 绕 $y$ 轴旋转所得旋转体的体积.
💡 答案与解析
解法 1 (1) 取自变量微元 $\left\lbrack {x,x + \mathrm{d}x}\right\rbrack$ ,相应的弧长微元 $\mathrm{d}s =$ $\sqrt{1 + {f}^{\prime 2}\left( x\right) }\mathrm{d}x$ ,从而侧面积微元 $\mathrm{d}{P}_{y} = {2\pi x}\sqrt{1 + {f}^{\prime 2}\left( x\right) }\mathrm{d}x$ ,由此即得曲线绕 $y$ 轴旋转所成曲面的侧面积为
$$ {P}_{y} = {\int }_{a}^{b}{2\pi x}\sqrt{1 + {f}^{\prime 2}\left( x\right) }\mathrm{d}x. $$
(2)取自变量微元 $\left\lbrack {x,x + \mathrm{d}x}\right\rbrack$ ,相应的体积微元为 $\mathrm{d}{V}_{y} = {2\pi x}$ . $f\left( x\right) \mathrm{d}x$ ,从而
$$ {V}_{y} = {\int }_{a}^{b}{2\pi x} \cdot f\left( x\right) \mathrm{d}x. $$
解法 2 (1) 用古鲁金第一定理. 若曲线 $y = f\left( x\right) \left( {a \leq x \leq b}\right)$ 的重心横坐标为 $\bar{x}$ ,则
$$ {P}_{y} = {2\pi }\bar{x} \cdot {\int }_{a}^{b}\sqrt{1 + {f}^{\prime 2}}\mathrm{\;d}x = {2\pi }{\int }_{a}^{b}x\sqrt{1 + {f}^{\prime 2}}\mathrm{\;d}x. $$
(2)用古鲁金第二定理. 若所给平面图形的重心横坐标为 $\bar{x}$ ,则
$$ {V}_{y} = {2\pi }\bar{x} \cdot {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = {2\pi }{\int }_{a}^{b}{xf}\left( x\right) \mathrm{d}x. $$