📝 题目
例 2 计算曲面积分
$$ I = {\iint }_{S}{x}^{2}\mathrm{\;d}y\mathrm{\;d}z + {y}^{2}\mathrm{\;d}z\mathrm{\;d}x + {z}^{2}\mathrm{\;d}x\mathrm{\;d}y, $$
其中 $S$ 为锥面 ${x}^{2} + {y}^{2} = {z}^{2}\left( {0 \leq z \leq h}\right)$ 所示那部分的外侧.
💡 答案与解析
解法 ${1S}$ 不是封闭曲面,为此令 ${S}_{1} : z = h\left( {{x}^{2} + {y}^{2} \leq {h}^{2}}\right)$ ,取上侧, 则
$$ \left\{ {{\iint }_{S} + {\iint }_{{S}_{1}}}\right\} {x}^{2}\mathrm{\;d}y\mathrm{\;d}z + {y}^{2}\mathrm{\;d}z\mathrm{\;d}x + {z}^{2}\mathrm{\;d}x\mathrm{\;d}y $$
$$ = {\iiint }_{V}2\left( {x + y + z}\right) \mathrm{d}x\mathrm{\;d}y\mathrm{\;d}z = 2{\iiint }_{V}z\mathrm{\;d}x\mathrm{\;d}y\mathrm{\;d}z $$
$$ = 2{\int }_{0}^{h}\mathrm{\;d}z{\iint }_{{x}^{2} + {y}^{2} \leq {z}^{2}}z\mathrm{\;d}x\mathrm{\;d}y = 2{\int }_{0}^{h}\pi {z}^{3}\mathrm{\;d}z = \frac{\pi }{2}{h}^{4}, $$
由此得出
$$ I = {\iint }_{S}{x}^{2}\mathrm{\;d}y\mathrm{\;d}z + {y}^{2}\mathrm{\;d}z\mathrm{\;d}x + {z}^{2}\mathrm{\;d}x\mathrm{\;d}y $$
$$ = \frac{\pi }{2}{h}^{4} - {\iint }_{{S}_{1}}{x}^{2}\mathrm{\;d}y\mathrm{\;d}z + {y}^{2}\mathrm{\;d}z\mathrm{\;d}x + {z}^{2}\mathrm{\;d}x\mathrm{\;d}y $$
$$ = \frac{\pi }{2}{h}^{4} - {\iint }_{{x}^{2} + {y}^{2} \leq {h}^{2}}{h}^{2}\mathrm{\;d}x\mathrm{\;d}y = \frac{\pi }{2}{h}^{4} - \pi {h}^{4} = - \frac{\pi }{2}{h}^{4}. $$
解法 2 锥面的参数表示为 $x = r\cos \theta ,y = r\sin \theta ,z = r(0 \leq r \leq h$ , $0 \leq \theta \leq {2\pi })$ . 因
$$ \left\lbrack \begin{array}{lll} {x}_{r}^{\prime } & {y}_{r}^{\prime } & {z}_{r}^{\prime } \\ {x}_{\theta }^{\prime } & {y}_{\theta }^{\prime } & {z}_{\theta }^{\prime } \end{array}\right\rbrack = \left\lbrack \begin{array}{rrr} \cos \theta & \sin \theta & 1 \\ - r\sin \theta & r\cos \theta & 0 \end{array}\right\rbrack , $$
所以 $A = - r\cos \theta ,B = - r\sin \theta ,C = r$ . 由于取外侧,故法向量的第三个分量 $C$ 应小于零,所以积分前取负号. 于是
$$ I = - {\iint }_{\begin{matrix} {0 \leq r \leq h} \\ {0 \leq \theta \leq {2\pi }} \end{matrix}}\left\lbrack {-{r}^{2}{\cos }^{2}\theta \cdot r\cos \theta - {r}^{2}{\sin }^{2}\theta \cdot r\sin \theta + {r}^{2} \cdot r}\right\rbrack \mathrm{d}r\mathrm{\;d}\theta $$
$$ = {\int }_{0}^{h}\mathrm{\;d}r{\int }_{0}^{2\pi }{r}^{3}\left( {{\cos }^{3}\theta + {\sin }^{3}\theta - 1}\right) \mathrm{d}\theta $$
$$ = - {2\pi }{\int }_{0}^{h}{r}^{3}\mathrm{\;d}r = - \frac{\pi }{2}{h}^{4}. $$
解法 3 因 $z = \sqrt{{x}^{2} + {y}^{2}}$ ,所以 ${z}_{x}^{\prime } = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}},{z}_{y}^{\prime } = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$ . 法向量 $\mathbf{n} = \left( {{z}_{x}^{\prime },{z}_{y}^{\prime }, - 1}\right)$ 的第三个分量是负的,与指定的曲面侧一致,故公式前取正号. 这样有
$$ I = {\iint }_{{x}^{2} + {y}^{2} \leq {h}^{2}}\left\lbrack {{x}^{2} \cdot \frac{x}{\sqrt{{x}^{2} + {y}^{2}}} + {y}^{2}\frac{y}{\sqrt{{x}^{2} + {y}^{2}}} - \left( {{x}^{2} + {y}^{2}}\right) }\right\rbrack \mathrm{d}x\mathrm{\;d}y. $$
由对称性可看出
$$ I = - {\iint }_{{x}^{2} + {y}^{2} \leq {h}^{2}}\left( {{x}^{2} + {y}^{2}}\right) \mathrm{d}x\mathrm{\;d}y = - {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{h}{r}^{2} \cdot r\mathrm{\;d}r $$
$$ = - \frac{\pi }{2}{h}^{4}. $$