📝 题目
例 5 试计算积分
$$ K = {\iint }_{S}z\mathrm{\;d}\sigma $$
这里 $S$ 是一段螺旋面:
$$ \mathbf{r} = \left( {u\cos v,u\sin v,{bv}}\right) , $$
$$ 0 \leq u \leq a,0 \leq v \leq {2\pi }. $$
💡 答案与解析
解 直接计算得到
$$ {\mathbf{r}}_{u} = \left( {\cos v,\sin v,0}\right) , $$
$$ {\mathbf{r}}_{v} = \left( {-u\sin v,u\cos v,b}\right) , $$
$$ E = {\begin{Vmatrix}{\mathbf{r}}_{u}\end{Vmatrix}}^{2} = 1, $$
$$ F = \left( {{\mathbf{r}}_{u},{\mathbf{r}}_{v}}\right) = 0, $$
$$ G = {\begin{Vmatrix}{\mathbf{r}}_{v}\end{Vmatrix}}^{2} = {u}^{2} + {b}^{2}, $$
$$ \sqrt{{EG} - {F}^{2}} = \sqrt{{u}^{2} + {b}^{2}}. $$
因而
$$ K = {\iint }_{\begin{matrix} {0 \leq u \leq a} \\ {0 \leq v \leq {2\pi }} \end{matrix}}{bv}\sqrt{{u}^{2} + {b}^{2}}\mathrm{\;d}u\mathrm{\;d}v $$
$$ = {\pi }^{2}b\left( {a\sqrt{{a}^{2} + {b}^{2}} + {b}^{2}\ln \frac{a + \sqrt{{a}^{2} + {b}^{2}}}{b}}\right) . $$