📝 题目
例 3 试计算双曲抛物面 $z = {xy}$ 被围在圆柱面 ${x}^{2} + {y}^{2} = {a}^{2}$ 内的那一部分的面积.
💡 答案与解析
解 计算得
$$ p = \frac{\partial z}{\partial x} = y,\;q = \frac{\partial z}{\partial y} = x, $$
$$ \sqrt{{p}^{2} + {q}^{2} + 1} = \sqrt{{x}^{2} + {y}^{2} + 1}. $$
于是
$$ \sigma \left( S\right) = {\iint }_{{x}^{2} + {y}^{2} \leq {a}^{2}}\sqrt{{x}^{2} + {y}^{2} + 1}\mathrm{\;d}x\mathrm{\;d}y. $$
换极坐标计算得到
$$ \sigma \left( S\right) = {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{a}\sqrt{{r}^{2} + 1}r\mathrm{\;d}r $$
$$ = \frac{2\pi }{3}\left\lbrack {{\left( {a}^{2} + 1\right) }^{3/2} - 1}\right\rbrack . $$