📝 题目
例 1 求球面 ${x}^{2} + {y}^{2} + {z}^{2} = {a}^{2}$ 被柱面 ${x}^{2} + {y}^{2} = {ax}$ 截下部分的面积.
💡 答案与解析
解 记曲线 ${x}^{2} + {y}^{2} = {ax}$ 所围平面区域为 $D$ ,球面的方程为 $z =$ $\pm \sqrt{{a}^{2} - {x}^{2} - {y}^{2}}$ . 由对称性可只考虑 ${Oxy}$ 平面之上那部分,所以
$$ S = 2{\iint }_{D}\sqrt{1 + {\left( \frac{\partial z}{\partial x}\right) }^{2} + {\left( \frac{\partial z}{\partial y}\right) }^{2}}\mathrm{\;d}x\mathrm{\;d}y $$
$$ = 2{\iint }_{D}\frac{a}{\sqrt{{a}^{2} - {x}^{2} - {y}^{2}}}\mathrm{\;d}x\mathrm{\;d}y $$
$$ = 2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{\;d}\theta {\int }_{0}^{a\cos \theta }\frac{a}{\sqrt{{a}^{2} - {r}^{2}}}r\mathrm{\;d}r $$
$$ = 2{a}^{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left( {1 - \left| {\sin \theta }\right| }\right) \mathrm{d}\theta = 4{a}^{2}\left( {\frac{\pi }{2} - 1}\right) . $$