Question

求 高数帝做几道提：第一题：求曲面Z=x²+2y²及Z=6-2x²-y²所围的体

Answer 1

由Z=x²+2y²和Z=6-2x²-y²可得二者的交线为x²+y²=2由此可得
V=∫∫[(x²+2y²)-(6-2x²-y²)]ds
积分区域为D={(x,y)|x²+y²=2}