• 2022-07-26
    求由平面[tex=2.357x1.286]F20DA9b5PZyvxJH27l4LOQ==[/tex],[tex=2.357x1.286]+lfyPLkaB2aZzha73p3Bvg==[/tex],[tex=4.0x1.286]Y2PAOcQLlnse9p/I1rNCIQ==[/tex]所围成的柱体被平面z=0 及曲面[tex=6.571x1.286]nmLOx5DEdt6xe2G92ml5N65PiDCXf0JzGFgaCiGvhfU=[/tex]截得的立体体积 .
  • 举一反三