第1个回答 2019-04-09
解：由已知两曲面方程
z=(x²+y²)/4 与 z=6-2x²-y²
消去z得所围立体在xOy平面上的投影区域为
D={(x,y)|9x²+5y²≤24}
={(x,y)|x²/(24/9)+y²/(24/5)≤1}，
故所求立体的体积为
∫∫D [(6-2x²-y²)-(x²+y²)/4]dσ
=(1/4)∫∫D (24-9x²-5y²)dσ
=∫[0, √(24/5)]dy
∫[0, √(24-5y²)/3](24-9x²-5y²)dx
=∫[0, √(24/5)]dy
[(24-5y²)x-3x³][0, √(24-5y²)/3]
=(2/9)∫[0, √(24/5)]√(24-5y²)³dy
=(2/9)∫[0, π/2]√(24cos²t)³d[√(24/5)sint]
=(128/√5)∫[0, π/2][(cost)^4]dt
=(128/√5)(3/4)(1/2)(π/2)
=24√5π/5.本回答被提问者和网友采纳
z=(x²+y²)/4 与 z=6-2x²-y²
消去z得所围立体在xOy平面上的投影区域为
D={(x,y)|9x²+5y²≤24}
={(x,y)|x²/(24/9)+y²/(24/5)≤1}，
故所求立体的体积为
∫∫D [(6-2x²-y²)-(x²+y²)/4]dσ
=(1/4)∫∫D (24-9x²-5y²)dσ
=∫[0, √(24/5)]dy
∫[0, √(24-5y²)/3](24-9x²-5y²)dx
=∫[0, √(24/5)]dy
[(24-5y²)x-3x³][0, √(24-5y²)/3]
=(2/9)∫[0, √(24/5)]√(24-5y²)³dy
=(2/9)∫[0, π/2]√(24cos²t)³d[√(24/5)sint]
=(128/√5)∫[0, π/2][(cost)^4]dt
=(128/√5)(3/4)(1/2)(π/2)
=24√5π/5.本回答被提问者和网友采纳
第2个回答 2019-04-09
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