é¢ç®æç¹é®é¢ï¼è¦æ±V(a)çæå¼ï¼aåºè¯¥ä¸ºéåºé´ï¼èä¸æ¯å¼åºé´ãå设0=<a<=1ï¼è§£é¢å¦ä¸ï¼
说æï¼^ââ表示次æ¹
解ï¼y=x^2ä¸y=a^2交ç¹ï¼(a,âa)
y=x^2ä¸x=1交ç¹ï¼(1,1)
V1(a)=â«(0,a)Ï(a^4-x^4)dx
=Ï(a^4x-1/5x^5)|(0,a)
=Ï[(a^4·a-1/5a^5)-(a^4·0-1/5·0^5)]
=4/5Ïa^5
V2(a)=â«(a,1)Ï(x^4-a^4)dx
=Ï(1/5x^5-a^4x)|(a,1)
=Ï[(1/5·1^5-a^4·1)-(1/5a^5-a^4·a)]
=Ï[1/5-a^4-1/5a^5+a^5]
=Ï/5-Ïa^4+4/5Ïa^5
V(a)=V1+V2
=4/5Ïa^5+Ï/5-Ïa^4+4/5Ïa^5
=8/5Ïa^5-Ïa^4+Ï/5
V'(a)=8Ïa^4-4Ïa^3
=4Ïa^3(a-1)
V'(a)<0
4Ïa^3(a-1)<0
0<a<1
åè°éååºé´ï¼(0,1)
æå°å¼ï¼V(a)min=V(1)
=8/5Ï·1^5-Ï·1^4+Ï/5
=8/5Ï-Ï+Ï/5
=4/5Ï
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