问题标题：
数学微积分中的积分求旋转的体积只要答案即可Findthevolumeofthesolidobtainedbyrotatingtheregionboundedbyy=x^2,y=0,x=5,andaboutthey-axis.求绕y轴旋转后的体积.Findthevolumeofthesolidformedbyrotatin
问题描述：
数学微积分中的积分求旋转的体积只要答案即可
Findthevolumeofthesolidobtainedbyrotatingtheregionboundedbyy=x^2,y=0,x=5,and about
the y-axis.求绕y轴旋转后的体积.
Findthevolumeofthesolidformedbyrotatingtheregionenclosedby
y=e^x+5,y=0,x=0,x=0.5, aboutthey-axis.求绕y轴旋转后的体积.
Aballofradius15hasaroundholeofradius3drilledthroughitscenter.
Findthevolumeoftheresultingsolid.
Findthevolumeofthesolidobtainedbyrotatingtheregionboundedbythe
givencurvesaboutthespecifiedaxis.y=0,y=(cos6x),x=0,x=pi/12, abouttheaxisy=-6.
有一个答案算一个,多多益善.
郭北苑回答：
it'seasiertousey-axisasvariable. Therangeis0to5forx,0to25fory;y=x^2,x=√y
V=∫₀²⁵π[5²-(√y)²]dy=π(25y-y²/2)|₀²⁵=625π/2
Justconsiderthefirstquadrantinaplane. Acircleofraduius15isexpressedasy=√(15²-x²);theholecanbeconsideredasliney=3;3=√(15²-x²),x²=216
theyinerceptat(√216,3)
Thevolumeisdoubletheresultfromrotatingtheregionaboutthex-axis
V=2∫π[(15²-x²)-3²]dx 0to√216
=2π(216x-x³/3) 0to√216
=(1064√54)π/3
x=0,y=1;x=π/12,y=cos(π/2)=0
Theinnerradiusofthesolidisr=0-(-6)=6;theouterradiusofthesolidisR=cos(6x)-(-6)=6+cos(6x)
V=∫π(((6+cos(6x))²-6²)dx 0toπ/12
=π∫[12cos(6x)+cos²(6x)]dx=π∫[12cos(6x)+cos²(6x)]dx
=π∫[12cos(6x)+1/2+(1/2)cos(12x)]dx
=π[x/2+2sin(6x)+(1/24)cos(12x)] 0toπ/12
=π(π/12+2-1/24)-π(0+0+1/24)
=π(2+π/12)
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