双纽线的旋转体体积
张俏武:双纽线的旋转体体积
双纽线的旋转体体积
2,22222y,xtan,,,,,x,y,ax,y摘要:本文推导出了双纽线绕直线 (0,,),2旋转所围的旋转体的体积公式(从而给出了戈衍三最近结果的一般性结论(
关键词:双扭线,三角函数,换元积分法(
中图分类号:O172(
The Volume for Body of Rotation of Lemniscates
ZHANG Qiao-Wu
Abstract: In this paper, it is given to deduce volume formulae for body of rotations of lemniscates
222222y,xtan,,,,,x,y,ax,y around a line and of tained the general conclusion of GE Yan-San recentl results
Keywords:lemniscates; trigonometric function; integral of interchange elements
2000MSC:51M25
1(引言
戈衍三于2007年在[2]中推导出了双纽线的
标准
excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载
方程
222222,,,,x,y,ax,y (1) 分别绕OX轴、OY轴与直线y,x旋转所围的旋转体的体积公式(自然人们要问:双纽线(1)绕一条过原点的直线旋转所围成的旋转体的体积如何计算呢,
,,,y,xtan,0,,本文考虑双纽线(1)绕直线旋转所围成的旋转体的体积, 得到下列,,,2,,
结果:
,,,定理1.1设,,则双纽线(1)绕直线 ,,max{0,,},,min{,,},,,,,,424
,,,0,,y,xtan,,V旋转所围成的旋转体的体积为 ,,2,,
44325532,, V,,acos,(4cos,,3)cos,cos2,,cos,cos2,,,acos,(,5cos,33
13332,,, ,3)cos,cos,2,cos,cos,2,,acos,(4cos,,1)cos,cos2,2
1
湛江师范学院2008届优秀毕业论文选编
13,,,coscos2,acosln|2cos,cos2|,ln|2cos,,,,,,,
22
4432553cos2|asin(4sin3)sincos2sincos2a ,,,,,,,,,,,,,,,,,33
123332 ,,,sin,(,5sin,,3)sin,cos2,,sin,cos2,,,asin,(4sin,,1)2
13,,,,,sincos2,sincos2,asinarcsin(2sin),arcsin(2sin),,,,,,,,(
22
,,y,xy,xtan, , y,0y,xtan,,,0当时,;当时,变为;当时, ,,,,42
xy,xtan,,cot,x,0,变为则这三种特殊情况是戈衍三的结果( y
y,0推论1.2 双纽线(1)绕轴旋转所围的旋转体的体积为 VoxOX
3,a2,,V,2ln(1,2),( ,,OX43,,
123V,a,推论1.3 双纽线(1)绕轴旋转所成曲面包围的体积为( oyx,0VOYOY42
123推论1.4 双纽线(1)绕直线旋转所围的旋转体的体积V为Va( y,x,,y,xy,x4
2(定理1.1的证明
为了证明定理1.1,首先证明下面引理(
引理2.1
112222x,1dx,x2x,1,ln2x,2x,1,C, ,222
111222322,ln2x,2x,1,C, x2x,1dx,x2x,1,x2x,1,416162
1114252322 x2x,1dx,x2x,1,x2x,1,x2x,1,64864
12,ln2x,2x,1,C(
642
证明:由[3]的附录III积分
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
(第305页)得
,,11111222,,x,dx,xx,,lnx,x,,C; 1,,,22222,,
2
张俏武:双纽线的旋转体体积
,,11111122222,,. xx,dx,x(2x,)x,,lnx,x,,C2,,,282242,,
,,121112222,,所以 2x,1dx,2x,dx,xx,,lnx,x,,2C1,,,,22222,,1122,x2x,1,ln2x,2x,1,C( 222
,,1211112222222,,x2x,1dx,2xx,dx,x(2x,)x,,lnx,x,,2C2,,,,282242,,
1113222,x2x,1,x2x,1,ln2x,2x,1,C ( 416162
42的结果证明如下( x2x,1dx,
5535422222222(4x,4x,1)2x,1dx,(2x,1)dx,x(2x,1),x(2x,1)(4x)dx因为 ,,,2
5242222, ,x(2x,1),20x2x,1dx,10x2x,1dx,,
5174222222x2x,1dx,x(2x,1),x2x,1dx,2x,1dx所以 ,,,2412
517111,23222 ,x(2x,1),x2x,1,x2x,1,ln2x,2412416162,
1,1,222,,2x,1,x2x,1 ,ln2x,2x,1,,C,,3,,2,22,11152322 ,x2x,1,x2x,1,x2x,164864
12,ln2x,2x,1,C(
642
11221,2xdx,x1,2x,arcsin2x,C引理2.2 , ,222
11122322x1,2xdx,x1,2x,x1,2x,arcsin2x,C , ,416162
1114252322 x1,2xdx,x1,2x,x1,2x,x1,2x,64864
3
湛江师范学院2008届优秀毕业论文选编
1,arcsin2x,C(
642
证明: 由[3]附录III积分表(第306页)得
,,111122,,,xdx,x,x,arcsin2x,C, 1,,,2222,,
1111,,2222. x,xdx,x(2x,)1,2x,arcsin2x,C,,2,2824,,
,,1211222,,,2,xdx,x,x,arcsin2x,2C所以 1,2xdx1,,,,2222,,
11222,x1,2x,arcsin2x,C x1,2xdx,222
,,121112222,,,2x,xdx,x(2x,),x,arcsin2x,2C 2,,,28224,,
111322,x1,2x,x1,2x,arcsin2x,C( 416162
42的结果证明如下( x1,2xdx,
5355224422222,x(1,2x),x(1,2x)(,4x)dx因为 (4x,4x,1)1,2xdx,(1,2x)dx,,,2
5242222 , ,x(1,2x),20x1,2xdx,10x1,2xdx,,
所以
517142222222x1,2xdx,x(1,2x),x1,2xdx,x1,2xdx ,,,241224
51711,23222 x(12x)x12xx12x,,,,,,,2412416,
,,,11112 ,arcsin2x,,,x1,2x,arcsin2x,,C3,,,24216222,,,
111125232,x1,2x,arcsin2x,C( ,x1,2x,x1,2x64648642
22引理2.3 ,( sin3,,sin,(4cos,,1)cos3,,cos,(1,4sin,)
4
张俏武:双纽线的旋转体体积
证明: 由三角函数倍角公式容易证得
2,,,22222,,,,yxx,y,ax,y ,tan0,,定理1.1的证明:将和化为极坐标方程为,,,,2,,
,,,4222222,,, , 0,,和,即和 r,ar(cos,,sin,) r,acos2, ,,,,,2,,
2,,,,,,2222222y,xtan,,,,, 0,,x,y,ax,y 0,,(则绕 即绕r,acos2, ,,,,,,22,,,,
,,,22 , 0,,旋转所围的旋转体的体积与绕X轴旋转所围的旋转体r,acos2(,,,),,,,,2,,
的体积等积(
又第二、四象限部分旋转时被第一、
4 三象限旋转部分所覆盖,则可以不考
2 虑(又第三象限部分旋转时的体积相
等,所以全双纽线旋转所围的旋转体 -5 5 的体积等于第一象限部分旋转时的 V-2
-4 体积的两倍( V1
,,r,acos2(,,,) r,0下面求函数在第一象限部分的变化范围(因为
,,,,cos2(,,,),022,于是 ,从而(设 ,,,,,,,,,,,,,2244
,,,,,,,,,min,,,,max0,,,, ,,,,,,,,,,424,,,,
(,,,,,,,)r,acos2(,,,)则在第一象限部分的变化范围是( ,,r,0
x,rcos,,acos2(,,,)cos,现在考虑第一象限部分旋转时的体积:关于,求导数
,,,,,,1sin2(),,, ,,,,,,xa2cos,cos2(,,)sin,,,2,cos2(),,,,
,,,,,,,,,sin2(,)cos,cos2(,)sinsin(3,2),a,,a(
cos2,(,,)cos2(,,,)
22,,,,,3,,2,,03,,2,,,x,0令,得或,即或(由双纽线方程 ,,,,33
22,,,,22知,当和时,所求旋转体的体积是相等的( r,acos2(,,,),,,,33
5
湛江师范学院2008届优秀毕业论文选编
3222,,,2x,acos2(,)cos,acos,( y,rsin,,acos2(,,,)sin,,333
,,y,0,,0,,0令 ,得或即或,所以或时,旋转x,0,,,,x,acos2,,,,,44
体的体积为第一象限部分旋转时体积的两倍(根据旋转体的体积公式得
332,2,22 acos acos2233V,2V,2,ydx,,ydx 121,, 0 cos2a,
2, 223 ,,,2,,acos2,(,,)sin, dacos2(,,,)cos,,, ,,
2, 223 ,,,2,,acos2,(,,)sin, dacos2(,,,)cos,,, ,,
,,,22,,,,,2acos2,(,,)sin, dacos2(,,,)cos, , ,,,
,, ,,,,,,sin(32)22,, ,,,,,2acos2,(,)sin, a d,,,, ,,,,cos2(),,,,
,,,32( (2) ,2,acos2(,,,)sin,sin(3,,2,) d,, ,,,
2下面对(,)式中的进行整理( sin,sin(3,,2,)
22 sin,sin(3,,2,),sin(,,,,,)sin(3,,3,,,)
2,,,,,sin(,,,)cos,,cos(,,,)sin,sin3(,,,)cos,,cos3(,,,)sin, 2222,,,sin(,,,)cos,,2sin,cos,sin(,,,)cos(,,,),cos(,,,)sin,
22 ,,, cos,sin(,,,),,,,4cos(,,,),1,sin,cos(,,,)4cos(,,,),3
222222 ,sin(,,,),,,,,sin(,,,)cos,,cos(,,,)sin,cos,4cos(,,,),1,2sin,222222,2sin,cos,cos(,,,),,,,4cos(,,,),3,cos(,,,)sin(,,,)cos,,,cos(,,,)22222 ,sin,,,,,sin,[4cos(,,,),3],2sin,cos,sin(,,,)4cos(,,,),1
222222 ,sin(,,,),,,,,cos,,cos,cos(,,,),sin,cos(,,,)cos,4cos(,,,),1222222 ,2sin,cos,cos(,,,),,,,4cos(,,,),3,cos(,,,)cos,sin(,,,,),sin,222222 ,sin,sin(,,,),,,,,,sin,1,4sin(,,,) ,2sin,cos,sin(,,,)3,4sin(,,,)
22243 ,sin(,,,),,,,4cos,(sin,,cos,),8sin,cos,cos(,,,),4cos,,cos,
6
张俏武:双纽线的旋转体体积
222232 ,(,cos,,sin,),6sin,cos,,,,cos(,,,) ,cos,,cos(,,,),4sin,(cos,,2242232 ,sin,),8sin,cos,,,,sin(,,,),sin,(cos,,sin,),4sin,,6sin,cos,23 ,sin(,,,) ,sin,,
24223 ,sin(,,,),,4cos,(3,4cos,)cos(,,,) ,cos,(12cos,,7)cos(,,,),cos,
24223( ,cos(,,,),,4sin,(4sin,,3)sin(,,,),sin,(,12sin,,7)sin(,,,) ,sin,
将整理后的结果代入(2)得
,,,3242 ,,V,,2acos2,(,,)sin(,,,)4cos,(3,4cos,)cos(,,,),cos,(12cos,, ,,,
2324 ,7)cos(,,,),cos,,,,cos(,,,)4sin,(4sin,,3)sin(,,,),sin,
223 ,(,12sin,,7)sin(,,,),sin,,,d,
,,,3242 ,,,2acos2,(,,) sin(,,,)4cos,(3,4cos,)cos(,,,),cos,(12cos,, ,,,
,,,2332 ,,,7)cos,(,,),cos,d,(,,),2,acos2(,,,)cos(,,,)4sin,(4sin,, ,,,
4223 ,3)sin(,,,),sin,(,12sin,,7)sin(,,,),sin,,d(,,,)
,,,32242 ,,,,2a2cos,(,,),14cos,(3,4cos,)cos(,,,),cos,(12cos,,7), ,,,
,,,23322 ,,,cos,(,,),cos,dcos(,,,),2,a1,2sin(,,,)4sin,(4sin,,3), ,,,
4223 ,sin(,,,),sin,(,12sin,,7)sin(,,,),sin,,dsin(,,,)
cos, cos,3242322 ,,8acos,(4cos,,3)u2u,1du,2,acos,(12cos,,7)u,, cos cos,,
cos, sin,2332324 ,2u,1du,,acos,2u,1du,8,asin,(4cos,,3)v,, cos sin,,
sin,2322233,1,2vdv,,2asin,(12sin,,7)v1,2vdv,2,asin, ,,sin
sin,2,1,2vdv ,,sin
cos,111,3252322,8,acos,(4cos,,3)u2u,1,u2u,1,u2u,1ln2u ,, cos,64864,
cos,111,232322,212cos(12cos7)2121,u,du,,a,,,uu,,uu,, ,,, cos,,416642,
7
湛江师范学院2008届优秀毕业论文选编
cos,1,233223,,,ln2u,2u,1du,,2acos,u2u,1,2u,1du,8,asin, ,,,, cos,,,2,
sin,1111,252322(4sin3)121212,,,v,v,v,v,x,v, ,, sin,64864642,
sin,11,32322,,arcsin2vdv,2,asin,(12sin,,7)v1,2v,v1,2v ,, sin,416,
sin,111,,,332 arcsin2vdv2,asin,v12varcsin2v dv,,,,,,,, sin,216222,,,
cos,44325232 ,,,,acos,(4cos,,3)u2u,1,,acos,(,5cos,,3) cos,33
3,, cos coscos,a,132322,,,,21cos(4cos1)21,uu,,a,uu,, ,,, cos cos,,222
, cos sin,223252,,,, ,ln2u,2u,1,,asin,(4sin,,3)v1,2v sin,,,,,3 cos,
sin, sin,413232322,, ,,,,asin,(,5sin,,3)v1,2v,,asin,(4sin,,1)v1,2v sin sin,,32
sin,13,,,asinarcsin2v,, sin,22
443252523 ,,,,acos,(4cos,,3)cos,2cos,,1,cos,2cos,,1,,acos,33
12323232 ,,,(,5cos,,3)cos,2cos,,1,cos,2cos,,1,,acos,(4cos,,1)2
12232,,,a,cos,2cos,,1,cos,2cos,,1,,cos,ln2cos,2cos,,1 ,,22
4232525, ,,ln2cos,,2cos,,1,,asin,(4sin,,3)sin,1,2sin,,sin,,,3
42323232, ,,,1,2sin,,,asin,(,5sin,,3)sin,1,2sin,,sin,1,2sin,3
1132223,,asin(4sin,1),sin1,2sin,sin1,2sin,asin,,,,,,,,, 222,,,arcsin(2sin,),arcsin(2sin,)
44325532,, ,,acos,(4cos,,3)cos,cos2,,cos,cos2,,,acos,(,5cos,33
8
张俏武:双纽线的旋转体体积
13332 ,,,,3)cos,cos,2,cos,cos,2,,acos,(4cos,,1)cos,cos2,2
13,,,coscos2,acosln|2cos,cos2|,ln|2cos,,,,,,,
22
4432553cos2|asin(4sin3)sincos2sincos2a ,,,,,,,,,,,,,,,,,33
123332 ,,,sin,(,5sin,,3)sin,cos2,,sin,cos2,,,asin,(4sin,,1)2
13,,,,,sincos2,sincos2,asinarcsin(2sin),arcsin(2sin),,,,,,,,(
22
定理1.1证明完毕(
3(推论1.2-1.4的证明
, 推论1.2 的证明: 当时,则,(由定理1.1得 ,,0,,0,,4
V,limV OX,,0
,,,,4432553,,,,acos0(4cos0,3)coscos2(),cos0cos0,,acos0 ,,3443,,
,,,,123332,,,(,5cos0,3)coscos2(),cos0cos0,,acos0(4cos0,1) ,,442,,
,,,1,,,,3,,,,coscos2(),coscos0,acos0ln2cos,cos2() ,,,,,444422,,,
,,4,,3255,,,ln2cos0,cos0,asin0(4sin0,3)sincos2(),sin0 ,,,344,,
,,,4,,3233,,,,cos0,asin0(,5sin0,3)sincos2(),sin0cos0 ,,,,344,,,
,,,,11323,,,,asin0(4sin0,1)sincos2(),sin0cos0,,asin0 ,,24422,,
,,,,arcsin(2sin),arcsin(2sin0) ,,4,,
3,a441333,a(4,3)(0,1),a(,5,3)(0,1),a(4,1)(0,1),(ln|1,0| ,,,33222
9
湛江师范学院2008届优秀毕业论文选编
3a,4333,ln|2,1|),0,0,0,0,a,a,(,ln|2,1|) ,,3222
3,a2,,,2ln(1,2),( ,,43,,
,,,,0 推论1.3 的证明:当时,,(由定理1.1得 ,,,,,24
V,limVOY,,,2
,,,,,,4432553,,acos(4cos3)cos0cos0cos()cos2()a,,,,,,,, ,,322443,,
,,1,,,,,2333,,,cos(,5cos,3)cos0cos0,cos(,)cos2(,),acos(4 ,,,224422,,
,,,1,,,,23,,, ,cos,1)cos0cos0,cos(,)cos2(,),cosln2cos0a,,,,244222,,,,,,,4325,, ,,cos0,ln2cos(,),cos2(,),,asin,(4sin,,3)sin0cos0,,443,,
,4,,,,53233,,sin(,)cos2(,),asin(,5sin,3)sin0cos0,sin(,) ,,,,,4434,,
,,,1,,,32,,,,cos2(,),sin(4sin,1)sin0cos0,sin(,)cos2(,)a ,,,,,,4244,,,1,,,,3a,sinarcsin(2sin(0)),arcsin(2sin(,)) ,,,2422,,
,111,,3323aa,,,,,,,0,0,0,0,0,0,0,,a0,arcsin(,1)( ,,2224222,,
,,, 推论1.4 的证明:当时,则,(由定理1.1得 ,,,,,,,444
V,V y,x
,,,,,,,,4432553,,acos(4cos3)coscos2()cos()cos2()a,,,,,,,, ,,34444443,,
,,1,,,,,,,2333,,cos(5cos3)coscos2()cos()cos2()cos,,,,,,,a ,,,44444424,,
10
张俏武:双纽线的旋转体体积
,,,1,,,,,,23,,, ,(4cos,1)coscos2(),cos(,)cos2(,),cosln2a,,,,44444422,,,
,4,,,,,,32, cos,cos2(),ln2cos(,),cos2(,),asin(4sin,3),,4444344,
,,4,,,,,,,,55323,,,sincos2(),sin(,)cos2(,),asin(,5sin,3)sin ,,,,44443444,,,
,,1,,,,,,,332,,,cos2(),sin(,)cos2(,),sin(4sin,1)sincos2()a ,,,44424444,,
,1,,,,,,,3,sincos2(),sin(,)cos2(,),asinarcsin(2sin) ,,,44444422,,
,,1,,3 ,arcsin2sin(,),0,0,0,0,0,0,0,,a(2arcsin1),,,,44,,,
123a( ,,4
参考文献
[1] I. S. Gradsbteyn, and L. M. Ryzbik. Table of integrals, series, and products
[M]. Singapore: Elsevier Pte Ltd. 2004. [2] 戈衍三. 一些经典数学问
题
快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题
的另类解算[M]. 北京: 北京理工大学出版社. 2007年. [3] 华东师范大学数学系. 数学分析[M], 第三版. 北京: 高等教育出版社.2001年. [4] 杨志和. 微积分[M]. 北京: 高等教育出版社. 2002年.
[5] 阎占立. 微积分[M]. 北京: 高等教育出版社. 2002年.
[6] 齐民友. 微积分[M]. 北京: 高等教育出版社. 2003年.
[7] 吴烔圻. 数学专业英语[M].北京:高等教育出版社. 2005年.
[8] 王沫然. MATLAB与科学计算[M],第二版.北京:电子工业出版社.2007年.
校稿人:张俏武
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