微积分全英微积分全英 (73).pdf

《微积分全英微积分全英 (73).pdf》由会员分享，可在线阅读，更多相关《微积分全英微积分全英 (73).pdf（14页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、Exercises Lessons（II）for Double Integral22121001(|2)3yyy eyedy=212013yy de=2122yDx edxdy=1is the part of where 0.Example 1Show that if is a positive integer,then 1lim0.pnn=Please calculatewhere22,yDx edxdy(,)|1,11.Dx yxyx=22yDx edxdy212002yyedyx dx=210323yey dy=21101(|)3yee=+11(12)3e=Example 2Show t

2、hat if is a positive integer,then 1lim0.pnn=Please calculatewhere(),Dx xy dxdy+222(,)|2,.Dx yxyxy=+is symmetric about-axis.0Ddxdyxy=Hence,2()DDx xy dxdyx dxdy+=2210222xxxxdyd=122202(2)xx dxx=1122400222xx dxx dx=Example 2Show that if is a positive integer,then 1lim0.pnn=Please calculatewhere(),Dx xy

3、dxdy+222(,)|2,.Dx yxyxy=+122402204sincos2xx dxttdt=401cos4812tdt=14015x dx=2()45Dx xy dxdy+=Example 3Show that if is a positive integer,then 1lim0.pnn=Please calculate22110().xyyedyedxx2222111111000()yyyxyxyedyedxdydxdydxeexx=2210010(1)yxxedxdxeyyy d=222011001xyyyye dye dxe d=+1(1)2e=Example 4Show t

4、hat if is a positive integer,then 1lim0.pnn=22The plane is bounded by,circle 2 and axis,find.=+=DDyxxyyyxyd2sin3204cos sinDxdrdryd=44242 sincos sin4d=4623s2in|=712=6221()32=Example 5Show that if is a positive integer,then 1lim0.pnn=5The plane is bounded by sin,and 1find(1).2DDyx xyxydxdy=，5(1)Dxydxd

5、y152sin2(1)xdxxydy=62211(sin1 sin)66xxxx dx=+611,sin,sin are.odd function66xxxx62211(sinsin)066xxxx dx+=522(1)(1)Dxydxdydx=Hence,Example 6Show that if is a positive integer,then 1lim0.pnn=The plane region is bounded by a curve 1cos(0)and a polar axisplease calculate.DDrxyd=+，3sincosDDxydrdrd=041sinc

6、o(1cos4)s d=+401(1cos)11(1co(coss)4)d=+5612216()45615=+=Example 7Show that if is a positive integer,then 1lim0.pnn=2222 is a bounded region bounded by lines 1,and,calculate.DxxyyDyyxyxdxdyxy=+about-axysymis metrDy220Dxydxdyxy=+22222222DDxxyydxdydxdyyxyxxy=+312224sin204(cossin)rddrrr=4234csc21()2d=34

7、4cot1()22|=12=Example 8Show that if is a positive integer,then 1lim0.pnn=222Please calculate(1),where(,)|2.DxdxdyDx yxyy+=+22(1)(21)DDxdxdyxxdxdy+=+32sin2002=cosDrx dxdyddr420=sincos4d 201cos2sin 22d=220011sin 2cos2 sin 222dd=0c1()441os4d=(21)0Dxdxdy+=+=25(1)4Dxdxdy+=Example 9Show that if is a posit

8、ive integer,then 1lim0.pnn=1130Find 1.yIdyx dx=+11301yIdyx dx=+213001xdxx dy=+12301xx dx=+2(2 21)9=Example 10Show that if is a positive integer,then 1lim0.pnn=1121330112Find.xxxxIdxydydxydydxydy+=+311yyyIdydx=31ydy=32112|y=4=Summary The choice of integral order Transformation formula of Cartesian coordinates and polar coordinates The integral method of substitution for one variable function Exercises Lessons(II)