微積分（定積分應(yīng)用，反常積分）彩版

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1、?旋轉(zhuǎn)體的體積1.由連續(xù)曲線y?f(x),yy?f(x)直線x?a,x?b及x軸圍成的曲邊梯形繞x軸oaaxbbx旋轉(zhuǎn)一周圍成的旋轉(zhuǎn)體體積.b2Vx???[f(x)]dxya2.連續(xù)曲線段x??(y()c?y?d)d繞y軸旋轉(zhuǎn)一周圍成的立體體積為yx??(y)d2cV??[?(y)]dyy?cox?套筒法y求由連續(xù)曲線y?f(x),y?f(x)直線x?a,x?b及x軸圍成的曲邊梯形繞y軸旋轉(zhuǎn)一周圍成的旋轉(zhuǎn)體體積.oaxx?dxbx[x,x?dx](a?x?b)?V?dV?2?xdS?2?xf(x)dxbb2Vy??a2?xf(x)dxVx???[f(x)]dxa3.設(shè)由擺線x?a(

2、t?sint,)y?a1(?cost)(a?)0的一拱與x軸所圍平面圖形.求面積S,Vx,Vy.(P275.5(4))解:dS?ydxy?a1(?cost)?a1(?costd)t2??S?a2?1(?cost)2dtox2?ax0t2?u???4a2sin4tdt224?8a?sinudu020??16a22sin4udu?16a2?3?1?2???3?a04223.設(shè)由擺線x?a(t?sint,)y?a1(?cost)(a?)0的一拱與x軸所圍平面圖形.求面積S,Vx,Vy.(P275.5(4))2y解:dVx??ydx22??a1(?cost)?a1(?costd)tox2?

3、ax?32??3V?a1(cost)dtx?0tu??332?6t224a3sin6udu?2asindt????020?532652a3?2?a?sinudu??03.設(shè)由擺線x?a(t?sint,)y?a1(?cost)(a?)0的一拱與x軸所圍平面圖形.求面積S,Vx,Vy.(P275.5(4))解:dVy?2?xydxy?32???2V2?a(tsint1()cost)dty?0ox2?ax33?6?ay繞y?2a旋轉(zhuǎn)體積V32a2V??2(a)?2?a3?2??2ox?a2?ax??2(ay)dx02332?2?8?a??a1(?cost)1(?costd)t72a3??

4、?0§4.3平面曲線的弧長定義:若在弧AB上任意作內(nèi)接折線,當(dāng)折線段的最大邊長?→0時,折線的長度趨向于一個確定的極限,則稱此極限為曲線弧AB的弧長,yMMi?1ins?limMM?i?1i??0i?1B?Mn并稱此曲線弧為可求長的.A?M0ox定理:任意光滑曲線弧都是可求長的.(證明略)?曲線的弧長(1)曲線弧由直角坐標(biāo)方程給出:y?f(x)y?f(x)(a?x?b)yds弧長元素(弧微分):22ds?(dx)?(dy)2?1?y?dxoaxx?dxbx因此所求弧長bb22s??1?y?dx??1?f?(xd)xaa?曲線的弧長(2)曲線弧由參數(shù)方程給出:?x??(t)()???

5、t???y??(t)弧長元素(弧微分):2222ds?(dx)?(dy)???(t)???(td)t因此所求弧長s???2(t)??2(t)dt?????曲線的弧長(3)曲線弧由極坐標(biāo)方程給出:r?r(?)(?????)令x?r(?)cos?,y?r(?)sin?,則得弧長元素(弧微分):2222ds?[x?(?)]?[y?(?)]d??r(?)?r?(?d)?因此所求弧長(自己驗證)s??r2?r?2?(?)(?)d??例10.兩根電線桿之間的電線,由于其本身的重量,下垂成懸鏈線.懸鏈線方程為yc1x?xy?(e?e)(?a?x?a)2求這一段弧長.?aoax解:ds1y2dx?

6、???1?1(ex?e?x)2dx?1(ex?e?xd)x421aax?xx?xa?a?s???a(e?ed)x??0(e?ed)x?e?e2?x?a(t?sint)例11.計算擺線?(a?)0一拱?y?a1(?cost)0(?t?2?)的弧長.ydx2dy2解:ds?(dt)?(dt)dto2?ax?a21(?cost)2a22t?sindtt?a1(2?costd)t?2asindt22?t?t?2??s??2asindt?2a??2cos??8a02?2?0例12.求阿基米德螺線r?a?(a?)0相應(yīng)于0≤?≤2?一段的弧長.?2?a解:s?r2?r?2oxd(?)(?)d?

7、2222?a??ad??a1??d?2?2?s?a?1??d?0??212?2??a1???ln??1????22??02a2?a?1?4??ln(2??1?4?)2§4.4定積分在物理上的應(yīng)用?變力沿直線所作的功設(shè)物體在連續(xù)變力F(x)作用下沿x軸從x＝a移動到x?b,力的方向與運動方向平行,則變力所做的功為bW??F(x)dxaaxx?dxbx在[a,b]上任取子區(qū)間[x,x?dx]，在其上所作的功為?W?dW?F(x)dx功元素例1.設(shè)有一直徑為20m的半球形水