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16.4-bo'lim : Vektor maydonlarining chiziqli integrallari

8. Quyidagi egri chiziqlarning har biri uchun \(\vec F\left( {x,y} \right) = 3\,\vec i + \left( {xy - 2x} \right)\vec j\) ni \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) ni baholang.

  1. \(C\) - soat miliga teskari aylanish bilan 4 radiusning boshida joylashgan aylananing yuqori yarmi.
  2. \(C\) - soat yo'nalishi bo'yicha aylanish bilan 4 radiusning boshida joylashgan aylananing yuqori yarmi.

a \(C\) - soat miliga teskari aylanish bilan 4 radiusning boshida joylashgan aylananing yuqori yarmi.
Yechim

Keling, muammoning ushbu qismi uchun egri chiziqning qisqacha chizmasidan boshlaylik.

A sketch of the curve \({C_1}\). It is the upper half of the circle of radius 4 centered at the origin, running between \(\left( {4,0} \right)\) and \(\left( { - 4,0} \right)\), and the arrows show that it is traveled in the counter clockwise direction, from right to left over the top.

Ushbu muammoning ikkinchi qismiga kelganimizdan keyin aniq bo'ladigan sabablarga ko'ra keling, bu egri chiziqni \(C\) o'rniga \({C_1}\) deb ataymiz. Mana, \({C_1}\) parametrizatsiyasi.

\({C_1}:\,\,\vec r\left( t \right) = \left\langle {4\cos \left( t \right),4\sin \left( t \right)} \right\rangle \hspace{0.25in}\,\,\,\,\,\,\,0 \le t \le \pi \)

2-qadam

Mana bu egri chiziq uchun integrali.

\[\begin{align*}\int\limits_{{{C_1}}}{{\vec F\centerdot d\vec r}} & = \int_{0}^{\pi }{{\left\langle {3,\left( {4\cos \left( t \right)} \right)\left( {4\sin \left( t \right)} \right) - 2\left( {4\cos \left( t \right)} \right)} \right\rangle \centerdot \left\langle { - 4\sin \left( t \right),4\cos \left( t \right)} \right\rangle \,dt}}\\ & = \int_{0}^{\pi }{{ - 12\sin \left( t \right) + 64\sin \left( t \right){{\cos }^2}\left( t \right) - 32{{\cos }^2}\left( t \right)\,dt}}\\ & = \int_{0}^{\pi }{{ - 12\sin \left( t \right) + 64\sin \left( t \right){{\cos }^2}\left( t \right) - 16\left( {1 + \cos \left( {2t} \right)} \right)\,dt}}\\ & = \left. {\left( {12\cos \left( t \right) - \frac{{64}}{3}{{\cos }^3}\left( t \right) - 16t - 8\sin \left( {2t} \right)} \right)} \right|_0^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{56}}{3} - 16\pi = - 31.5988}}\end{align*}\]


b \(C\) - soat yo'nalishi bo'yicha aylanish bilan 4 radiusning boshida joylashgan doiraning yuqori yarmi.
Yechim

Endi, oldingi qismda qilganimizdek, keling, bu egri chiziqni \(C\) o‘rniga \({C_2}\) deb “qayta nomlaymiz”.

Keyinchalik, bu egri chiziq qarama-qarshi yo'nalishdagi birinchi qadamdagi egri chiziq ekanligini unutmang. Boshqacha qilib aytganda, bizda bu \({C_2} = - {C_1}\) mavjud. Mana, to'liqlik uchun \({C_2}\) ning qisqacha eskizi.

A sketch of the curve \({C_2}\). It is the same upper half of the circle of radius 4 centered at the origin, but the arrows now show that it is traveled in the clockwise direction, from left to right over the top, which is the opposite direction from \({C_1}\).
2-qadam

Endi, bu nuqtada biz integralni baholash uchun ikki xil usuldan foydalanishimiz mumkin.

Birinchi usul, agar biz egri chiziq yo'nalishini o'zgartirsak, bu turdagi chiziqli integralning qiymati faqat belgilarni o'zgartirishi haqidagi faktdan foydalaning. Ushbu faktni ushbu qismdagi egri chiziq va birinchi qismdagi egri chiziq o'rtasidagi munosabat bilan birgalikda ishlatib, i.e. \({C_2} = - {C_1}\) , chiziqli integrali shunchaki,

\[\int\limits_{{{C_2}}}{{\vec F\centerdot d\vec r}} = \int\limits_{{ - {C_1}}}{{\vec F\centerdot d\vec r}} = - \int\limits_{{{C_1}}}{{\vec F\centerdot d\vec r}} = \require{bbox} \bbox[2pt,border:1px solid black]{{16\pi - \frac{{56}}{3} = 31.5988}}\]

E'tibor bering, yuqoridagi birinchi tenglik belgisi faqat ikkita egri chiziq o'rtasidagi munosabatni tan oldi. Ikkinchi tenglik belgisi - biz qaydlardagi faktdan foydalanganmiz.

Bu muammoni hal qilishning "oson" usuli. Shu bilan bir qatorda, biz egri chiziqni parametrlashimiz va chiziqli integralni to'g'ridan-to'g'ri hisoblashimiz mumkin. Biz buni muammoning qolgan qismida qanday qilishimizni ko'rsatish uchun qilamiz.

3-qadam

Mana bu egri chiziq uchun parametrlash.

\({C_2}:\,\,\vec r\left( t \right) = \left\langle { - 4\cos \left( t \right),4\sin \left( t \right)} \right\rangle \hspace{0.25in}\,\,\,\,\,\,\,0 \le t \le \pi \)

4-qadam

Endi biz faqat chiziq integralini hisoblashimiz kerak.

\[\begin{align*}\int\limits_{{{C_2}}}{{\vec F\centerdot d\vec r}} & = \int_{0}^{\pi }{{\left\langle {3,\left( { - 4\cos \left( t \right)} \right)\left( {4\sin \left( t \right)} \right) - 2\left( { - 4\cos \left( t \right)} \right)} \right\rangle \centerdot \left\langle {4\sin \left( t \right),4\cos \left( t \right)} \right\rangle \,dt}}\\ & = \int_{0}^{\pi }{{12\sin \left( t \right) - 64\sin \left( t \right){{\cos }^2}\left( t \right) + 32{{\cos }^2}\left( t \right)\,dt}}\\ & = \int_{0}^{\pi }{{12\sin \left( t \right) - 64\sin \left( t \right){{\cos }^2}\left( t \right) + 16\left( {1 + \cos \left( {2t} \right)} \right)\,dt}}\\ & = \left. {\left( { - 12\cos \left( t \right) + \frac{{64}}{3}{{\cos }^3}\left( t \right) + 16t + 8\sin \left( {2t} \right)} \right)} \right|_0^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{16\pi - \frac{{56}}{3} = 31.5988}}\end{align*}\]

Demak, bu qismning chiziqli integrali biz kutganimizdek, belgidan tashqari birinchi qismdagi chiziqli integrali bilan bir xil qiymatga ega edi.