16.3-bo'lim: Chiziqli integrallar - II qism
Oldingida yoy uzunligiga nisbatan chiziqli integrallarni ko'rib chiqdik. Ushbu bo'limda biz \(x\) va/yoki \(y\) ga nisbatan chiziqli integrallarni ko'rib chiqmoqchimiz.
Oxirgi bo'limda bo'lgani kabi biz parametrlash bilan \(C\) ikki o'lchovli egri chizig'idan boshlaymiz,
\[x = x\left( t \right)\hspace{0.25in}y = y\left( t \right)\hspace{0.25in}a \le t \le b\]\(x\) va \(y\) ga nisbatan chiziqli integrallar
\(f\) ning \(x\) ga nisbatan chiziqli integrali is,
\[\int\limits_{C}{{f\left( {x,y} \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right)} \right)x'\left( t \right)\,dt}}\]\(f\) ning \(y\) ga nisbatan chiziqli integrali is,
\[\int\limits_{C}{{f\left( {x,y} \right)\,dy}} = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right)} \right)y'\left( t \right)\,dt}}\]E'tibor bering, bu ikki va chiziq integrali o'rtasidagi yoy uzunligi bo'yicha (oldingi bo'limdan) yagona notatsion farq bu differentsialdir. Ularda \(dx\) yoki \(dy\) bor, yoy uzunligiga nisbatan chiziq integrali esa \(ds\) ga ega. Shunday qilib, chiziqli integrallarni baholashda siz noto'g'ri chiziqli integral bilan ishlamaslik uchun birinchi navbatda qaysi differentsialga ega ekanligingizga e'tibor bering.
Ushbu ikkita integral ko'pincha birga paydo bo'ladi va shuning uchun bizda bu holatlar uchun quyidagi stenografiya mavjud.
Birlashtirilgan chiziqli integrallar uchun stenografiya (2D)
Keling, ushbu turdagi chiziqli integralning misolini tez ko'rib chiqaylik.
Yechim
Bu erda egri chiziqning parametrizatsiyasi.
\[\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,2} \right\rangle + t\left\langle {1,4} \right\rangle = \left\langle {t,2 + 2t} \right\rangle \,\hspace{0.25in}0 \le t \le 1\]Chiziqli integrali,
\[\begin{align*}\int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\, + \,y{x^2}\,dx}} & = \int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\,}} + \int\limits_{C}{{y{x^2}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\sin \left( {\pi \left( {2 + 2t} \right)} \right)\left( 2 \right)\,dt}} + \int_{{\,0}}^{{\,1}}{{\left( {2 + 2t} \right){{\left( t \right)}^2}\left( 1 \right)\,dt}}\\ & = - \left. {\frac{1}{\pi }\cos \left( {2\pi + 2\pi \,t} \right)} \right|_0^1 + \left. {\left( {\frac{2}{3}{t^3} + \frac{1}{2}{t^4}} \right)} \right|_0^1\\ & = \frac{7}{6}\end{align*}\]Oldingida biz chiziqli integral uchun egri chiziq yo'nalishini yoy uzunligiga nisbatan o'zgartirish integral qiymatini o'zgartirmasligini ko'rdik. Keling, \(x\) va/yoki \(y\) ga nisbatan chiziqli integrallar bilan nima sodir bo'lishini ko'rib chiqaylik.
Yechim
Shunday qilib, biz shunchaki egri chiziq yo'nalishini o'zgartirdik. Mana yangi parametrlash.
\[\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {1,4} \right\rangle + t\left\langle {0,2} \right\rangle = \left\langle {1 - t,4 - 2t} \right\rangle \,\hspace{0.25in}0 \le t \le 1\]Bu holda chiziqli integrali,
\[\begin{align*}\int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\, + \,y{x^2}\,dx}} & = \int\limits_{C}{{\sin \left( {\pi y} \right)\,dy\,}} + \int\limits_{C}{{y{x^2}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\sin \left( {\pi \left( {4 - 2t} \right)} \right)\left( { - 2} \right)\,dt}} + \int_{{\,0}}^{{\,1}}{{\left( {4 - 2t} \right){{\left( {1 - t} \right)}^2}\left( { - 1} \right)\,dt}}\\ & = - \left. {\frac{1}{\pi }\cos \left( {4\pi - 2\pi t} \right)} \right|_0^1 - \left. {\left( { - \frac{1}{2}{t^4} + \frac{8}{3}{t^3} - 5{t^2} + 4t} \right)} \right|_0^1\\ & = - \frac{7}{6}\end{align*}\]Shunday qilib, egri chiziqning yo'nalishini o'zgartirish bizga boshqa qiymatni yoki hech bo'lmaganda birinchi misoldagi qiymatning qarama-qarshi belgisini oldi. Aslida, bu har doim bunday chiziqli integrallar bilan sodir bo'ladi.
Egri chiziq yo'nalishini teskari aylantirish
Agar \(C\) har qanday egri chiziq bo'lsa,
\[\int\limits_{{ - C}}{{f\left( {x,y} \right)\,dx}} = - \int\limits_{C}{{f\left( {x,y} \right)\,dx}}\hspace{0.25in}\hspace{0.25in}{\rm{and}}\hspace{0.25in}\,\,\,\,\,\,\int\limits_{{ - C}}{{f\left( {x,y} \right)\,dy}} = - \int\limits_{C}{{f\left( {x,y} \right)\,dy}}\]Ushbu ikkita integralning qo'shma shakli bilan biz olamiz:
\[\int\limits_{{ - C}}{{Pdx\, + Q\,dy}} = - \int\limits_{C}{{Pdx\, + Q\,dy}}\]Biz bu integrallarni uch o'lchovli egri chiziqlar ustida ham bajarishimiz mumkin. Bu holda biz uchinchi integralni olamiz (\(z\) ga nisbatan) va uchta integral bo'ladi.
\(x\), \(y\) va \(z\) ga nisbatan chiziqli integrallar
bu erda \(C\) egri chizig'i bilan parametrlangan
\[x = x\left( t \right)\hspace{0.25in}y = y\left( t \right)\hspace{0.25in}z = z\left( t \right)\hspace{0.25in}a \le t \le b\]Ikki o'lchovli versiyada bo'lgani kabi, bu uchtasi ko'pincha birga bo'ladi, shuning uchun biz bu erda ishlatadigan stenografiya:
Birlashtirilgan chiziqli integrallar uchun stenografiya (3D)
Keling, misol bilan ishlaylik.
Yechim
Shunday qilib, bizda allaqachon egri chiziq parametrlangan, shuning uchun integralni baholashdan boshqa qiladigan ko'p narsa yo'q.
\[\begin{align*}\int\limits_{C}{{y\,dx + x\,dy + z\,dz}} &= \int\limits_{C}{{y\,dx}} + \int\limits_{C}{{x\,dy}} + \int\limits_{C}{{z\,dz}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\sin t\left( { - \sin t} \right)\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{\cos t\left( {\cos t} \right)\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{{t^2}\left( {2t} \right)\,dt}}\\ & = - \int_{{\,0}}^{{\,2\pi }}{{{{\sin }^2}t\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{{{\cos }^2}t\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{2{t^3}\,dt}}\\ & = - \frac{1}{2}\int_{{\,0}}^{{\,2\pi }}{{\left( {1 - \cos \left( {2t} \right)} \right)\,dt}} + \frac{1}{2}\int_{{\,0}}^{{\,2\pi }}{{\left( {1 + \cos \left( {2t} \right)} \right)\,dt}} + \int_{{\,0}}^{{\,2\pi }}{{2{t^3}\,dt}}\\ & = \left. {\left( { - \frac{1}{2}\left( {t - \frac{1}{2}\sin \left( {2t} \right)} \right) + \frac{1}{2}\left( {t + \frac{1}{2}\sin \left( {2t} \right)} \right) + \frac{1}{2}{t^4}} \right)} \right|_0^{2\pi }\\ & = 8{\pi ^4}\end{align*}\]