16.2-bo'lim : Chiziqli integrallar - I qism
9. Quyidagi egri chiziqlarning har biri uchun \( \displaystyle \int\limits_{C}{{4{x^2}\,ds}}\) ni baholang.
- \(C\) aylananing 1 dagi 2 radiusning boshiga markazlashgan qismidir.st soat yo'nalishi bo'yicha aylanadigan kvadrant.
- \(C\) - \(\left( {0,2} \right)\) dan \(\left( {2,0} \right)\) gacha bo'lgan chiziq segmenti.
Yechim
Keling, muammoning ushbu qismi uchun egri chiziqning qisqacha chizmasidan boshlaylik.
Mana bu egri chiziq uchun parametrlash.
\(C:\,\,\vec r\left( t \right) = \left\langle {2\sin \left( t \right),2\cos \left( t \right)} \right\rangle \hspace{0.25in}\,\,\,\,\,\,\,0 \le t \le \frac{1}{2}\pi \)2-qadam
Mana bu egri chiziq uchun integrali.
\[\begin{align*}\int\limits_{C}{{4{x^2}\,ds}} &= \int_{0}^{{\frac{1}{2}\pi }}{{4{{\left( {2\sin \left( t \right)} \right)}^2}\sqrt {{{\left( {2\cos \left( t \right)} \right)}^2} + {{\left( { - 2\sin \left( t \right)} \right)}^2}} \,dt}}\\ & = \int_{0}^{{\frac{1}{2}\pi }}{{16{{\sin }^2}\left( t \right)\sqrt {4{{\cos }^2}\left( t \right) + 4{{\sin }^2}\left( t \right)} \,dt}}\\ & = \int_{0}^{{\frac{1}{2}\pi }}{{16{{\sin }^2}\left( t \right)\sqrt 4 \,dt}}\\ & = \int_{0}^{{\frac{1}{2}\pi }}{{32\left( {\frac{1}{2}} \right)\left( {1 - \cos \left( {2t} \right)} \right)\,dt}}\\ & = \left. {16\left( {t - \frac{1}{2}\sin \left( {2t} \right)} \right)} \right|_0^{\frac{1}{2}\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{8\pi }}\end{align*}\]b \(C\) - \(\left( {0,2} \right)\) dan \(\left( {2,0} \right)\) gacha bo'lgan chiziq segmenti.
Yechim
Keling, muammoning ushbu qismi uchun egri chiziqning qisqacha chizmasidan boshlaylik.
Demak, bu qismda bizda \(\left( {0,2} \right)\) dan \(\left( {2,0} \right)\) gacha bo'lgan boshqa egri chiziq mavjud. Ushbu egri chiziq birinchi qismdagi egri chiziq bilan bir xil boshlang'ich va tugash nuqtasiga ega bo'lishiga qaramay, chiziqli integrali bir xil qiymatga ega bo'lishini kutish uchun hech qanday sabab yo'q. Shuning uchun biz ishni ko'rib chiqishimiz va chiziqli integraldan nimani olishimizni ko'rishimiz kerak.
Biz egri chiziqni parametrlashimiz kerak, shuning uchun bu haqda g'amxo'rlik qilaylik.
\(C:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,2} \right\rangle + t\left\langle {2,0} \right\rangle = \left\langle {2t,2 - 2t} \right\rangle \hspace{0.25in}\,\,\,\,\,\,\,0 \le t \le 1\)E'tibor bering, biz bu egri chiziqning tenglamasini topishimiz mumkin edi, lekin ikkita nuqta orasidagi chiziq segmentining vektor shaklini ishlatish juda oson tuyuldi.
2-qadam
Endi biz faqat chiziq integralini hisoblashimiz kerak.
\[\begin{align*}\int\limits_{C}{{4{x^2}\,ds}} & = \int_{0}^{1}{{4{{\left( {2t} \right)}^2}\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} \,dt}}\\ & = \sqrt 8 \int_{0}^{1}{{16{t^2}\,dt}}\\ & = \left. {2\sqrt 2 \left( {\frac{{16}}{3}{t^3}} \right)} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32\sqrt 2 }}{3}}}\end{align*}\]Shunday qilib, ushbu qismning boshida ta'kidlanganidek, egri chiziqning bir xil boshlang'ich va tugash nuqtalariga qaramay, chiziqli integralning qiymati birinchi qismdagi chiziqli integralning qiymati bilan bir xil emas edi. E'tibor bering, bir xil boshlang'ich va tugash nuqtalari bo'lgan ikkita chiziqli integral bir xil qiymatga ega bo'lishi mumkin, ammo biz bunday bo'lishini kuta olmaymiz va shuning uchun ishni bajarishimiz kerak.