16.2-bo'lim : Chiziqli integrallar - I qism
7. \( \displaystyle \int\limits_{C}{{4y - x\,ds}}\) ni baholang, bunda \(C\) 3 radiusning boshi markazida \(\displaystyle\left( {\frac{3}{{\sqrt 2 }},\frac{3}{{\sqrt 2 }}} \right)\) dan \(\displaystyle\left( { - \frac{3}{{\sqrt 2 }}, - \frac{3}{{\sqrt 2 }}} \right)\) gacha bo‘lgan aylananing yuqori qismi soat miliga teskari aylanishda, so‘ngra \(\displaystyle\left( { - \frac{3}{{\sqrt 2 }}, - \frac{3}{{\sqrt 2 }}} \right)\) dan \(\displaystyle\left( {4, - \frac{3}{{\sqrt 2 }}} \right)\) gacha bo‘lgan chiziq segmenti, o‘z navbatida \[\begin{align*}\int\limits_{{{C_1}}}{{4y - x\,ds}} &= \int_{{\frac{1}{4}\pi }}^{{\frac{5}{4}\pi }}{{\left[ {4\left( {3\sin \left( t \right)} \right) - 3\cos \left( t \right)} \right]\sqrt {{{\left( { - 3\sin \left( t \right)} \right)}^2} + {{\left( {3\cos \left( t \right)} \right)}^2}} \,dt}}\\ & = \int_{{\frac{1}{4}\pi }}^{{\frac{5}{4}\pi }}{{\left[ {12\sin \left( t \right) - 3\cos \left( t \right)} \right]\sqrt {9{{\sin }^2}\left( t \right) + 9{{\cos }^2}\left( t \right)} \,dt}}\\ & = 3\int_{{\frac{1}{4}\pi }}^{{\frac{5}{4}\pi }}{{12\sin \left( t \right) - 3\cos \left( t \right)\,dt}}\\ & = \left. {3\left( { - 12\cos \left( t \right) - 3\sin \left( t \right)} \right)} \right|_{\frac{1}{4}\pi }^{\frac{5}{4}\pi } = \underline {\frac{{90}}{{\sqrt 2 }}} \end{align*}\]K dan keyin \(\displaystyle\left( {4, - \frac{3}{{\sqrt 2 }}} \right)\) gacha bo‘lgan chiziq segmenti. \(\left( {4,4} \right)\). Yo'nalish uchun quyidagi eskizga qarang.
Yechim
Muammoni hal qilish uchun har bir egri chiziqni quyidagicha belgilaymiz:
Endi bu egri chiziqlarning har birini parametrlashtiramiz.
\(\displaystyle {C_1}:\,\,\vec r\left( t \right) = \left\langle {3\cos \left( t \right),3\sin \left( t \right)} \right\rangle \hspace{0.25in}\,\,\,\,\,\,\,\frac{1}{4}\pi \le t \le \frac{5}{4}\pi \)\(\displaystyle {C_2}:\,\,\vec r\left( t \right) = \left\langle {t, - \frac{3}{{\sqrt 2 }}} \right\rangle \hspace{0.25in} - \frac{3}{{\sqrt 2 }} \le t \le 4\)
\(\displaystyle {C_3}:\,\,\vec r\left( t \right) = \left\langle {4,t} \right\rangle \hspace{0.25in} - \frac{3}{{\sqrt 2 }} \le t \le 4\)
\({C_1}\) uchun chegaralarni topish juda oson. Boshlang'ich nuqtada, \(\left( {\frac{3}{{\sqrt 2 }},\frac{3}{{\sqrt 2 }}} \right)\), bizda bo'lishi kerakligini bilamiz,
\[3\cos \left( t \right) = 3\sin \left( t \right) = \frac{3}{{\sqrt 2 }}\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( t \right) = \sin \left( t \right) = \frac{1}{{\sqrt 2 }}\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\tan \left( t \right) = 1\]\({C_1}\) ning boshlang'ich nuqtasi 1da bo'lgani uchunst kvadrant bizda \(t = \frac{1}{4}\pi \) bo'lishi kerakligini bilamiz.
Biz \(\left( { - \frac{3}{{\sqrt 2 }}, - \frac{3}{{\sqrt 2 }}} \right)\) yakuniy nuqtaning shunga o'xshash argumentini qilishimiz mumkin. Bu nuqta 3-banddard kvadrant va yana bizda \(\tan \left( t \right) = 1\) bo'ladi. Shuning uchun bizda bu nuqtada \(t = \frac{5}{4}\pi \) bo'lishi kerak.
\({C_2}\) va \({C_3}\) uchun biz shunchaki chiziqlar tenglamalaridan foydalandik, chunki ular mos ravishda gorizontal va vertikal edi va mos ravishda \(x\) va \(y\) o'sishini kuzatdik. Ular uchun chegaralar biroz yoqimsiz, lekin biz ularni qanday qilsak ham, ularda tartibsiz raqamlar bo'lishi haqida hech narsa qila olmaymiz.
2-qadam
OK, endi biz ushbu egri chiziqlarning har biri bo'ylab chiziqli integralni hisoblashimiz kerak. Ushbu bo'limdagi birinchi bir nechta muammolardan farqli o'laroq, biz integratsiya bosqichidan oldin kattalik va integralni topdik, biz shunchaki to'g'ridan-to'g'ri integralga o'tamiz va u erda barcha ishlarni qilamiz.
Bu erda har bir egri chiziq bo'ylab integral mavjud.
\[\begin{align*}\int\limits_{{{C_1}}}{{4y - x\,ds}} &= \int_{{\frac{1}{4}\pi }}^{{\frac{5}{4}\pi }}{{\left[ {4\left( {3\sin \left( t \right)} \right) - 3\cos \left( t \right)} \right]\sqrt {{{\left( { - 3\sin \left( t \right)} \right)}^2} + {{\left( {3\cos \left( t \right)} \right)}^2}} \,dt}}\\ & = \int_{{\frac{1}{4}\pi }}^{{\frac{5}{4}\pi }}{{\left[ {12\sin \left( t \right) - 3\cos \left( t \right)} \right]\sqrt {9{{\sin }^2}\left( t \right) + 9{{\cos }^2}\left( t \right)} \,dt}}\\ & = 3\int_{{\frac{1}{4}\pi }}^{{\frac{5}{4}\pi }}{{12\sin \left( t \right) - 3\cos \left( t \right)\,dt}}\\ & = \left. {3\left( { - 12\cos \left( t \right) - 3\sin \left( t \right)} \right)} \right|_{\frac{1}{4}\pi }^{\frac{5}{4}\pi } = \underline {\frac{{90}}{{\sqrt 2 }}} \end{align*}\] \[\begin{align*}\int\limits_{{{C_2}}}{{4y - x\,ds}} &= \int_{{ - \,\,\frac{3}{{\sqrt 2 }}}}^{4}{{\left[ {4\left( { - \frac{3}{{\sqrt 2 }}} \right) - t} \right]\sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2}} \,dt}}\\ & = \int_{{ - \,\,\frac{3}{{\sqrt 2 }}}}^{4}{{ - \frac{{12}}{{\sqrt 2 }} - t\,dt}}\\ & = \left. {\left( { - \frac{{12}}{{\sqrt 2 }}t - \frac{1}{2}{t^2}} \right)} \right|_{ - \,\,\frac{3}{{\sqrt 2 }}}^4 = \underline { - \frac{{95}}{4} - \frac{{48}}{{\sqrt 2 }}} \end{align*}\] \[\begin{align*}\int\limits_{{{C_3}}}{{4y - x\,ds}} &= \int_{{ - \,\,\frac{3}{{\sqrt 2 }}}}^{4}{{\left[ {4\left( t \right) - 4} \right]\sqrt {{{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}} \,dt}}\\ & = \int_{{ - \,\,\frac{3}{{\sqrt 2 }}}}^{4}{{4t - 4\,dt}}\\ & = \left. {\left( {2{t^2} - 4t} \right)} \right|_{ - \,\,\frac{3}{{\sqrt 2 }}}^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{7 - \frac{{12}}{{\sqrt 2 }}}}\end{align*}\]3-qadam
Yaxshi, bu muammoni hal qilish uchun biz qilishimiz kerak bo'lgan narsa to'liq chiziqli integralni olish uchun ushbu egri chiziqlar ustiga chiziqli integrallarni qo'shishdir.
\[\int\limits_{C}{{4y - x\,ds}} = \left( {\frac{{90}}{{\sqrt 2 }}} \right) + \left( { - \frac{{95}}{4} - \frac{{48}}{{\sqrt 2 }}} \right) + \left( {7 - \frac{{12}}{{\sqrt 2 }}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{67}}{4} + \frac{{30}}{{\sqrt 2 }} = 4.4632}}\]Esda tutingki, biz har bir alohida chiziqli integral natijasiga uning qayerdan kelganini ko'rsatish uchun qavs qo'yamiz va ular umuman kerak emas.