16.2-bo'lim : Chiziqli integrallar - I qism
6. \( \displaystyle \int\limits_{C}{{16{y^5}\,ds}}\) ni baholang, bu erda \(C\) \(x = {y^4}\) ning \(y = 0\) dan \(y = 1\) gacha bo'lgan qismi, keyin \(\left( {1,1} \right)\) dan \(\left( {1, - 2} \right)\) gacha bo'lgan chiziq segmenti, o'z navbatida \(\left( {1, - 2} \right)\) dan MATHBLOCK ga bo'lgan chiziq segmenti. Yo'nalish uchun quyidagi eskizga qarang.
Yechim
Muammoni hal qilish uchun har bir egri chiziqni quyidagicha belgilaymiz:
Endi bu egri chiziqlarning har birini parametrlashtiramiz.
\({C_1}:\,\,\vec r\left( t \right) = \left\langle {{t^4},t} \right\rangle \hspace{0.25in}0 \le t \le 1\)\({C_2}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {1,1} \right\rangle + t\left\langle {1, - 2} \right\rangle = \left\langle {1,1 - 3t} \right\rangle \hspace{0.25in}0 \le t \le 1\)
\({C_3}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {1, - 2} \right\rangle + t\left\langle {2,0} \right\rangle = \left\langle {1 + t,2t - 2} \right\rangle \hspace{0.25in}0 \le t \le 1\)
\({C_2}\) uchun chiziq tenglamasi oʻrniga ikki nuqta orasidagi chiziq segmenti uchun vektor shaklidan foydalanishga toʻgʻri keldi (bu, albatta, ancha sodda), chunki yoʻnalish kamayib boruvchi \(y\) yoʻnalishida edi va integralimizdagi chegaralar kichikdan kattagacha boʻlishi kerak. Agar biz xohlasak, chiziq tenglamasidan foydalanishga imkon beradigan ikkita yo'nalish uchun chiziq qanday integral bo'lishi haqida eslatmadagi faktdan foydalanishimiz mumkin edi. Biz buni chiziq segmenti uchun vektor shakli bilan ishlash amaliyoti uchun shunday qilishga qaror qildik va natija bilan shug'ullanish unchalik qiyin emas va chegaralar "yaxshiroq".
Shuni ham yodda tutingki, \({C_3}\) uchun biz chiziq tenglamasini echishimiz mumkin edi va bu yo'nalish \(x\) ortib borayotgan yo'nalishda bo'lgani uchun foydalanishimiz mumkin edi. Biroq, ikkita nuqta orasidagi chiziq segmenti uchun vektor shaklini ishlatish juda oson, shuning uchun biz uning o'rniga foydalandik.
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.
\[\begin{align*}\int\limits_{{{C_1}}}{{16{y^5}\,ds}} & = \int_{0}^{1}{{16{{\left( t \right)}^5}\,\sqrt {{{\left( {4{t^3}} \right)}^2} + {{\left( 1 \right)}^2}} \,dt}} = \int_{0}^{1}{{16{t^5}\,\sqrt {16{t^6} + 1} \,dt}}\\ & = \left. {\frac{1}{9}{{\left( {16{t^6} + 1} \right)}^{\frac{3}{2}}}} \right|_0^1 = \underline {\frac{1}{9}\left( {{{17}^{\frac{3}{2}}} - 1} \right) = 7.6770} \end{align*}\] \[\begin{align*}\int\limits_{{{C_2}}}{{16{y^5}\,ds}} &= \int_{0}^{1}{{16{{\left( {1 - 3t} \right)}^5}\,\sqrt {{{\left( 0 \right)}^2} + {{\left( { - 3} \right)}^2}} \,dt}} = \int_{0}^{1}{{48{{\left( {1 - 3t} \right)}^5}\,\,dt}}\\ & = \left. { - \frac{8}{3}{{\left( {1 - 3t} \right)}^6}} \right|_0^1 = \underline { - 168} \end{align*}\] \[\begin{align*}\int\limits_{{{C_3}}}{{16{y^5}\,ds}} &= \int_{0}^{1}{{16{{\left( {2t - 2} \right)}^5}\,\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} \,dt}} = \int_{0}^{1}{{16\sqrt 5 {{\left( {2t - 2} \right)}^5}\,\,dt}}\\ & = \left. {\frac{{4\sqrt 5 }}{3}{{\left( {2t - 2} \right)}^6}} \right|_0^1 = \underline { - \frac{{256\sqrt 5 }}{3} = - 190.8111} \end{align*}\]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.
\[\int\limits_{C}{{16{y^5}\,ds}} = \left( {\frac{1}{9}\left( {{{17}^{\frac{3}{2}}} - 1} \right)} \right) + \left( { - 168} \right) + \left( { - \frac{{256\sqrt 5 }}{3}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 351.1341}}\]Esda tutingki, biz har bir alohida chiziqli integral natijasiga uning qayerdan kelganini ko'rsatish uchun qavs qo'yamiz va ular umuman kerak emas.