16.3-bo'lim : Chiziqli integrallar - II qism
5. \( \displaystyle \int\limits_{C}{{2{x^2}\,dy - xy\,dx}}\) ni baholang, bu erda \(C\) \(\left( {1, - 5} \right)\) dan \(\left( { - 2, - 3} \right)\) gacha bo'lgan chiziq segmenti va keyin \(y = 1 - {x^2}\) ning \(x = - 2\) dan \(x = 2\) ga bo'lgan qismi, o'z navbatida \(\left( {2, - 3} \right)\) 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.
\({C_1}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {1, - 5} \right\rangle + t\left\langle { - 2, - 3} \right\rangle = \left\langle {1 - 3t, - 5 + 2t} \right\rangle \hspace{0.25in}0 \le t \le 1\)\({C_2}:\,\,\vec r\left( t \right) = \left\langle {t,1 - {t^2}} \right\rangle \hspace{0.25in} - 2 \le t \le 2\)
\({C_3}:\,\,\vec r\left( t \right) = \left\langle {t, - 3} \right\rangle \hspace{0.25in}2 \le t \le 4\)
E'tibor bering, \({C_1}\) uchun ikkita nuqta orasidagi chiziq segmenti uchun vektor shaklidan foydalanishimiz kerak edi, chunki belgilangan yo'nalish kamayuvchi \(x\) yo'nalishida edi va shuning uchun chiziq tenglamasi ishlamaydi, chunki chiziqli integral chegaralari kichikroqdan kattaroq qiymatlarga o'tishi kerak.
Biz faqat \({C_3}\) uchun chiziq tenglamasidan foydalandik, chunki buni qilish juda oddiy va chegaralar ham etarlicha yaxshi edi.
2-qadam
Endi har bir egri chiziq uchun chiziqli integralini hisoblashimiz kerak.
\[\begin{align*}\int\limits_{{{C_1}}}{{2{x^2}\,dy - xy\,dx}} &= \int\limits_{{{C_1}}}{{2{x^2}\,dy}} - \int\limits_{{{C_1}}}{{xy\,dx}}\\ & = \int_{0}^{1}{{2{{\left( {1 - 3t} \right)}^2}\left( 2 \right)\,dt}} - \int_{0}^{1}{{\left( {1 - 3t} \right)\left( { - 5 + 2t} \right)\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{4{{\left( {1 - 3t} \right)}^2} - 3\left( {6{t^2} - 17t + 5} \right)\,dt}}\\ & = \left. {\left( { - \frac{4}{9}{{\left( {1 - 3t} \right)}^3} - 3\left( {2{t^3} - \frac{{17}}{2}{t^2} + 5t} \right)} \right)} \right|_0^1 = \underline {\frac{{17}}{2}} \end{align*}\] \[\begin{align*}\int\limits_{{{C_2}}}{{2{x^2}\,dy - xy\,dx}} &= \int\limits_{{{C_2}}}{{2{x^2}\,dy}} - \int\limits_{{{C_2}}}{{xy\,dx}}\\ & = \int_{{ - 2}}^{2}{{2{{\left( t \right)}^2}\left( { - 2t} \right)\,dt}} - \int_{{ - 2}}^{2}{{\left( t \right)\left( {1 - {t^2}} \right)\left( 1 \right)\,dt}}\\ & = \int_{{ - 2}}^{2}{{ - 3{t^3} - t\,dt}}\\ & = \left. {\left( { - \frac{3}{4}{t^4} - \frac{1}{2}{t^2}} \right)} \right|_{ - 2}^2 = \underline 0 \end{align*}\] \[\begin{align*}\int\limits_{{{C_3}}}{{2{x^2}\,dy - xy\,dx}} &= \int\limits_{{{C_3}}}{{2{x^2}\,dy}} - \int\limits_{{{C_2}}}{{xy\,dx}}\\ & = \int_{2}^{4}{{2{{\left( t \right)}^2}\left( 0 \right)\,dt}} - \int_{2}^{4}{{\left( t \right)\left( { - 3} \right)\left( 1 \right)\,dt}}\\ & = \int_{2}^{4}{{3t\,dt}}\\ & = \left. {\left( {\frac{3}{2}{t^2}} \right)} \right|_2^4 = \underline {18} \end{align*}\]Chiziqli integralni "standart" integralga o'tkazishda differensiallar bilan to'g'ri ishlashni unutmang.
Bundan tashqari, differensiallardan biri \({C_3}\) integralidagi birinchisi kabi nolga “baholansa” hayajonlanmang. Bu ba'zida sodir bo'ladi va bu sodir bo'lganda tashvishlanadigan narsa emas.
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.
\[ \displaystyle \int\limits_{C}{{2{x^2}\,dy - xy\,dx}} = \left( {\frac{{17}}{2}} \right) + \left( 0 \right) + \left( {18} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{53}}{2}}}\]Esda tutingki, biz har bir alohida chiziqli integral natijasiga uning qayerdan kelganini ko'rsatish uchun qavs qo'yamiz va ular umuman kerak emas.