16.4-bo'lim : Vektor maydonlarining chiziqli integrallari
6. \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) ni baholang, bunda \(\vec F\left( {x,y} \right) = xy\,\vec i + \left( {1 + 3y} \right)\vec j\) va \(C\) \(\left( {0, - 4} \right)\) dan \(\left( { - 2, - 4} \right)\) gacha bo‘lgan chiziq segmenti, keyin \(y = - {x^2}\) ning \(x = - 2\) dan \(x = 2\) ga bo‘lgan qismi, o‘z navbatida MATHBLOCK2 dan MATHBLOCK1 ga bo‘lgan chiziq segmenti. Quyidagi eskizga qarang.
Yechim
Muammoni hal qilish uchun har bir egri chiziqni quyidagicha belgilaymiz:
Har bir egri chiziqning parametrlanishi,
\({C_1}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0, - 4} \right\rangle + t\left\langle { - 2, - 4} \right\rangle = \left\langle { - 2t, - 4} \right\rangle \hspace{0.25in}0 \le t \le 1\)\({C_2}:\,\,\vec r\left( t \right) = \left\langle {t, - {t^2}} \right\rangle \hspace{0.25in} - 2 \le t \le 2\)
\({C_3}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {2, - 4} \right\rangle + t\left\langle {5,1} \right\rangle = \left\langle {2 + 3t, - 4 + 5t} \right\rangle \hspace{0.25in}0 \le t \le 1\)
2-qadam
Endi har bir egri chiziq uchun chiziqli integralini hisoblashimiz kerak. Ushbu bo'limdagi birinchi bir nechta masalalarda biz vektor funksiyani egri chiziq bo'ylab baholadik, parametrlashning hosilasini oldik va nuqta mahsulotini alohida hisobladik. Ushbu muammo uchun biz barcha ishlarni integralning o'zida qilamiz.
Bu erda har bir egri chiziq uchun integrali berilgan.
\[\begin{align*}\int\limits_{{{C_1}}}{{\vec F\centerdot d\vec r}} & = \int_{0}^{1}{{\left\langle {\left( { - 2t} \right)\left( { - 4} \right),1 + 3\left( { - 4} \right)} \right\rangle \centerdot \left\langle { - 2,0} \right\rangle \,dt}}\\ & = \int_{0}^{1}{{ - 16t\,dt}} = \left. {\left( { - 8{t^2}} \right)} \right|_0^1 = \underline { - 8} \end{align*}\] \[\begin{align*}\int\limits_{{{C_2}}}{{\vec F\centerdot d\vec r}} & = \int_{{ - 2}}^{2}{{\left\langle {\left( t \right)\left( { - {t^2}} \right),1 + 3\left( { - {t^2}} \right)} \right\rangle \centerdot \left\langle {1, - 2t} \right\rangle \,dt}}\\ & = \int_{{ - 2}}^{2}{{5{t^3} - 2t\,dt}} = \left. {\left( {\frac{5}{4}{t^4} - {t^2}} \right)} \right|_{ - 2}^2 = \underline 0 \end{align*}\] \[\begin{align*}\int\limits_{{{C_3}}}{{\vec F\centerdot d\vec r}} & = \int_{0}^{1}{{\left\langle {\left( {2 + 3t} \right)\left( { - 4 + 5t} \right),1 + 3\left( { - 4 + 5t} \right)} \right\rangle \centerdot \left\langle {3,5} \right\rangle \,dt}}\\ & = \int_{0}^{1}{{45{t^2} + 69t - 79\,dt}} = \left. {\left( {15{t^3} + \frac{{69}}{2}{t^2} - 79t} \right)} \right|_0^1 = \underline { - \frac{{59}}{2}} \end{align*}\]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.
\[\int\limits_{C}{{\vec F\centerdot d\vec r}} = \left( { - 8} \right) + \left( 0 \right) + \left( { - \frac{{59}}{2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{75}}{2}}}\]Esda tutingki, biz har bir alohida chiziqli integral natijasiga uning qayerdan kelganini ko'rsatish uchun qavs qo'yamiz va ular umuman kerak emas.