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16.4-bo'lim : Vektor maydonlarining chiziqli integrallari

1. \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) ni baholang, bu erda \(\vec F\left( {x,y} \right) = {y^2}\,\vec i + \left( {3x - 6y} \right)\vec j\) va \(C\) \(\left( {3,7} \right)\) dan \(\left( {0,12} \right)\) gacha bo'lgan chiziq segmentidir.

Yechim

Bu erda ko'rsatilgan muammo bayonotida ko'rsatilgan yo'nalish bilan \(C\) ning tezkor eskizi.

A sketch of the curve \(C\).  It is the straight line segment between \(\left( {3,7} \right)\) and \(\left( {0,12} \right)\), and the arrows show that it is traveled from \(\left( {3,7} \right)\) up to \(\left( {0,12} \right)\).
2-qadam

Keyinchalik, egri chiziqni parametrlashimiz kerak.

\[\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {3,7} \right\rangle + t\left\langle {0,12} \right\rangle = \left\langle {3 - 3t,7 + 5t} \right\rangle \,\,\,\,\,\,\,\,0 \le t \le 1\]
3-qadam

Ushbu chiziqli integralni baholash uchun bizga vektor maydonining nuqta mahsuloti (egri chiziq bo'ylab baholanadi) va parametrlashning hosilasi kerak bo'ladi.

Mana egri chiziq bo'ylab baholangan vektor maydoni (i.e. vektor maydoniga parametrlashdan \(x\) va \(y\) ni ulang).

\[\vec F\left( {\vec r\left( t \right)} \right) = {\left( {7 + 5t} \right)^2}\,\vec i + \left( {3\left( {3 - 3t} \right) - 6\left( {7 + 5t} \right)} \right)\vec j = {\left( {7 + 5t} \right)^2}\,\vec i + \left( { - 33 - 39t} \right)\vec j\]

Parametrlashning hosilasi,

\[\vec r'\left( t \right) = \left\langle { - 3,5} \right\rangle \]

Nihoyat, vektor maydonining nuqta mahsuloti va parametrlashning hosilasi.

\[\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) = - 3{\left( {7 + 5t} \right)^2} - 5\left( {33 + 39t} \right)\]
4-qadam

Endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash.

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{0}^{1}{{ - 3{{\left( {7 + 5t} \right)}^2} - 5\left( {33 + 39t} \right)\,dt}}\\ & = \left. {\left[ { - \frac{1}{5}{{\left( {7 + 5t} \right)}^3} - 165t - \frac{{195}}{2}{t^2}} \right]} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{1079}}{2}}}\end{align*}\]