Keyinchalik, egri chiziqni parametrlashimiz kerak.

\[\vec r\left( t \right) = \left\langle {t,{{\left( {t - 1} \right)}^2}} \right\rangle \,\,\,\,\,\,\,\,0 \le t \le 3\]

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).

\[\vec F\left( {\vec r\left( t \right)} \right) = {\left[ {{{\left( {t - 1} \right)}^2}} \right]^2}\,\vec i + \left( {{{\left( t \right)}^2} - 4} \right)\vec j = {\left( {t - 1} \right)^4}\,\vec i + \left( {{t^2} - 4} \right)\vec j\]

Parametrlashning hosilasi,

\[\vec r'\left( t \right) = \left\langle {1,2\left( {t - 1} \right)} \right\rangle \]

Nihoyat, vektor maydonining nuqta mahsuloti va parametrlashning hosilasi.

\[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) & = {\left( {t - 1} \right)^4}\left( 1 \right) + \left( {{t^2} - 4} \right)\left( {2t - 2} \right)\\ & = {\left( {t - 1} \right)^4} + 2{t^3} - 2{t^2} - 8t + 8\end{align*}\]

Integralni bajarish uchun nuqta mahsulotini soddalashtirganingizga ishonch hosil qiling!

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

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{0}^{3}{{{{\left( {t - 1} \right)}^4} + 2{t^3} - 2{t^2} - 8t + 8\,dt}}\\ & = \left. {\left[ {\frac{1}{5}{{\left( {t - 1} \right)}^5} + \frac{1}{2}{t^4} - \frac{2}{3}{t^3} - 4{t^2} + 8t} \right]} \right|_0^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{171}}{{10}}}}\end{align*}\]