16.4-bo'lim : Vektor maydonlarining chiziqli integrallari
2. \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) ni baholang, bunda \(\vec F\left( {x,y} \right) = \left( {x + y} \right)\,\vec i + \left( {1 - x} \right)\vec j\) va \(C\) \(\displaystyle \frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1\) ning 4-banddagi qismidir.th soat miliga teskari aylanish bilan kvadrant.
Yechim
Bu erda ko'rsatilgan muammo bayonotida ko'rsatilgan yo'nalish bilan \(C\) ning tezkor eskizi.
2-qadam
Keyinchalik, egri chiziqni parametrlashimiz kerak.
\[\vec r\left( t \right) = \left\langle {2\cos \left( t \right),3\sin \left( t \right)} \right\rangle \,\,\,\,\,\,\,\,\frac{3}{2}\pi \le t \le 2\pi \]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).
\[\vec F\left( {\vec r\left( t \right)} \right) = \left( {2\cos \left( t \right) + 3\sin \left( t \right)} \right)\,\vec i + \left( {1 - 2\cos \left( t \right)} \right)\vec j\]Parametrlashning hosilasi,
\[\vec r'\left( t \right) = \left\langle { - 2\sin \left( t \right),3\cos \left( t \right)} \right\rangle \]Nihoyat, vektor maydonining nuqta mahsuloti va parametrlashning hosilasi.
\[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) & = \left( {2\cos \left( t \right) + 3\sin \left( t \right)} \right)\left( { - 2\sin \left( t \right)} \right) + \left( {1 - 2\cos \left( t \right)} \right)\left( {3\cos \left( t \right)} \right)\\ & = - 4\cos \left( t \right)\sin \left( t \right) - 6{\sin ^2}\left( t \right) + 3\cos \left( t \right) - 6{\cos ^2}\left( t \right)\\ & = - 4\cos \left( t \right)\sin \left( t \right) - 6\left[ {{{\sin }^2}\left( t \right) + {{\cos }^2}\left( t \right)} \right] + 3\cos \left( t \right)\\ & = - 2\sin \left( {2t} \right) + 3\cos \left( t \right) - 6\end{align*}\]Integralni bajarish uchun nuqta mahsulotini soddalashtirganingizga ishonch hosil qiling! Bunday holda, bu integralning birinchi atamasini "soddalashtirish" uchun sinus uchun ikki burchakli formuladan foydalanishni anglatadi.
4-qadam
Endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash.
\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{{\frac{3}{2}\pi }}^{{2\pi }}{{ - 2\sin \left( {2t} \right) + 3\cos \left( t \right) - 6\,dt}}\\ & = \left. {\left[ {\cos \left( {2t} \right) + 3\sin \left( t \right) - 6t} \right]} \right|_{\frac{3}{2}\pi }^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{5 - 3\pi }}\end{align*}\]