16.4-bo'lim : Vektor maydonlarining chiziqli integrallari
4. \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) ni baholang, bu erda \(\vec F\left( {x,y,z} \right) = {{\bf{e}}^{2x}}\,\vec i + z\left( {y + 1} \right)\vec j + {z^3}\,\vec k\) va \(C\) \(\vec r\left( t \right) = {t^3}\,\vec i + \left( {1 - 3t} \right)\vec j + {{\bf{e}}^t}\,\vec k\) tomonidan \(0 \le t \le 2\) uchun berilgan.
Yechim
Xo'sh, bu muammo uchun bizga muammo bayonotida egri chiziqning parametrizatsiyasi berilgan, shuning uchun biz bu muammo uchun tashvishlanishga hojat yo'q va biz chiziq integralini baholash uchun zarur bo'lgan ishga o'tishimiz mumkin. Bu shuni anglatadiki, 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).
\[\vec F\left( {\vec r\left( t \right)} \right) = {{\bf{e}}^{2{t^{\,3}}}}\,\vec i + {{\bf{e}}^t}\left( {1 - 3t + 1} \right)\vec j + {\left( {{{\bf{e}}^t}} \right)^3}\,\vec k = {{\bf{e}}^{2{t^{\,3}}}}\,\vec i + {{\bf{e}}^t}\left( {2 - 3t} \right)\vec j + {{\bf{e}}^{3t}}\,\vec k\]Parametrlashning hosilasi,
\[\vec r'\left( t \right) = 3{t^2}\,\vec i - 3\vec j + {{\bf{e}}^t}\,\vec k\]Nihoyat, vektor maydonining nuqta mahsuloti va parametrlashning hosilasi.
\[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) & = {{\bf{e}}^{2{t^{\,3}}}}\left( {3{t^2}} \right) + {{\bf{e}}^t}\left( {2 - 3t} \right)\left( { - 3} \right) + {{\bf{e}}^{3t}}\left( {{{\bf{e}}^t}} \right)\\ & = 3{t^2}{{\bf{e}}^{2{t^{\,3}}}} - 3{{\bf{e}}^t}\left( {2 - 3t} \right) + {{\bf{e}}^{4t}}\end{align*}\]Integralni bajarish uchun nuqta mahsulotini soddalashtirganingizga ishonch hosil qiling!
2-qadam
Endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash.
\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{0}^{2}{{3{t^2}{{\bf{e}}^{2{t^{\,3}}}} - 3{{\bf{e}}^t}\left( {2 - 3t} \right) + {{\bf{e}}^{4t}}\,dt}}\\ & = \left. {\left[ {\frac{1}{2}{{\bf{e}}^{2{t^{\,3}}}} - 3{{\bf{e}}^t}\left( {5 - 3t} \right) + \frac{1}{4}{{\bf{e}}^{4t}}\,dt} \right]} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{57}}{4} + 3{{\bf{e}}^2} + \frac{1}{4}{{\bf{e}}^8} + \frac{1}{2}{{\bf{e}}^{16}}}}\end{align*}\]E'tibor bering, integraldagi ikkinchi atama qismlar bo'yicha integratsiyani o'z ichiga oladi (buni qanday qilish kerakligini eslaysizmi?). Tasdiqlash va natijalarni biroz soddalashtirganimizni ta'kidlash uchun qismlar bo'yicha integratsiyani sizga qoldiramiz.
Bundan tashqari, bu erda "tartibsiz" javobdan hayajonlanmang! Bunday javoblarni vaqti-vaqti bilan olasiz.