16.6-bo'lim : Konservativ vektor maydonlari
8. \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) ni baholang, bunda \(C\) aylananing koordinata markazida joylashgan, radiusi 2 bo‘lgan qismidir.st soat miliga teskari aylanish va \(\displaystyle \vec F\left( {x,y} \right) = \left( {2xy - 4 - \frac{1}{2}\sin \left( {\frac{1}{2}x} \right)\sin \left( {\frac{1}{2}y} \right)} \right)\,\vec i + \left( {{x^2} + \frac{1}{2}\cos \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right)} \right)\vec j\) bilan kvadrant.
Yechim
Ushbu muammoni hal qilishning ikki yo'li mavjud: qiyin va oson. Qiyin yo'l - egri chiziqli chiziqli integralni va berilgan vektor maydonini ko'rish va to'g'ridan-to'g'ri chiziqli integralni hisoblashni boshlash (ehtimol, bu holatda juda qiyin). Eng oson yo'li - vektor maydonining konservativ yoki yo'qligini tekshirish va u potentsial funktsiyani topish va oldingi bo'limda ko'rgan chiziqli integrallar uchun asosiy teoremadan foydalanishdir.
Shunday qilib, keling, oson yo'ldan boramiz va vektor maydoni konservativ yoki yo'qligini tekshiramiz.
\[\begin{align*}P & = 2xy - 4 - \frac{1}{2}\sin \left( {\frac{1}{2}x} \right)\sin \left( {\frac{1}{2}y} \right) & \hspace{0.25in}{P_y} & = 2x - \frac{1}{4}\sin \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right)\\ Q & = {x^2} + \frac{1}{2}\cos \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right) & \hspace{0.25in}{Q_x} & = 2x - \frac{1}{4}\sin \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right)\end{align*}\]Shunday qilib, biz \({P_y} = {Q_x}\) va shuning uchun vektor maydoni konservativ ekanligini ko'rishimiz mumkin.
2-qadam
Endi biz faqat vektor maydoni uchun potentsial funktsiyani topishimiz kerak. Bu safar biz bu tafsilotlarni biroz tezroq va oldingi muammolarga qaraganda kamroq tushuntirish bilan ko'rib chiqamiz.
Birinchidan, \(P\)-ni \(x\)-ga integratsiya qilaylik.
\[\begin{align*}f\left( {x,y} \right) & = \int{{P\,dx}}\\ & = \int{{2xy - 4 - \frac{1}{2}\sin \left( {\frac{1}{2}x} \right)\sin \left( {\frac{1}{2}y} \right)\,dx}}\\ & = {x^2}y - 4x + \cos \left( {\frac{1}{2}x} \right)\sin \left( {\frac{1}{2}y} \right) + h\left( y \right)\end{align*}\]Endi \(y\) ga nisbatan farqlang va \(Q\) ga tenglashtiring.
\[{f_y} = {x^2} + \frac{1}{2}\cos \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right) + h'\left( y \right) = {x^2} + \frac{1}{2}\cos \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right) = Q\hspace{0.25in} \Rightarrow \hspace{0.25in} h'\left( y \right) = 0\]\(h\left( y \right)\) ni hal qilish \(h\left( y \right) = c\) ni beradi va shuning uchun bu vektor maydoni uchun potentsial funktsiya:
\[f\left( {x,y} \right) = {x^2}y - 4x + \cos \left( {\frac{1}{2}x} \right)\sin \left( {\frac{1}{2}y} \right) + c\]3-qadam
Endi biz potentsial funktsiyaga egamiz, biz oddiygina chiziqli integrallar uchun asosiy teoremadan foydalanishimiz mumkin:
\[ \int\limits_{C}{{\vec F\centerdot d\vec r}} = f\left( {{\mbox{end point}}} \right) - f\left( {{\mbox{start point}}} \right)\]Muammo bayonotidan bilamizki, \(C\) 1 dagi 2 radiusli doiraning qismidir.st soat miliga teskari aylanish bilan kvadrant. Shuning uchun \(C\) ning boshlang'ich nuqtasi \(\left( {2,0} \right)\) va \(C\) ning tugash nuqtasi \(\left( {0,2} \right)\) hisoblanadi.
U holda integral,
\[\int\limits_{C}{{\vec F\centerdot d\vec r}} = f\left( {0,2} \right) - f\left( {2,0} \right) = \left( {\sin \left( 1 \right) + c} \right) - \left( { - 8 + c} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\sin \left( 1 \right) + 8 = 8.8415}}\]