17.1-bo'lim : jingalak va farqlanish
1. \(\vec F = {x^2}y\,\vec i - \left( {{z^3} - 3x} \right)\vec j + 4{y^2}\vec k\) uchun \({\mathop{\rm div}\nolimits} \vec F\) va \({\mathop{\rm curl}\nolimits} \vec F\) ni hisoblang.
Yechim
Keling, avval farqni hisoblab chiqamiz va formulani ko'rib chiqishdan boshqa qiladigan ko'p narsa yo'q.
\[{\mathop{\rm div}\nolimits} \vec F = \nabla \centerdot \vec F = \frac{\partial }{{\partial x}}\left( {{x^2}y} \right) + \frac{\partial }{{\partial y}}\left( {3x - {z^3}} \right) + \frac{\partial }{{\partial z}}\left( {4{y^2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{2xy}}\]Ehtiyot bo'ling, har qanday vektor komponentlari (2nd bu holda komponent!). Shoshilish va ularni sog'inish oson.
2-qadam
Jingalak biroz ko'proq ishlaydi, lekin baribir faqat formula ishlaydi, shuning uchun bu erda jingalak.
\[\begin{align*}{\mathop{\rm curl}\nolimits} \vec F & = \nabla \times \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\displaystyle \frac{\partial }{{\partial x}}}&{\displaystyle \frac{\partial }{{\partial y}}}&{\displaystyle \frac{\partial }{{\partial z}}}\\{{x^2}y}&{3x - {z^3}}&{4{y^2}}\end{array}} \right|\\ & = \frac{\partial }{{\partial y}}\left( {4{y^2}} \right)\vec i + \frac{\partial }{{\partial z}}\left( {{x^2}y} \right)\vec j + \frac{\partial }{{\partial x}}\left( {3x - {z^3}} \right)\vec k - \frac{\partial }{{\partial y}}\left( {{x^2}y} \right)\vec k - \frac{\partial }{{\partial x}}\left( {4{y^2}} \right)\vec j - \frac{\partial }{{\partial z}}\left( {3x - {z^3}} \right)\vec i\\ & = 8y\vec i + 3\vec k - {x^2}\vec k + 3{z^2}\vec i\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\left( {8y + 3{z^2}} \right)\vec i + \left( {3 - {x^2}} \right)\vec k}}\end{align*}\]Shunga qaramay, 2-dagi minus belgisini unutmangnd komponent.