Find divergence and Curl of f = (x2 + yz) i + (y2 + zx)j +
(z2 + xy) k.​

1 thought on “Find divergence and Curl of f = (x2 + yz) i + (y2 + zx)j +<br />(z2 + xy) k.​”

1. [tex]\textbf{Given:}[/tex]

   [tex]\mathsf{f=(x^2+yz)i+(y^2+zx)j+(z^2+xy)k}[/tex]

   [tex]\textbf{To find:}[/tex]

   [tex]\textsf{Divergence and curl of f}[/tex]

   [tex]\textbf{Solution:}[/tex]

   [tex]\textsf{Consider,}[/tex]

   [tex]\mathsf{f=(x^2+yz)i+(y^2+zx)j+(z^2+xy)k}[/tex]

   [tex]\underline{\mathsf{Divergence\;of\;f:}}[/tex]

   [tex]\mathsf{=\dfrac{\partial\,f_1}{\partial\,x}+\dfrac{\partial\,f_2}{\partial\,y}+\dfrac{\partial\,f_3}{\partial\,z}}[/tex]

   [tex]\mathsf{=\dfrac{\partial(x^2+yz)}{\partial\,x}+\dfrac{\partial(y^2+zx)}{\partial\,y}+\dfrac{\partial(z^2+xy)}{\partial\,z}}[/tex]

   [tex]\mathsf{=2x+2y+2z}[/tex]

   [tex]\implies\boxed{\mathsf{Divergence\;of\;f=2(x+y+z)}}[/tex]

   [tex]\underline{\mathsf{Curl\;f:}}[/tex]

   [tex]=\left|\begin{array}{ccc}i&j&k\\\dfrac{\partial}{\partial\,x}&\dfrac{\partial}{\partial\,y}&\dfrac{\partial}{\partial\,z}\\f_1&f_2&f_3\end{array}\right|[/tex]

   [tex]\mathsf{=\left(\dfrac{\partial\,f_3}{\partial\,y}-\dfrac{\partial\,f_2}{\partial\,z}\right)i-\left(\dfrac{\partial\,f_3}{\partial\,x}-\dfrac{\partial\,f_1}{\partial\,z}\right)j+\left(\dfrac{\partial\,f_2}{\partial\,x}-\dfrac{\partial\,f_1}{\partial\,y}\right)k}[/tex]

   [tex]\mathsf{=\left(\dfrac{\partial(z^2+xy)}{\partial\,y}-\dfrac{\partial(y^2+zx)}{\partial\,z}\right)i-\left(\dfrac{\partial(z^2+xy)}{\partial\,x}-\dfrac{\partial(x^2+yz)}{\partial\,z}\right)j+\left(\dfrac{\partial(y^2+zx)}{\partial\,x}-\dfrac{\partial(x^2+yz)}{\partial\,y}\right)k}[/tex]

   [tex]\mathsf{=(x-x)i+(y-y)j+(z-z)k}[/tex]

   [tex]\mathsf{=(0)i+(0)j+(0)k}[/tex]

   [tex]\implies\boxed{\mathsf{Curl\,f=\overrightarrow{0}}}[/tex]

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