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Question Number 27 by user1 last updated on 25/Jan/15

If f=x^2 zi−2y^3 z^2 j+xy^2 zk. Find div f, curl f, at(1, −1, 1).

$$\mathrm{If}\:\boldsymbol{\mathrm{f}}={x}^{\mathrm{2}} {z}\boldsymbol{\mathrm{i}}−\mathrm{2}{y}^{\mathrm{3}} {z}^{\mathrm{2}} \boldsymbol{\mathrm{j}}+{xy}^{\mathrm{2}} {z}\boldsymbol{\mathrm{k}}.\:\mathrm{Find}\:{div}\:\boldsymbol{\mathrm{f}},\:{curl}\:\boldsymbol{\mathrm{f}},\: \\ $$$${at}\left(\mathrm{1},\:−\mathrm{1},\:\mathrm{1}\right). \\ $$

Answered by user1 last updated on 03/Nov/14

div f=Σi∙(∂f/∂x)=▽∙f =(i(∂/∂x)+j(∂/∂y)+k(∂/∂z))∙(x^2 zi−2y^3 z^2 j+xy^2 zk) =(∂/∂x)(x^2 z)−(∂/∂y)(2y^3 z^2 )+(∂/∂z)(xy^2 z) =2xz−6y^2 z^2 +xy^2 = −3 at (1, −1, 1) curl f=▽×f= determinant ((( i),( j),( k)),((∂/∂x),( (∂/∂y)),( (∂/∂z))),((x^2 z),(−2y^3 z^2 ),(xy^2 z))) =i(2xyz+4y^3 z)+j(x^2 −y^2 z)+k(0−0) = −6i at (1, −1, 1)

$$\:\mathrm{div}\:\boldsymbol{\mathrm{f}}=\Sigma\boldsymbol{\mathrm{i}}\centerdot\frac{\partial\boldsymbol{\mathrm{f}}}{\partial{x}}=\bigtriangledown\centerdot\boldsymbol{\mathrm{f}} \\ $$$$\:=\left(\boldsymbol{\mathrm{i}}\frac{\partial}{\partial{x}}+\boldsymbol{\mathrm{j}}\frac{\partial}{\partial{y}}+\boldsymbol{\mathrm{k}}\frac{\partial}{\partial{z}}\right)\centerdot\left({x}^{\mathrm{2}} {z}\boldsymbol{\mathrm{i}}−\mathrm{2}{y}^{\mathrm{3}} {z}^{\mathrm{2}} \boldsymbol{\mathrm{j}}+{xy}^{\mathrm{2}} {z}\boldsymbol{\mathrm{k}}\right) \\ $$$$\:=\frac{\partial}{\partial{x}}\left({x}^{\mathrm{2}} {z}\right)−\frac{\partial}{\partial{y}}\left(\mathrm{2}{y}^{\mathrm{3}} {z}^{\mathrm{2}} \right)+\frac{\partial}{\partial{z}}\left({xy}^{\mathrm{2}} {z}\right) \\ $$$$\:=\mathrm{2}{xz}−\mathrm{6}{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{xy}^{\mathrm{2}} \\ $$$$\:=\:−\mathrm{3}\:\:\mathrm{at}\:\left(\mathrm{1},\:−\mathrm{1},\:\mathrm{1}\right)\: \\ $$$$\:\mathrm{curl}\:\boldsymbol{\mathrm{f}}=\bigtriangledown×\boldsymbol{\mathrm{f}}=\begin{vmatrix}{\:\:\:\boldsymbol{\mathrm{i}}}&{\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{j}}}&{\:\:\:\boldsymbol{\mathrm{k}}}\\{\frac{\partial}{\partial{x}}}&{\:\:\:\:\:\frac{\partial}{\partial{y}}}&{\:\:\frac{\partial}{\partial{z}}}\\{{x}^{\mathrm{2}} {z}}&{−\mathrm{2}{y}^{\mathrm{3}} {z}^{\mathrm{2}} }&{{xy}^{\mathrm{2}} {z}}\end{vmatrix} \\ $$$$\:=\boldsymbol{\mathrm{i}}\left(\mathrm{2}{xyz}+\mathrm{4}{y}^{\mathrm{3}} {z}\right)+\boldsymbol{\mathrm{j}}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} {z}\right)+\boldsymbol{\mathrm{k}}\left(\mathrm{0}−\mathrm{0}\right) \\ $$$$\:=\:\:−\mathrm{6}\boldsymbol{\mathrm{i}}\:\:\mathrm{at}\:\:\left(\mathrm{1},\:−\mathrm{1},\:\mathrm{1}\right) \\ $$

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