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Consider an arbitrary vector field $F$

$$\eqalign{F&=F_1\hat{i}+F_2\hat{j}+F_3\hat{k}\\ &=F_{C_1}\hat{e}_\rho+F_{C_2}\hat{e}_{\phi}+F_{C_3}\hat{e}_{z}\\ &=F_{S_1}\hat{e}_r+F_{S_2}\hat{e}_{\theta}+F_{S_3}\hat{e}_{\phi}}$$

So the divergence of $F$ in cartesian,cylindical and spherical coordinates is:

$$\nabla \cdot F=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$$ $$=\frac{1}{\rho}\frac{\partial \left(\rho F_{ C_1}\right)}{\partial \rho}+\frac{1}{\rho}\frac{\partial F_{C_2}}{\partial \phi}+\frac{\partial F_{C_3}}{\partial z}$$ $$=\frac{1}{r^2}\frac{\partial\left( r^2 F_{ S_1}\right)}{\partial r}+\frac{1}{r \\sin {\theta}}\frac{\partial \left( \\sin {\theta} F_{S_2}\right)}{\partial \theta}+\frac{1}{r \\sin {\theta}}\frac{\partial F_{S_3}}{\partial \phi}$$

U\sin g the appropiate form find the divergence of the following vector fields. Indicate whether the vector field is compressible or not.

(i) $F(x,y,z)=(x^2+y\\sin {z})\hat{i}+(z^3+e^{3\\cos {z}-y}-y)\hat{j}+z(1-2x)\hat{k}$

(ii)$F(\rho,\phi,z)=\\cos {\left(\rho \phi\right)}\hat{e}_\rho+\\sin {\left(z \phi\right)}\hat{e}_\phi+e^z \hat{e}_z$

(iii)$F(r,\theta,\phi)=\\sin {\left(2\theta\right)}\hat{e}_r+\cot{\left(\theta\right)}\hat{e}_\theta+r^2\hat{e}_\phi$

(iv) Take $F$ to be an arbitrary position vector. Find the divergence in each of coordinate system.

Question ii

Vector field $$ F(ρ,φ,z)=\cos (ρ φ) eˆ ρ + \sin (z φ) eˆ φ + ez e$$ The divergence is given by:

$$ \nabla \cdot F(ρ,φ,z)=\left(\frac{1}{ρ}\frac{\partial}{\partial ρ}(ρF_C1),\frac{1}{ρ}\frac{\partial F_C2}{\partial φ},\frac{\partial F_3}{\partial z}\right)$$

$$\frac{1}{ρ}\frac{\partial}{\partial ρ}(ρF_C1)=-\frac{1}{ρ} \sin ρ$$ $$\frac{1}{ρ}\frac{\partial F_C2}{\partial φ}=\frac{1}{ρ}\cos φ$$ $$\frac{\partial F_3}{\partial z} =e^z $$


Qn1 c.

Vector field $$ F(r,θ,φ)=\sin (2θ) eˆ r+ cot(θ) e ˆ θ + r^2 e ˆφ$$ The divergence is given by:

$$ \nabla \cdot F(r,θ,φ)=\left(\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 F_S1),\frac{1}{r \sin (θ)}\frac{\partial }{\partial θ}(\sin (θ) F_S2),\frac{1}{r \sin (θ)} \frac{\partial F_S3}{\partial φ}\right)$$

$$\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 F_S1)=0$$ $$\frac{1}{r \sin (θ)}\frac{\partial }{\partial θ}(\sin (θ) F_S2)=-\frac{1}{r \sin (θ)}csc^2 θ$$ $$\frac{1}{r \sin (θ)} \frac{\partial F_S3}{\partial φ}=0 $$ So the divergence is: $$ \nabla \cdot F(r,θ,φ)=-\frac{1}{r \sin (θ)}csc^2 θ$$

Qn1 iv)

Given an arbitrary vector $$ F(u,v,w)$$

a)The divergence in cartesian is given by:

$$ \nabla \cdot F(u,v,w)=\left(\frac{\partial F_1}{\partial u},\frac{\partial F_2}{\partial v},\frac{\partial F_3}{\partial w}\right)$$

b) The divergence in cylindrical is given by:

$$ \nabla \cdot F(u,v,w)=\left(\frac{1}{ρ}\frac{\partial}{\partial u}(ρF_C1),\frac{1}{u}\frac{\partial F_C2}{\partial v},\frac{\partial F_3}{\partial w}\right)$$

c) The divergence in cylindrical is given by:

$$ \nabla \cdot F(u,v,w)=\left(\frac{1}{u^2}\frac{\partial}{\partial u}(u^2 F_S1),\frac{1}{u \sin (v)}\frac{\partial }{\partial v}(\sin (v) F_S2),\frac{1}{u \sin (v)} \frac{\partial F_S3}{\partial w}\right)$$

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  • $\begingroup$ Welcome to mathSE! First look in which coordinates are writed the force fields. Then apply the formula they give you $\endgroup$
    – rlartiga
    Commented May 22, 2014 at 21:01
  • $\begingroup$ Hi, thanks. Which formular? $\endgroup$
    – user152775
    Commented May 22, 2014 at 21:03
  • $\begingroup$ The formula of the divegence in cartesian (first), cilindrical (second) and spherical (third one) $\endgroup$
    – rlartiga
    Commented May 22, 2014 at 21:04
  • $\begingroup$ Hi, can you please show the first one, then I will take from there? $\endgroup$
    – user152775
    Commented May 22, 2014 at 21:14
  • $\begingroup$ Hi I'm back still need help? $\endgroup$
    – rlartiga
    Commented May 23, 2014 at 2:04

1 Answer 1

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If you have a vector field of the form $$ F(x,y,z)=(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z))$$ The divergence is given by:

$$ \nabla \cdot F(x,y,z)=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$$ Then for the first case: $$\frac{\partial F_1}{\partial x}=2x$$ $$\frac{\partial F_2}{\partial y}=-1$$ $$\frac{\partial F_3}{\partial z}=1-2x$$ So the divergence is: $$ \nabla \cdot F(x,y,z)=(2x) + (-1) + (1-2x)=0$$

How the divergence is $0$ the field is not compressible.

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  • $\begingroup$ Got it! Thanks a lot. Any idea on what "arbitrary position vector" means? is it the co-od (0,0)? The last bit, No. iv) $\endgroup$
    – user152775
    Commented May 22, 2014 at 21:25
  • $\begingroup$ Suppose $F(u,v,w)$ then how you could calculate the divergence? $\endgroup$
    – rlartiga
    Commented May 22, 2014 at 21:50
  • $\begingroup$ what is F(u,v,w)? Give me an example of an arbitrary position vector? $\endgroup$
    – user152775
    Commented May 23, 2014 at 1:59

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