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