Question

ques

Answer 1

Do you have a question for us to answer? @Nishant_Garg

Answer 2

How can I prove that
\[(dV)_{cylindrical}=\rho d \rho d \varphi dz\]
From the quadratic equation of differentials,
\[(ds)^2=h_{1}^2(du)^2+h_{2}^2(dv)^2+h_{3}^2(dw)^2\]
For cartesian we have
\[(ds)^2=(dx)^2+(dy)^2+(dz)^2\]
and
Thus I can intuitively tell that
\[dx=h_{1}du \space \space ; \space \space dy=h_{2}dv \space \space ; \space \space dz=h_{3}dw\]
\[dV=dxdydz\]
So we get
\[dV=h_{1}h_{2}h_{3}dudvdw\]
Therefore
\[(dV)_{cylindrical}=\rho d\rho d\varphi dz\]
But my intuition can be wrong, is this good enough?

Answer 3

@ganeshie8

Answer 4

Also check the following
\[\nabla^2\phi=\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial \phi}{\partial \rho})+\frac{1}{\rho^2}\frac{\partial^2 \phi}{\partial \varphi^2}+\frac{\partial^2\phi}{\partial z^2}\]
\[\nabla^2\phi=\frac{1}{\rho}(\frac{\partial \phi}{\partial \rho}+\rho\frac{\partial^2\phi}{\partial \rho^2})+\frac{1}{\rho^2}\frac{\partial^2 \phi}{\partial \varphi^2}+\frac{\partial^2 \phi}{\partial z^2}\]
\[\nabla^2\phi=\frac{1}{\rho}\frac{\partial \phi}{\partial \rho}+\frac{\partial^2\phi}{\partial \rho^2}+\frac{1}{\rho^2}\frac{\partial^2 \phi}{\partial \varphi^2}+\frac{\partial^2 \phi}{\partial z^2}\]
\[\nabla^2\phi=(\frac{1}{\rho}\frac{\partial }{\partial \rho}+\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho^2}\frac{\partial^2 }{\partial \varphi^2}+\frac{\partial^2}{\partial z^2})\phi\]
\[\implies\nabla^2 \equiv\frac{1}{\rho}\frac{\partial }{\partial \rho}+\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho^2}\frac{\partial^2 }{\partial \varphi^2}+\frac{\partial^2}{\partial z^2}\]
and
\[\nabla^2\phi=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial \phi}{\partial r})+\frac{1}{r^2\sin(\theta)}\frac{\partial}{\partial \theta}(\sin(\theta)\frac{\partial \phi}{\partial \theta})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2 \phi}{\partial \varphi^2}\]\[\nabla^2\phi=\frac{1}{r^2}(2r\frac{\partial \phi}{\partial r}+r^2\frac{\partial^2\phi}{\partial r^2})+\frac{1}{r^2\sin(\theta)}(\cos(\theta)\frac{\partial \phi}{\partial \theta}+\sin(\theta)\frac{\partial^2 \phi}{\partial \theta^2})+\]\[...+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2\phi}{\partial \varphi^2}\]
\[\nabla^2\phi=\frac{2}{r}\frac{\partial \phi}{\partial r}+\frac{\partial^2\phi}{\partial r^2}+\frac{1}{r^2}(\cot(\theta)\frac{\partial \phi}{\partial \theta}+\frac{\partial^2 \phi}{\partial \theta^2})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2 \phi}{\partial \varphi^2}\]
\[\nabla^2\phi=[\frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}(\cot(\theta)\frac{\partial}{\partial \theta}+\frac{\partial^2}{\partial \theta^2})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2}]\phi\]
\[\nabla^2\equiv\frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}(\cot(\theta)\frac{\partial}{\partial \theta}+\frac{\partial^2}{\partial \theta^2})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2}\]

Answer 5

for the first question, one would ordinarily use a simple geometric proof; or a Jacobian (linked) if you want to mechanically to go from rectangular to cylindrical
http://www.usciences.edu/~lvas/Calc3/Triple_substitution.pdf


i say this only because i personally see no intuition in connecting arc length:
ie \((ds)^2=(dx)^2+(dy)^2+(dz)^2\)
with volume

maybe there is one but i don't see it.

Answer 6

Then the question is why does
dxdydz becomes |J|dudvdw?