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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    Do you have a question for us to answer? @Nishant_Garg

  2. anonymous
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    @ganeshie8

  4. anonymous
    • one year ago
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    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 \[\nabla^2\phi=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial \phi}{\partial r})+\frac{1}{r^2\sin(\theta)}\frac{\partial}{\partial \theta}(\sin(\theta)\frac{\partial \phi}{\partial \theta})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2 \phi}{\partial \varphi^2}\]\[\nabla^2\phi=\frac{1}{r^2}(2r\frac{\partial \phi}{\partial r}+r^2\frac{\partial^2\phi}{\partial r^2})+\frac{1}{r^2\sin(\theta)}(\cos(\theta)\frac{\partial \phi}{\partial \theta}+\sin(\theta)\frac{\partial^2 \phi}{\partial \theta^2})+\]\[...+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2\phi}{\partial \varphi^2}\] \[\nabla^2\phi=\frac{2}{r}\frac{\partial \phi}{\partial r}+\frac{\partial^2\phi}{\partial r^2}+\frac{1}{r^2}(\cot(\theta)\frac{\partial \phi}{\partial \theta}+\frac{\partial^2 \phi}{\partial \theta^2})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2 \phi}{\partial \varphi^2}\] \[\nabla^2\phi=[\frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}(\cot(\theta)\frac{\partial}{\partial \theta}+\frac{\partial^2}{\partial \theta^2})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2}]\phi\] \[\nabla^2\equiv\frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}(\cot(\theta)\frac{\partial}{\partial \theta}+\frac{\partial^2}{\partial \theta^2})+\frac{1}{r^2\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2}\]

  5. IrishBoy123
    • one year ago
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    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.

  6. anonymous
    • one year ago
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    Then the question is why does dxdydz becomes |J|dudvdw?

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