I have seen a picture for $dV$ so that $dV = r^{2} \sin(\theta)\,dr\,d\theta\,d\phi$. But how can I deduce things like $dA$ and $dV$? In a simpler coordinate (not sure about the name), $dA = r \,dr\,d\phi$, again no idea how to derive it (just given to me to memorize it). So explain what is $dA$ in polar coordinates?
dA in polar coordinates?
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polar-coordinates
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0I do not know the rigorous version of it, but since this comes from a physics background, this is mostly done by visualizing. dV is then the infinitesimal change in volume, which can be visualized as the volume of an element bound by $dr$ an arc $r\sin\theta d\theta$ and another $rd\phi$. For dA it depends which surface you're taking, $dA=dr (rsin\theta d\phi)$ or $dA=dr (r d\theta)$ or $dA=(rd\theta) (r\sin\theta d\phi)$ if you keep $\theta$, $r$ fixed or $\phi$ fixed. – 2011-03-08
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1The area element $dA = r dr d\phi$ is from polar coordinates. The boundaries are the radial $dr$ and the circumferential $rd\phi$ – 2011-03-08
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0@Approximist: there is an err. I think it must be $r \sin(\theta) d\theta$, $dr$ and $r d\phi$. N.b. the term $r^{2}$. The last part about $dA$ got me confused. Could you explain it with general surfaces? – 2011-03-08
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0just a note, if you write $dA=r dr d\phi$ then you mean $\phi$ to be the polar angle. In the formula for dV you wrote $\phi$ would be the azimuthal angle. (@user error corrected) – 2011-03-08
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0There is no general expression for surface elements $dA$ (atleast, one like $dV$), since these depend on the orientation of the surface. You simply have to analyze the geometry for any given case If you are integrating over the surface of a sphere, for instance, then r is constant, whereas $\theta$ and $\phi$ change. That is the last expression for $dA$ I wrote in previous comment. On the other hand, if the surface lies in the xy plane, say, so that $\theta$ is constant ($\pi /2$) while $r$ and $\phi$ vary, then dA is the first of the dA's I wrote in comment. – 2011-03-08
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0@Approximist: I don't know but I feel that rcollyer is on the right track. But you have very good points like *"For dA it depends which surface you're taking, dA=dr(rsinθdϕ) or dA=dr(rdθ) or dA=(rdθ)(rsinθdϕ) if you keep θ, r fixed or ϕ fixed."* which I am still unsure, could you write them to an answer? – 2011-03-08
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0@user there is no general expression, but yes, a general sytematic method – 2011-03-08
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1@Approximist: technically there are some very nice general expressions for the element of surface area $dA$ for an arbitrarily parametrized surface (or even a $k$-manifold in Euclidean space). The element of $k$-dimensional volume looks like $\sqrt{Det(J^tJ)}dx_1\cdots dx_k$ where $J$ is the Jacobian matrix (matrix of partial derivatives) of the function that parametrizes the manifold. $J^t$ is the transpose of $J$. – 2011-03-08
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0@Ryan yes I realized that a bit later :) thank you. – 2011-03-08