I Cylindrical coordinates -Curvilinear

chwala
Gold Member
Messages
2,825
Reaction score
413
TL;DR Summary
Why are the coordinates seemingly used when the symmetry is around ##z## axis? Any particular reason why not ##x## or ##y##. In transforming from Cartesian to cylindrical form; I can see that ##z## is not considered when determining ##r##.
Can we also use ##x## and ##z## assuming that the symmetry is about ##y##? Sorry using phone to type ...will put this into context later. I hope my query is clear enough.
Why are the coordinates seemingly used when the symmetry is around ##z## axis? Any particular reason why not ##x## or ##y##. In transforming from Cartesian to cylindrical form; I can see that ##z## is not considered when determining ##r##.
Can we also use ##x## and ##z## assuming that the symmetry is about ##y##? Sorry using phone to type ...will put this into context later. I hope my query is clear enough.
 
Physics news on Phys.org
The directions of the cartesian axes in space are not absolute, but can be chosen to fit the geometry of the particular problem; by convention in axisymmetric geometries the z-axis is placed on the axis of symmetry, giving an obvious extension of plane polar coordinates (x,y) = (r \cos \theta, r \sin \theta).

It is clearly possible to set \begin{split}<br /> x &amp;= w \\<br /> y &amp;= u \cos v \\<br /> z &amp;= u \sin v\end{split}<br /> or \begin{split}<br /> x &amp;= u \sin v \\<br /> y &amp;= w \\<br /> z &amp;= u \cos v \end{split} if you want.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top