Compensation factor for converting dy dx to cylindrical coordinates?

[WK95]

Homework Statement

What is the compensation factor for converting dy dx to cylindrical coordinates?

Homework Equations

None that I know of besides the bottom ones as part of the attempt

The Attempt at a Solution

So I know that the conversion formulas for going from Cartesian (x,y,z) coordinates to Cylindrical (r, θ, z) coordinates are as follows.
$x^2 + y^2 = r^2$
$x = r cos(θ)$
$y = r sin(θ)$
$z = z$
"When evaluating an integral, the integrating factors have a slightly different formula." - My professor regarding these formulas at the above question.

 

[STEMucator]

Don't forget your jacobian.

 

[WK95]

Don't forget your jacobian.

Since my class pretty much rests on the idea of giving us stuff we haven't really covered, I don't know how to use a Jacobian. Be right back, researching that.

 

[STEMucator]

Since my class pretty much rests on the idea of giving us stuff we haven't really covered, I don't know how to use a Jacobian. Be right back, researching that.

You accidentally posted another thread I see. What is it you don't understand? I've already told you what you were missing.

This should help you : http://en.wikipedia.org/wiki/Jacobian

 

[WK95]

You accidentally posted another thread I see. What is it you don't understand? I've already told you what you were missing.

This should help you : http://en.wikipedia.org/wiki/Jacobian

Thanks. I know how to find the Jacobian Determinant now, even the 3D ones. But I still don't know how to apply them to this problem to find a compensation factor.

 

[vanhees71]

What is a compensation factor? I've never heard this in connection with integrals.

You find the area element in arbitrary coordinates (q^1,q^2) by using the Jacobian determinant, because this gives the area of the surface elements spanned by the coordinate lines,
d^2 \vec{x}=\mathrm{det} \frac{\partial (x^1,x^2)}{\partial (q^1,q^2)} \mathrm{d}x^1 \mathrm{d} x^2,
where
\frac{\partial (x^1,x^2)}{\partial (q^1,q^2)} = \begin{pmatrix}<br /> \partial x^1/\partial q^1 &amp; \partial x^1/\partial q^2\\<br /> \partial x^2/\partial q^1 &amp; \partial x^2/\partial q^2<br /> \end{pmatrix}<br />
For polar coordinates you can even read it off from the geometry of the coordinate lines which are radial straight lines and circles around the origin. Obviously this gives immediately
\mathrm{d^2} \vec{x}=\mathrm{d}r \mathrm{d} \varphi r,
which you can easily check using the Jacobian.

 

[HallsofIvy]

One way to approach this problem is purely "geometric": Imagine two points with coordinates (rcos(\theta), r(sin(theta), z) and ((r+dr)cos(\theta+ d\theta), (r+ dr)sin(\theta+ d\theta), z+dz). Drawing the appropriate "coordinate lines" (lines from (0, 0, z) to (rcos(\theta), r(sin(theta), z), (rcos(\theta), r(sin(theta), z+dz), ((r+ dr)cos(\theta+d\theta), (r+dr)sin(theta+d\theta), z+dz), the cylinders at r and r+ dr, we have a curved figure that is almost a rectangular solid. The "height" is dz, the "width" is dr and the smaller "length" (the curved sides) is rd\theta (the larger is (r+ dr)d\theta). The difference between those two lengths won't matter in the limit so we can take as the volume the product of those: (dz)(dr)(r+ dr)d\theta= rdrd\theta dz+ (dr)^2d\theta dz. As those all go to 0, the term with dr^2 will go to 0 faster than the first term so in the limit we have rdrd\theta dz.

(Of course, since "cylindrical coordinates" is just "polar coordinates" with z added, the differential of volume is just the differential of area in polar coorrdinates, rdrd\theta with "dz" multiplied.)

 

[STEMucator]

I'm assuming what your professor is trying to show you is this :

$\int \int \int_R f(x,y,z) dxdydz = \int \int \int_{R'} f(r, \theta, z) |J| drdθdz$

This equivalence is actually based off a much simpler relationship. Try convincing yourself of this first ( and then after convincing yourself, you will be able to show a much more general case and then even extend it to higher dimensions yourself ). This method is a good way to understand this, but if you don't know about curves and such it might not be the best.

Suppose that $R$ and $R'$ are bounded, connected regions such that Green's Theorem holds. Then :

Area of a region R = $\int \int_R 1dxdy = \int \int_{R'} |J|dudv$ where |J| is the Jacobian.

How do we show this?

Suppose the boundary of $R$ is a smooth curve $C$ given by $P(t) = (x(t), y(t))$ for $t \in [a,b]$ such that :

$\int_C x(t) dy = \int \int_R 1dxdy$

Now, take the boundary of $R'$ to be another smooth curve $C'$ oriented so that the transformation $u = h(x,y), v = k(x,y)$
maps $C → C'$.

Then we have ( Note I'm using the chain rule here and the subscripts for the functions are the derivatives of that function with respect to that variable ) :

$\int_C xdy = \int_{a}^{b} x(t)y'(t) dt = \int_{a}^{b} f(u,v) \frac{∂}{∂t} (g(u,v)) dt = \int_{C'} f(u,v) g_u(u,v) du + f(u,v) g_v(u,v) dv$

Continuing ( I'll drop the (u,v)'s for the functions to make this easier to read ) :

$\int_{C'} fg_u du + fg_v dv = ± \int \int_{R'} \frac{∂}{∂u}(fg_v) - \frac{∂}{∂v} (fg_u) dudv = ± \int \int_{R'} f_ug_v - f_vg_u dudv = ± \int \int_{R'} J dudv = \int \int_{R'} |J| dudv$

Hence the result is shown. Now using this result, you can show how to compute the area for any general function in any dimension.

 

[WK95]

One way to approach this problem is purely "geometric": Imagine two points with coordinates (rcos(\theta), r(sin(theta), z) and ((r+dr)cos(\theta+ d\theta), (r+ dr)sin(\theta+ d\theta), z+dz). Drawing the appropriate "coordinate lines" (lines from (0, 0, z) to (rcos(\theta), r(sin(theta), z), (rcos(\theta), r(sin(theta), z+dz), ((r+ dr)cos(\theta+d\theta), (r+dr)sin(theta+d\theta), z+dz), the cylinders at r and r+ dr, we have a curved figure that is almost a rectangular solid. The "height" is dz, the "width" is dr and the smaller "length" (the curved sides) is rd\theta (the larger is (r+ dr)d\theta). The difference between those two lengths won't matter in the limit so we can take as the volume the product of those: (dz)(dr)(r+ dr)d\theta= rdrd\theta dz+ (dr)^2d\theta dz. As those all go to 0, the term with dr^2 will go to 0 faster than the first term so in the limit we have rdrd\theta dz.

(Of course, since "cylindrical coordinates" is just "polar coordinates" with z added, the differential of volume is just the differential of area in polar coorrdinates, rdrd\theta with "dz" multiplied.)

Thanks. I just looked through your post and it looks like I got the answer right.