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
$$\frac{\partial (x^1,x^2)}{\partial (q^1,q^2)} = \begin{pmatrix} \partial x^1/\partial q^1 & \partial x^1/\partial q^2\\ \partial x^2/\partial q^1 & \partial x^2/\partial q^2 \end{pmatrix}$$
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.