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Cartesian to polar cordinates

  • #1
If we expres cartesian cordinates in polar coordinates we get:

x=r*cos(theta)
y=r*sin(theta)

let's differentiate those 2 eqs:

dx= dr cos(theta) -r* d(theta) * sin(theta)
dy=dr sin(theta) + r* d(theta) * cos(theta)

why isn't dx*dy= r* dr* d(theta) ( like when taking the jacobian , or when doing geometric interpretation)?
 
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Answers and Replies

  • #2
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##x## and ##y## may depend on ##r## and ##\theta.## Shouldn't it be partial derivative?
 
  • #3
PeroK
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If we expres cartesian cordinates in polar coordinates we get:

x=r*cos(theta)
y=r*sin(theta)

let's differentiate those 2 eqs:

dx= dr cos(theta) -r* d(theta) * sin(theta)
dy=dr sin(theta) + r* d(theta) * cos(theta)

why isn't dx*dy= r* dr* d(theta) ( like when taking the jacobian , or when doing geometric interpretation)?
When you say "differentiate", what are you differentiating with respect to?
 
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  • #4
Zondrina
Homework Helper
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If we expres cartesian cordinates in polar coordinates we get:

x=r*cos(theta)
y=r*sin(theta)

let's differentiate those 2 eqs:

dx= dr cos(theta) -r* d(theta) * sin(theta)
dy=dr sin(theta) + r* d(theta) * cos(theta)

why isn't dx*dy= r* dr* d(theta) ( like when taking the jacobian , or when doing geometric interpretation)?
I assume you are speaking in the context of integration. When transforming to polar co-ordinates, it can be shown:

$$\iint_D f(x,y) \space dA = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space dA'$$

Where ##dA' = J_{r , \theta} (x, y) \space dr d \theta##. We have to multiply the function by the Jacobian ##J## whenever an invertible transformation is used to transform an integral. It turns out:

$$J_{r , \theta} (x, y) = x_r y_{\theta} - x_{\theta} y_r = r$$

You can work this result out yourself by taking the partials of the transformation. Hence we can write:

$$\iint_D f(x,y) \space dA = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space dA' = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space J_{r , \theta} (x, y) \space dr d \theta = \iint_{D'} f(r \cos(\theta), r \sin(\theta)) \space r \space dr d \theta$$

Where the order of integration is still to be decided. It is very important to multiply by the volume element ##J## because you want to preserve the result of the integral (the transformation is one to one and onto).
 
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  • #5
##x## and ##y## may depend on ##r## and ##\theta.## Shouldn't it be partial derivative?
Imagine a third variable t, and we know x,y, r, theta are functions of t. So when you write the total derivative of x to respect to t we get:

dx/dt = d(partial)x/d(partial)r * dr/dt + d(partial)x/d(partial)(theta) * d(theta)/dt, and if you multiply by dt this equation , you get the above one.
 
  • #6
Zondrina
Homework Helper
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I'm not sure what a total derivative has to do with this. Why don't we see how a change of variables is going to affect a double integral?

Suppose there is a rectangle ##R## in the arbitrary ##uv##-plane. Suppose further the lower left corner of the rectangle is at the point ##(u_0, v_0)##, and the dimensions of the rectangle are ##\Delta u## for the width and ##\Delta v## for the height respectively.

Now lets define an invertible transformation ##T: (u, v) \rightarrow (x, y)## such that the rectangle ##R## in the ##uv##-plane can be mapped to a region ##R'## in the ##xy##-plane. This transformation from the ##uv##-plane to the ##xy##-plane will be given by some ##x = g(u, v)## and ##y = h(u, v)##. For example, the lower left corner of ##R## can be mapped to the boundary of ##R'## by using ##T## like so:

$$T(u_0, v_0) = (x_0, y_0)$$
$$x_0 = g(u_0, v_0), \space y_0 = h(u_0, v_0)$$

Now, define a vector function ##\vec r(u, v)## to be the position vector of the image of the point ##(u, v)##:

$$\vec r(u, v) = g(u, v) \hat i + h(u, v) \hat j$$

Note the equation for the bottom side of the rectangle ##R## in the ##uv##-plane is given by ##v = v_0##, and the equation for the left side of the rectangle ##R## in the ##uv##-plane is given by ##u = u_0##. The image of the bottom side of ##R## in the ##xy##-plane is given by ##\vec r(u , v_0)##, and the image of the left side of ##R## in the ##xy##-plane is given by ##\vec r(u_0 , v)##.

The tangent vector at ##(x_0, y_0)## to the image ##\vec r(u , v_0)## is given by:

$$\vec r_u(u, v) = g_u(u_0, v_0) \hat i + h_u(u_0, v_0) \hat j = x_u \hat i + y_u \hat j$$

Similarly, the tangent vector at ##(x_0, y_0)## to the image ##\vec r(u_0 , v)## is given by:

$$\vec r_v(u, v) = g_v(u_0, v_0) \hat i + h_v(u_0, v_0) \hat j = x_v \hat i + y_v \hat j$$

We can approximate the region ##R'## in the ##xy##-plane by a parallelogram determined by the secant vectors:

$$\vec a = \vec r(u_0 + \Delta u, v_0) - \vec r(u_0, v_0) ≈ \Delta u \vec r_u$$
$$\vec b = \vec r(u_0, v_0 + \Delta v) - \vec r(u_0, v_0) ≈ \Delta v \vec r_v$$

So to determine the area of ##R'##, we must determine the area of the parallelogram formed by the secant vectors. So we compute:

$$\Delta A_{R'} ≈ \left| (\Delta u \vec r_u) \times (\Delta v \vec r_v) \right| = \left| \vec r_u \times \vec r_v \right| (\Delta u \Delta v)$$

Where we have pulled out ##\Delta u \Delta v## because it is constant. Computing the magnitude of the cross product we obtain:

$$\left| \vec r_u \times \vec r_v \right| = x_u y_v - x_v y_u$$

So we may write:

$$\Delta A_{R'} ≈ [x_u y_v - x_v y_u] (\Delta u \Delta v)$$

Where ##x_u y_v - x_v y_u## can be determined by evaluating ##g_u(u_0, v_0), h_v(u_0, v_0), g_v(u_0, v_0)##, and ##h_u(u_0, v_0)## respectively.

Now that we have formalized all of that, divide the region ##R## in the ##uv##-plane into infinitesimally small rectangles ##R_{ij}##. The images of the ##R_{ij}## in the ##xy##-plane are represented by the rectangles ##R_{ij}'##. Applying the approximation ##\Delta A_{R'}## to each ##R_{ij}'##, we can approximate the double integral of a function ##f## over ##R'## like so:

$$\iint_{R'} f(x, y) dA ≈ \sum_{i = 1}^m \sum_{j = 1}^n f(x_i, y_j) \Delta A_{R'} ≈ \sum_{i = 1}^m \sum_{j = 1}^n f(g(u_i, v_j), h(u_i, v_j)) [x_u y_v - x_v y_u] (\Delta u \Delta v)$$

Notice this looks like a typical Riemann sum. Now as ##m \to \infty## and ##n \to \infty##, the double sum converges to a double integral over ##R##:

$$\displaystyle \lim_{m \to \infty} \displaystyle \lim_{n \to \infty} \sum_{i = 1}^m \sum_{j = 1}^n f(g(u_i, v_j), h(u_i, v_j)) [x_u y_v - x_v y_u] (\Delta u \Delta v) = \iint_R f(g(u, v), h(u, v)) [x_u y_v - x_v y_u] \space dudv$$

Where we usually write the Jacobian ##J = [x_u y_v - x_v y_u]##. So we can finally conclude:

$$\iint_{R'} f(x,y) \space dA = \iint_{R} f(g(u, v), h(u, v)) \space J \space dudv$$

This argument for an arbitrary ##(u, v)## space applies to polar ##(r, \theta)## space as well. In fact, this argument will apply for any kind of other invertible transformation ##T##.
 

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