# Cartesian to polar cordinates

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)?

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
$x$ and $y$ may depend on $r$ and $\theta.$ Shouldn't it be partial derivative?

PeroK
Homework Helper
Gold Member
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?

Last edited by a moderator:
Zondrina
Homework Helper
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).

Last edited by a moderator:
• ShayanJ and RUber
$x$ 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.

Zondrina
Homework Helper
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 $T$.

• RUber and ShayanJ