# Polar Integration

1. Oct 8, 2012

### ruzfactor

Hi

I have a function [e.g. f(r)] which I want to integrate over r and θ. What would be the integration form? Which one is correct?

∫∫f(r) drdθ OR ∫∫f(r) rdrdθ

Please explain. Also, can it be said as area integration as well like the one in cartesian coordinate?

2. Oct 8, 2012

### haruspex

it depends what you mean by integrating over r and theta. Taken at face value, that just means drdθ. But if you mean that you want to integrate over an area using polar coordinates then it's rdrdθ. The reason is that you can carve up an area into small annular sectors, each running over a range of r to r+dr and θ to θ+dθ. The dimensions of such an area element are dr by rdθ, and approximate a rectangle, so the area of the element is rdrdθ.

3. Oct 8, 2012

### ruzfactor

Thanks. I have a function of r. For example at theta=0, the value of the function have different value at different r values (e.g. r=0 to r=a). I want to evaluate this function over 2∏. So I thought first integrating the function about r and then theta. That is where I am confused, whether to use ∫∫f(r) drdθ or ∫∫f(r) rdrdθ. Please see the attachment where my problem is explained in a pic.

Also, I thought conversion from cartesian coordinate (dxdy) using jacobian makes it rdrdθ in polar coordinate.

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4. Oct 9, 2012

### haruspex

Clearly you are trying to integrate over an area, so rdrdθ is the correct form.
The Jacobian arises from coordinate transformations where the two coordinate systems describe the same region. In Cartesian, an element of area is dxdy; in polar it is rdrdθ.

5. Oct 9, 2012

### zmp3

In any double integration problem, you can usually start with either. In this situation, you'd be better off starting with the integration of the radius first then integrating with respect to theta.

6. Oct 10, 2012

### haruspex

The OP was not trying to choose between drdθ and dθdr, but between drdθ and rdrdθ.

7. Oct 10, 2012

### HallsofIvy

If your problem is just "given $f(r,\theta)$, integrate it", then you would have $\int\int f(r, \theta)drd\theta$. If, however, the problem is "given $f(r,\theta)$, integrate over a given area in the plane", then you would have $\int\int f(r,\theta) r drd\theta$ because "$rdrd\theta$" is the "differential of area" in polar coordinates. In particular, if you are converting $\int\int f(x,y) dxdy$ to polar coordinates, because dxdy is the "differential of area" in Cartesian coordinates, it would become $\int\int f(r,\theta)r drd\theta$.

8. Oct 10, 2012

### ruzfactor

Thanks. Actually the problem says that, integrate f(r,θ) in radial and circumferential direction. So it is a bit confusing. I guess rdrdθ could be used depending on my problem.

Here, the function f(r,θ) is the mode shape function of a circular plate.

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