Question

Problem set 7, Qn. 3B-1: Why is it dA taken as drdθ and not r.drdθ? Thanks.

Answer 1

The scaling factor "r" in some polar integrals results from the change of variables f(x,y) to f(r,θ) where dA=dx dx in the x-y domain maps to dA=r dr dθ in the r-θ domain. Replay the Lecture on Jacobians. The integrals in 3B-1 originate in the r,θ domain so you only integrate over dr dθ. in fact, should you change variables for these problems into the x-y domain, you would need to use dA = (1/(SQRT(X^2+Y^2)^0.5)) dx dy to replace dA = dr dθ.

Answer 2

Thanks. So, is it correct that when we do an integration of the form Integral of (1.dr dθ), we are not getting the volume below the surface f(r,θ)=1, as dr.dθ doesn't represent an infinitesimal area ?

Answer 3

Guess what? I'm only on the next Unit after Problem Set 7.

To answer your latest post, you do get the "Volume with thickness 1" (commonly referred to as Area)" since "dr dθ" DOES represent an infinitesimal area (in the transformed domain). Your new bounds of integration are in terms of r,θ rather than x,y. Even though the region of integration does not change, those bounds ARE different because the coordinate systems are different, and it is the Jacobean scaling factor that applies the required multiple so that the solutions in both domains are identical. The Jacobean calculation from Cartesian to Polar yields "r".

I find the calculation of these new bounds of integration to be the most difficult of the 3 steps in changing variables.

But to return to your original post, your 3B-1 functions and regions can both be expressed in r,θ so you need only calculate the r,θ bounds of integration, then use dA=dr dθ. No conversions necessary.

Why make work. :-)

Answer 4

Thanks again. But, consider a circle of radius 3. If we take integral (1.dr dθ), with bounds r: 0 to 3 and θ: 0 to 2pi, we get the answer as 6pi instead of 9pi. Doesn't that mean we need to take dA as r dr dθ to get the area? ( Integration with r dr dθ does give the correct answer.

Answer 5

OK, I see where you are having a problem. When you integrate a circle in polar coordinates, the function you are integrating is f(r)=r, not 1. In xy, you would use \[\int\limits\limits_{}^{} \int\limits\limits_{R}^{}\sqrt{x^{2}+y ^{2}}dx dy \]Likewise, in rθ you set it up as \[\int\limits_{}^{} \int\limits\limits_{R}^{} f(r)dr d \theta = \int\limits\limits_{0}^{2\pi} \int\limits\limits_{0}^{3} (r) dr d \theta \]

Here, f(r,θ)=r and dA= dr dθ. "r" is the function f(r,θ), NOT a factor in dA.

Answer 6

Thanks again. I realised that I need to learn more on polar coordinates; hence tried reading up some more on this. But, unfortunately I still don't get your explanation. Can you suggest some book, from which I can learn more? I use 'Calculus: Early Transcendentals' by Stewart. It is a very good book; but I couldn't find more on how to do double integrals with polar coordinates (directly, not changing from xy coordinates)

Answer 7

I cannot recommend a good book since I am only using the course material, and my original courses were completed in the 1960's. Not that calculus has changed a lot, but the books certainly have.

I can only restate that when you set up an integral with the same differential elements as the function variables themselves, i.e. f(r,θ) dr dθ, no scale factor (r) is required. But when you CHANGE VARIABLES, you have to understand that "dx dy" IS NOT EQUAL TO "dr dθ" so you CANNOT substitute "dr dθ in the original f(x,y) integral without including that scaling factor. Even though one thinks of differentials as infinitesimally small (and they are), they are NOT equal. They have different shapes and that shape has to be accounted for. You will see this in even greater fashion when you get to the cylindrical coordinate system. In that arena "dx dy dz" IS NOT EQUAL TO " dr dθ dz". Then in spherical coordinates, "dx dy dz" IS NOT EQUAL TO \[d \rho d \phi d \theta \]These differential elements are all "different sizes and shapes" and will have scale factors you need to apply as you construct the integral.

Don't lose sleep over this, a light bulb will light soon and you will be humored by what seemed so difficult. Good luck.