How to integrate a polar graph with respect to radius

[grahas]

How is this done? My textbook only specifies integrating polar graphs with respect to theta.

 

[BvU]

Hi,
You want to do what, precisely? Could you give a bit more context ?

I assume your textbook does something like integrate ##{1\over 2} r^2(\theta)\, d\theta## to find an area ?

 

[grahas]

Maby this will help
https://drive.google.com/file/d/0B1sPUbiSLzB0ZnBQbzJvbTZxaTA/view?usp=sharing
https://drive.google.com/file/d/0B1sPUbiSLzB0ZnBQbzJvbTZxaTA/view?usp=sharing

 

[BvU]

Doesn't help me. What am I looking at ?

 

[grahas]

Well to integrate with respect to r my best guess was to use trapezoids to estimate the area. The points on the trapezoid are calculated from intersection points with a circle of radius r and the thickness of the trapezoid is dr. The 2 intersection points from a circle form one base and then the circle of radius r + dr is used to calculate the other base and dr is used as the height.

 

[BvU]

Just to have a starting point: you want to find the area of the red outlined figure, for which you may have a parametric description in the form ##f(r,\theta) = 0 ## ?
Perhaps something like ##\left (\cos 2\theta\right )^2 - r = 0 ## ?

And the black lines are to show the approach ? ("In this case dr = 1 but ideally it would approach infinity" -- infinity probable means infinitesimal ?)

 

[grahas]

Not just find the area I want to make a graph of A vs r where A is on the y axis. Yes the black lines were my thoughts on how to solve the problem.

 

[BvU]

Well to integrate with respect to r my best guess was to use trapezoids to estimate the area.

No need for trapezoids. For infinitesimal dh rectangles are enough

The points on the trapezoid are calculated from intersection points with a circle of radius r

right. From symmetry, finding one point suffices.

and the thickness of the trapezoid is dr

Not correct. Can you see why not ?

 

[grahas]

No I'm not sure why dr wouldn't be the width of a rectangle.

 

[BvU]

Just look at your innermost trapezia

 

[BvU]

Which doesn't mean you can't integrate over your dr: it's just that you don't have trapezia or rectangles, but sections of annuli (i.e. y₁-y₂ and y₃-y₄ are curved)
(And then my dr and your dr coincide again )

 

[Hendrik Boom]

Integrating with respect to r, wouldn't you be dealing with rings or arcs instead of straight lines?

 

[grahas]

https://drive.google.com/open?id=0B1sPUbiSLzB0NWhseEg4alY4OEE

How could this be generalized to a formula that could be graphed?

 

[BvU]

Integrating with respect to r, wouldn't you be dealing with rings or arcs instead of straight lines?

Yes. My charcoal english used the term 'annuli'.

How could this be generalized to a formula that could be graphed?

Hehe, PF culture insists that you do the work and we help by asking, hinting etcetera

At the risk of violating these cultural rules: from your picture (beautiful ! ) you see that ##dA/8 = 2\pi \rho \theta(\rho) d\rho## so the answer to your question (the answer for which that you should have at least proposed an attempt ...) is $$A(r) = 8\,\int_0^r 2\pi\, \rho \,\theta(\rho) \, d\rho $$ and now it's up to you to work out the integrand and integrate it ...

 

[grahas]

I have been trying it for a while. I can't figure out how to convert the polar graph to a parametric one. I tried internet resources but they weren't to great. Is there any other way to solve this without converting it to parametric?

 

[BvU]

how to convert the polar graph to a parametric one

Isn't ##\left (\cos 2\theta\right )^2 - r = 0## good enough ? if you want to integrate over ##dr## all you need to do is work this around to a function ##\theta(r)##, something with an ##\arccos##, I suppose...

 

[grahas]

Still haven't been able to solve it. Sorry. I think I will just try to solve it numerically in python.

 

[grahas]

Well I got an answer but I am not positive it is correct. The bold black line represents the area accumulated of the top half of one leaf. An approximation of the area of the top half of the east leaf with a triangle gives a value of 0.075 and the bold black line has a value of 0.08. I think that the results of graph are important because it yields the random probability of finding a object in that region. The thought stems from a method of approximating the value of pi by randomly placing dots in a square with a circle circumscribed, then finding the ratio between the dots in and out of the circle but between the square. Please correct if I am wrong about the meaning of the results or if I did my math wrong. I was investigating the relationship between polar graphs, path of electron in an atom, and how the results compare to other theories.