Need help with Polar Coordinates, esp. for integration

[Legendre]

Any web resources regarding changing the variable of integration from cartesian to polar coordinates that goes beyond the basic :

x = r cos theta
y = r sin theta
r = sq rt (x^2 + y^2)I totally don't get how to find the limits of integration using polar coordinates and my undergrad textbooks appears to assume we already know it.

E.g. r = 2 sin 2 theta , first quadrant. How do we find the limits of integration for r?

Totally clueless. Tried googling but keep getting just the 3 definitions above. What is the general technique or algorithm?

 

[tiny-tim]

Hi Legendre!

(have a theta: θ and a square-root: √ and try using the X² tag just above the Reply box )

I totally don't get how to find the limits of integration using polar coordinates and my undergrad textbooks appears to assume we already know it.

E.g. r = 2 sin 2 theta , first quadrant. How do we find the limits of integration for r?

The limits when you use r and θ depend on what they are when you use x and y.

If it's for all x and y, then you use 0 ≤ r < ∞, and 0 ≤ θ ≤ 2π (or -π ≤ θ ≤ π if you prefer).

If it's for the first quadrant, then you use 0 ≤ r < ∞, and 0 ≤ θ ≤ π/2.

Whatever it is in x and y, you have to convert …

do you have an example that you want to try?

 

[rzaidan]

Hi Legendre
Since you work is in the first quardrant, then θ takes the values from 0 to π/2
and r takes the values from 0 _which corresponds to θ=0 i.e r= 2 sin (2 *0)=0 _
to r= 2 sin (2 θ) since r is a variable and the integral is of the form:
π/2 2sin(2θ) π/2
∫ ∫ dr dθ = ∫ 2sin(2θ) dθ
0 0 0
π/2
= 2(-(1/2)cos(2θ)] = -[cos π -cos 0]=-(-1-1)=2 units of area
0
Best Regards
Riad Zaidan

 

[Legendre]

do you have an example that you want to try?

r = 2 sin (2θ) in the first quadrant.

r takes the values from 0 _which corresponds to θ=0 i.e r= 2 sin (2 *0)=0 _
to r= 2 sin (2 θ) since r is a variable

How do we know the lower limit of integration for r is 0?
How do I know the upper limit of integration for r is 2 sin (2θ)?

Is this done by sketching the area? I have no idea how to begin sketching r = 2 sin (2θ)!


# Another question : How do you sketch first quadrant, limits of integration r = 1/cos θ and r = 0?


Note : I am a lot more eager to find out how we arrive at the answer than the answer itself.

Thanks for all the help! :)

 

[tiny-tim]

r = 2 sin (2θ) in the first quadrant.

(I take it you mean the interior of that curve?)

ok … since you're restricted to the first quadrant, θ is restricted to [0,π/2].

And r = 2 sin (2θ) is obviously single-valued for any particular θ (and r = 0 at θ = 0 and π/2) …

so r is restricted to [0, 2sin(2θ)] for each θ

So the limits are ∫₀^π/2∫₀^2sin(2θ)

(do you see that you have to pick the coordinates off one at a time? … in this case, do the θ limits first, then for a fixed θ, do the r limits)

Is this done by sketching the area? I have no idea how to begin sketching r = 2 sin (2θ)!

sin(2θ) obviously goes from 0 up to 1 and down to 0, so without sketching it you know it starts at the origin, goes away, and comes back (once).

# Another question : How do you sketch first quadrant, limits of integration r = 1/cos θ and r = 0?

1/cosθ goes from 1 upwards to ∞ … if you want to sketch it, just do a freehand curve through (in (x,y) coordinates) (1,0) and (0,∞)

 

[rzaidan]

Hi tiny-tim
thanks for the detailes
Riad Zaidan

 

[Legendre]

Thanks a lot everyone!

One more question :

Regarding the graph of r = 2 sin (2θ), I know how to sketch it if its y = 2 sin 2x but in polar coordinates, we're plotting r on the "x-axis" and θ on the "y-axis" (right??).

So doesn't this mean it is something like sketching x = 2 sin (2y) on the regular vertical y-axis and horizontal x-axis?

Can I simply treat r as the vertical axis and θ as the horizontal axis? Then r = 2 sin (2θ) is very easy to plot.

 

[Legendre]

And r = 2 sin (2θ) is obviously single-valued for any particular θ (and r = 0 at θ = 0 and π/2) …

so r is restricted to [0, 2sin(2θ)] for each θ

Pardon me but I don't quite get what you are saying.

Keeping it in the first quadrant. i.e. 0 <= θ <= π/2.

Are you saying that if r = f(θ) is single-valued for any particular θ, then r is restricted to the interval [f(0), f(θ)]?

What do I do with a general r = f(θ) if I know 0 <= θ <= π/2?

 

[tiny-tim]

Can I simply treat r as the vertical axis and θ as the horizontal axis? Then r = 2 sin (2θ) is very easy to plot.

No no no no noooooo!

Lines of constant r are circles, and lines of constant θ are straight lines through the origin ("spokes").

You need to go back to your book and to study this.

Alternatively, google "graph polar coordinates" and look at sites such as http://www.ies.co.jp/math/java/calc/sg_kyok/sg_kyok.html"

 

[Legendre]

No no no no noooooo!

Lines of constant r are circles, and lines of constant θ are straight lines through the origin ("spokes").

You need to go back to your book and to study this.

Alternatively, google "graph polar coordinates" and look at sites such as http://www.ies.co.jp/math/java/calc/sg_kyok/sg_kyok.html"

arghhhh...i hate polar coordinates lol.

yeah, of course we can't interchange r and θ. r = 2 sin (2θ) is a circle in (r,θ) but is a transformed sin curve in (θ,r). lol what was i thinking.

yes i need to go back to my book and study this. any books to recommend? i looked through Binmore & Davies (Calculus), Ostaszewski (Advanced Mathematical Methods), Sydsaeter (Further Mathematics for Economic Analysis) as well as two set of lecture notes on line but they all just touch on the basics briefly. any good basic calculus book that has a lot of materials on converting and integration in polar coordinates beyond the basics?


# regarding the site : very useful, thanks a lot!

 

[tiny-tim]

arghhhh...i hate polar coordinates lol.

?? Polar coordinates are fun when you get used to them!

… yes i need to go back to my book and study this. any books to recommend?

Sorry, I don't know.

You could try putting out an appeal in the Math & Science Learning Materials forum at https://www.physicsforums.com/forumdisplay.php?f=151".

But maybe it's best just to browse the internet until you find something you like.