What is Area in Polar Coordinates?

[MoradS]

Homework Statement

Find the area of the infinitismal region expressed in polar coordinates as lying between r and r+dr and between theta and theta+dtheta

Homework Equations

A= [integral] (1/2)r^2 d[theta]

The Attempt at a Solution

To be honest I solved many of this kind of equations when r is given as a function of theta, but I totally suck when it comes to doing the same problems with symbols...

Thanks in Advance.

 

[psholtz]

You'll end up w/ a double integral.. Something that looks like:

A = \int \int dA(r,\theta)

What is the differential element for the radius?

What is the differential element for the angle?

I don't want to give the whole answer away, but let's say that (this is somewhat sloppy notation, but it gets the point across):

dr(r,\theta) = f(r,\theta)dr

and

d\theta(r,\theta) = g(r,\theta)d\theta

so you'll have:

A = \int_\theta^{\theta+d\theta} \int_r^{r+dr} f(r,\theta) g(r,\theta) dr d\theta

Figure out what f and g should be (they're both extremely simple functions), and you'll have the answer.

 

[MoradS]

Thank you for your quick reply, but as simple as that looks, I still don't see it o_0

 

[psholtz]

Think about how "r" changes as "r" changes (I know that sounds silly)...

And think about how "theta" changes as "theta" changes.

So first think about r.. that's the simpler one.

Suppose we go out to a distance "r" (in polar coordinates) from the center of origin. Then we go just a bit further, to a distance r+dr.

By how much as r changed?

The answer is straightforward: by dr.

So that's simple.

In my rather awkward, silly notation above (which I'm now slightly regretting), we simply have:

f(r,\theta) = 1

That is, the differential for "r" is simply "dr"..

The differential for the angle is slightly different.

Suppose we have a circle of radius r=1.

Pick any point on the circle.

Travel 1/4 of the circumference around the circle.

How much distance have you traveled?

The answer is 2*pi/4, or pi/2.

Now... suppose we have a circle of radius r=2.

Pick any point on the circle.

Travel 1/4 of the circumference around the circle.

How much distance have you traveled?

This time the answer is 4*pi/4 or pi.

Sooo.. in BOTH cases you went around by pi/2 radians, but in the first case you traveled only pi/2, and in the second case you traveled pi.

Clearly, the size of the circle has something to do w/ how much distance you travel, for a "constant" "change" in the angle.

This argument I've made for theta = pi/2.

Make the same argument for theta = dtheta, and you'll have what the differential element for the angle is.

 

[MoradS]

ok so Generalizing this, Ill be moving by r^2 dtheta, true? so the function would end up to be r^2 dr dtheta?

 

[psholtz]

I'm thinking that in polar coordinates, you're looking at something more like:

dA = rdr d\theta

You should be able to integrate from there...

 

[psholtz]

Remember too, area must have units of "area".. that is to say "length*length"..

So "dA" must have units of length*length.

The unit of r is "length"

The unit of dr is "length"

The unit of dtheta is none, or dimensionless...

You can't have dA = r^2drd\theta, b/c then you'd be left w/ units of volume, not area.

 

[MoradS]

Alright , I hae ended up integrating and got (r*dr+dr^2/2)dtheta which seems to be incorrect.

 

[psholtz]

What you want to do is evaluate the double integral:

A = \int \int dA

where... dA is as I've given above, right?

 

[MoradS]

Thank you very much my friend, I got it now. I was so sleepy last night and trying stupid irrational stuff.