What is Area in Polar Coordinates?

In summary, the homework statement is to find the area of the infinitismal region expressed in polar coordinates as lying between r and r+dr and between theta and theta+dtheta. A= [integral] (1/2)r^2 d[theta] and the Attempt at a Solution states that to be honest I solved many of these 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... so I asked my friend for help and he was able to help me out. The differential element for the radius is dr and the differential element for the angle is d\theta. To be able to integrate
  • #1
MoradS
5
0

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.
 
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  • #2
You'll end up w/ a double integral.. Something that looks like:

[tex]A = \int \int dA(r,\theta)[/tex]

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):

[tex]dr(r,\theta) = f(r,\theta)dr[/tex]

and

[tex]d\theta(r,\theta) = g(r,\theta)d\theta[/tex]

so you'll have:

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

Figure out what f and g should be (they're both extremely simple functions), and you'll have the answer.
 
  • #3
Thank you for your quick reply, but as simple as that looks, I still don't see it o_0
 
  • #4
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:

[tex]f(r,\theta) = 1[/tex]

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.
 
  • #5
ok so Generalizing this, Ill be moving by r^2 dtheta, true? so the function would end up to be r^2 dr dtheta?
 
  • #6
I'm thinking that in polar coordinates, you're looking at something more like:

[tex]dA = rdr d\theta[/tex]

You should be able to integrate from there...
 
  • #7
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 [tex]dA = r^2drd\theta[/tex], b/c then you'd be left w/ units of volume, not area.
 
  • #8
Alright , I hae ended up integrating and got (r*dr+dr^2/2)dtheta which seems to be incorrect.
 
  • #9
What you want to do is evaluate the double integral:

[tex]A = \int \int dA[/tex]

where... dA is as I've given above, right?
 
  • #10
Thank you very much my friend, I got it now. I was so sleepy last night and trying stupid irrational stuff.
 

1. What is "Area in Polar Coordinates"?

"Area in Polar Coordinates" is a concept in mathematics that involves calculating the area of a shape or region on a polar coordinate system. This system uses polar coordinates (r, θ) to represent points in a plane, where r is the distance from the origin and θ is the angle formed between the positive x-axis and the line connecting the point to the origin.

2. How is the area in polar coordinates calculated?

The formula for calculating the area in polar coordinates depends on the shape of the region. For a circle with radius r, the area can be calculated using the formula A = πr². For a sector of a circle with central angle θ, the area can be calculated using the formula A = (1/2)θr². For more complex shapes, the area can be divided into smaller regions and the formulas for those regions can be used to find the total area.

3. Can the area be negative in polar coordinates?

No, the area in polar coordinates cannot be negative. Since the area represents a physical quantity, it must have a positive value. In polar coordinates, the area is always represented as a positive number, regardless of the shape or region being measured.

4. How does the area in polar coordinates compare to the area in Cartesian coordinates?

The area in polar coordinates can be converted to the area in Cartesian coordinates using the formula A = |J|dA, where J is the Jacobian, which represents the scaling factor between the two coordinate systems. In general, the area in polar coordinates may be easier to calculate for certain shapes, while the area in Cartesian coordinates is more straightforward for others.

5. What real-world applications use the concept of area in polar coordinates?

One example of a real-world application using the concept of area in polar coordinates is in radar technology. Radars use polar coordinates to detect and track objects, and the area of the radar's coverage can be calculated using the area in polar coordinates formula. Additionally, polar coordinates are often used in engineering and physics to analyze circular or rotational motion.

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