Polar integration - Length and Area of curve

In summary, the area of a polar curve is given by A=(1/2)∫ r2 d (theta), while the length is given by L=∫√(r2+r'2)d(theta). This is because when calculating the area, deviations from the circular section can be neglected, while this is not the case for calculating the length. This can also be seen in the different parametrizations for the two calculations. The area is parametrized by the radius vector multiplied by a parameter, while the length is parametrized by the radius vector itself.
  • #1
rahuljayanthb
13
0
The area of a polar curve is given by A=(1/2)∫ r2 d (theta).
this can be interpreted as δA= ∏r2δ(theta)/2∏ (treating the area element as the area of a sector of a circle with angle δ(theta).)
taking limit of δ(theta)→0,
dA= [STRIKE]∏[/STRIKE]r2 d(theta)/2[STRIKE]∏[/STRIKE]=1/2 (r2d(theta) )
there fore A=1/2∫r2 d(theta).

By the same logic, shouldn't length of a polar curve be L=∫[STRIKE]2∏[/STRIKE]r d(theta)/[STRIKE]2∏[/STRIKE]=∫r d(theta)?

the actual equation for curve length is L=∫√(r2+r'2)d(theta).

why is this approach giving me an incorrect equation in the second case but a correct one in the first?

ps. the equation for length of a curve can also be derived by partially differentiating the equation for area by r: ∂A/∂r=L. This is again consistent with looking at length of a curve as the rate of change of area.
 
Physics news on Phys.org
  • #2
The length of a curve is not, in general, the rate of change of the area.

As a simple example, the circumference of a square is 4l, whereas the area is l^2
 
  • #3
but where is the derivation of the length of curve wrong?
 
  • #4
It violates the Pythagorean theorem, for example.

You have, differentially, that the curve length ds equals:
[tex]ds=\sqrt{(dr)^{2}+(rd\theta)^{2}} [/tex]
 
  • #5
but i am treating every length element of the curve as a part of a circle of appropriate radius. so i am treating the curve as a conglomerate of numerous circular elements just in the area derivation.
 
  • #6
well, but that is precisely what you cannot do.

For areas, deviations from area of the circular section are negligible relative to the area for the circular section, while a similar negligibility is not present for the CURVE.

Basically, the reason why this works for areas is that an area of "tiny*tiny" can be neglated relative to "tiny", while for curves "tiny" can't be neglected relative to.."tiny"
 
  • Like
Likes 1 person
  • #7
Ok, what you consider here is a curve parametrized with the polar angle [itex]\theta[/itex] of polar coordinates in a plane. Your position vector reads
[tex]\vec{r}=\vec{r}(\theta)=r(\theta) \hat{r}(\theta).[/tex]
The arc element is
[tex]\mathrm{d} s=|\mathrm{d} \vec{r}|=\mathrm{d} \theta|r' \hat{r}+r \partial_{\theta} \hat{r}|=\mathrm{d} \theta \sqrt{r^2 +(r')^2},[/tex]
which indeed shows that your assumption about [itex]\mathrm{d} s[/itex] is wrong and the "actual equation" is right.

The area is another thing. Your formula gives the area swept out by the above radius vector. This area is thus parametrized by
[tex]\vec{r}(\lambda,\theta)=\lambda r(\theta) \hat{r}(\theta)[/tex]
with [itex]\lambda \in [0,1][/itex]. The area element is
[tex]\mathrm{d} F=\mathrm{d} \lambda \mathrm{d} \theta |\partial_\lambda \vec{r} \times \partial_{\theta} \vec{r}|=\mathrm{d} \lambda \mathrm{d} \theta |\vec{r}(\theta) \times \partial_{\theta} \vec{r}|=\mathrm{d} \lambda \mathrm{d} \theta \lambda r^2(\theta) |\hat{r} \times \hat{\theta}|=\lambda r^2(\theta).[/tex]
The total area swept out by the radius vector from angle [itex]\theta_1[/itex] to [itex]\theta_2[/itex] thus is
[tex]F=\int_0^{1} \mathrm{d} \lambda \int_{\theta_1}^{\theta_2} \mathrm{d} \theta \; \lambda r^2(\theta)=\frac{1}{2} \int_{\theta_1}^{\theta_2} \mathrm{d} \theta \; r^2(\theta).[/tex]
As you see, here one component of [itex]\partial_{\theta} \vec{r}[/itex] doesn't contribute, because of the cross product in the formula for the area element!
 

1. How do you calculate the length of a polar curve?

The length of a polar curve can be calculated using the formula L = ∫√(r^2 + (dr/dθ)^2)dθ, where r is the function representing the curve and dr/dθ is its derivative with respect to θ. This formula is derived from the Pythagorean theorem and provides an accurate measurement of the curve's length.

2. Can polar integration be used to find the area under a curve?

Yes, polar integration can be used to find the area under a polar curve. The formula for finding the area is A = ½∫r^2dθ, where r is the function representing the curve. This formula is derived from the formula for finding the area of a sector of a circle and can be used to find the area of any polar curve.

3. What is the difference between finding the length and area of a polar curve?

The main difference between finding the length and area of a polar curve is the formula used. As mentioned earlier, the length is calculated using the formula L = ∫√(r^2 + (dr/dθ)^2)dθ, while the area is calculated using A = ½∫r^2dθ. Additionally, the length represents the total distance along the curve, while the area represents the total amount of space enclosed by the curve.

4. Are there any other methods for finding the length and area of a polar curve?

Yes, there are other methods for finding the length and area of a polar curve, such as using the arc length formula and the shoelace formula. However, these methods may not be as efficient as using polar integration and may not provide as accurate results.

5. Can polar integration be used for any polar curve?

Yes, polar integration can be used for any polar curve as long as the curve can be represented by a function in the form of r = f(θ). This includes both closed and open curves. However, for some curves, the calculation of length or area may be more complex and require advanced techniques.

Similar threads

  • Calculus
Replies
29
Views
443
Replies
2
Views
136
Replies
2
Views
114
Replies
2
Views
2K
Replies
33
Views
3K
Replies
4
Views
143
Replies
5
Views
2K
Replies
3
Views
172
Replies
3
Views
1K
  • Calculus
Replies
15
Views
2K
Back
Top