# Polar integration - Length and Area of curve

1. Nov 4, 2013

### rahuljayanthb

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.

2. Nov 4, 2013

### arildno

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. Nov 4, 2013

### rahuljayanthb

but where is the derivation of the length of curve wrong?

4. Nov 4, 2013

### arildno

It violates the Pythagorean theorem, for example.

You have, differentially, that the curve length ds equals:
$$ds=\sqrt{(dr)^{2}+(rd\theta)^{2}}$$

5. Nov 4, 2013

### rahuljayanthb

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. Nov 4, 2013

### arildno

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"

7. Nov 4, 2013

### vanhees71

Ok, what you consider here is a curve parametrized with the polar angle $\theta$ of polar coordinates in a plane. Your position vector reads
$$\vec{r}=\vec{r}(\theta)=r(\theta) \hat{r}(\theta).$$
The arc element is
$$\mathrm{d} s=|\mathrm{d} \vec{r}|=\mathrm{d} \theta|r' \hat{r}+r \partial_{\theta} \hat{r}|=\mathrm{d} \theta \sqrt{r^2 +(r')^2},$$
which indeed shows that your assumption about $\mathrm{d} s$ 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
$$\vec{r}(\lambda,\theta)=\lambda r(\theta) \hat{r}(\theta)$$
with $\lambda \in [0,1]$. The area element is
$$\mathrm{d} F=\mathrm{d} \lambda \mathrm{d} \theta |\partial_\lambda \vec{r} \times \partial_{\theta} \vec{r}|=\mathrm{d} \lambda \mathrm{d} \theta |\vec{r}(\theta) \times \partial_{\theta} \vec{r}|=\mathrm{d} \lambda \mathrm{d} \theta \lambda r^2(\theta) |\hat{r} \times \hat{\theta}|=\lambda r^2(\theta).$$
The total area swept out by the radius vector from angle $\theta_1$ to $\theta_2$ thus is
$$F=\int_0^{1} \mathrm{d} \lambda \int_{\theta_1}^{\theta_2} \mathrm{d} \theta \; \lambda r^2(\theta)=\frac{1}{2} \int_{\theta_1}^{\theta_2} \mathrm{d} \theta \; r^2(\theta).$$
As you see, here one component of $\partial_{\theta} \vec{r}$ doesn't contribute, because of the cross product in the formula for the area element!