Polar integration - Length and Area of curve

[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.

 

[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

 

[rahuljayanthb]

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

 

[arildno]

It violates the Pythagorean theorem, for example.

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

 

[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.

 

[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"

 

[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!