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Polar integration - Length and Area of curve

  1. Nov 4, 2013 #1
    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. jcsd
  3. Nov 4, 2013 #2


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    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
  4. Nov 4, 2013 #3
    but where is the derivation of the length of curve wrong?
  5. Nov 4, 2013 #4


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    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]
  6. Nov 4, 2013 #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.
  7. Nov 4, 2013 #6


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    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"
  8. Nov 4, 2013 #7


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    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!
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