Arc length in polar coordinates: Why isn't $dS=r×d\theta$ As a sort of exercise, I tried to derive the formula for arc length in polar coordinates, using the following logic:
$$dS = r(\theta)d\theta\\
\implies S=\int r(\theta)d\theta$$
However, it turns out the formula is
$$S = \int \sqrt {r^2+\left(\frac {dr}{d\theta}\right)^2}d\theta$$
I could follow the derivation for the correct formula, but why is mine wrong?
Thanks
 A: Your first formula is incorrect. For instance, consider the line $y=x$ and you want to get the length of the line segment from $x=0$ to $x=1$. The length is $\sqrt2$, and the equation in polar coordinates is $\theta=\dfrac{\pi}4$. If we use the first formula, we get the length to be $0$.
In terms of $x$, $y$, we have
$$S = \int \sqrt{(dx)^2+(dy)^2} = \int \sqrt{1+y'^2} dx$$
Setting $x=r \cos(t)$ and $y=r\sin(t)$, we get
$$dx = dr \cos(t) - r \sin(t)dt$$
and
$$dy = dr \sin(t) + r \cos(t)dt$$
Hence,
$$(dx)^2 + (dy)^2 = (dr)^2 + (rdt)^2$$
A: Imagine a segment of the curve along a radius from the origin of your polar co-ordinates. That increases the arc length without changing $\theta$ at all and $rd\theta=0$ for this segment. So you need to take into account the radial component.
A: I just realized where I went wrong. The length of an arc is the radius of curvature times the angle, not the radial distance between the arc and the origin times the angle. Thanks to all the other answers
A: Interesting topic.
First to @mathguy, $\int r d\theta = r θ$ is wrong. Because usually $r$ is a function of $\theta$. so $\int r d\theta =\int r(\theta)d\theta$.
The reason why the concept of "curve length $r d\theta$" can apply to area intergration in polar coordinates $Area=\int \frac{1}{2} r (r d\theta)$ is that the radial component of the length of the curve doesn't contribute to the area at all, so it is safe to use $r d\theta$ as "effective curve length" while the radial component of curve length is completely dropped off in area integration.
