# Area of a band in $\mathbb{R}^2$

If I have a continuous, and smooth curve $\mathcal{C}$, length $\ell$, in $\mathbb{R}^2$ and at each point on the curve I were to draw a line segment, length $d$, normal to the curve centered at the point; would the area covered by all the line segments be $d\cdot\ell$ provided that no two line segments intersect with each other?

Also: if this is true, can this be generalized to more dimensions?

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Sure. Let $d$ or $l$ be $0$. –  Jeremy Dec 18 '13 at 5:00
How do you define a normal of a continuous curve? If you curve is smooth, then you can do what you describe, but the area will not be $dl$ in general. –  Andrey Sokolov Dec 18 '13 at 5:10
If you draw a line segment at each point, then the only way no two line segments will intersect is if all are parallel, i.e., if $C$ is a straight line. In that one case, the area will indeed be $d\cdot\ell$. –  mjqxxxx Dec 18 '13 at 6:00
I am going to add smooth as a criterion. Also, I think that as long as the length d/2 is less than the radius of curvature, no two segments will intersect. –  davik Dec 19 '13 at 0:28

To give a concrete example, let $C$ be a circle of radius $r$ (and let $d < r$ if you choose the inward-facing normal). Then the region swept out by the normal lines is an annulus, and you can compute its area for both the inward- and outward-facing normals; you'll see that you don't get the answer $d\ell$ in either case.
The normal is supposed to be “centered at that point”: I think davik means that the normal of equal length is drawn in both directions to achieve the band. And in that case it's true that the area is $dl$ provided the normals are not too long. –  Michael Hoppe Dec 18 '13 at 10:09