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## Calculating the area of a sector April 5, 2011

Posted by Phi. Isett in Uncategorized.
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Here’s a nice way to calculate the area of a sector.

sectorArea

Take your sector of a disk (it could be the whole disk but for now it’s better to picture only a partial sector with 2 radial “edges”) of radius R.  Now, foliate it into circular arcs having radii $0 < r < R$, starting each arc at the same edge of the sector.  Now, straighten each of those circular arcs so that they are perpendicular to the edge from which they begin.  The circular arc of radius $r$ has length $r \theta$ where $\theta$ is the angle of the sector.

After straightening, you end up with a triangle of base $R$ and of height $R \theta$ (this is because $R \theta$ is the length of the outermost circular arc).

Punch-line (not obvious): the map preserves area.  So you have the formula:

$A = \frac{R^2 \theta}{2}$

So why does bending the circles preserve area?

This part requires much more detail, but is still interesting.  Imagine chopping the triangle up into extremely small squares with sides perpendicular to the edges of the triangle.  I claim that the area of each of those little squares will be preserved.  If you look at one of those little (“infinitesimal”) squares $\partial_x \wedge \partial_y$ with (bottom) horizontal edge $\partial_x$ and (left) vertical edge $\partial_y$ inside of the triangle and try to map it back to the sector, you don’t end up with a rectangle but rather you get a parallelogram $\partial_x^\# \wedge \partial_y^\#$ with an oblique angle.  (At least, if your square is small enough, back in the sector it really does look like a parallelogram.  At the end of the post, I’ll say how to make this heuristic argument fully rigorous.)

When viewed in the sector, the edge which used to be vertical, $\partial_y^{\#}$, now points in the angular direction and has the same length $|\partial_y|$ (this is because vertical line segments in the triangle correspond to circular arcs in the sector).  But the horizontal edge, upon being mapped back to the sector, does not point in the radial direction, but instead makes an oblique angle with $\partial_y^\#$.  Fortunately, we never change the area of a parallelogram by shifting one edge in a direction parallel to the other.  So after shifting $\partial_x^\#$ in the direction of $\partial_y^\#$, we see that it is only the component of $\partial_x^\#$  perpendicular to $\partial_y^\#$ (i.e. the “radial” component rather than the “angular” component) that matters.  To see what this radius is, look at where $\partial_x$ lives in the triangle.  It has edges on two different vertical lines $x = x_0$ and $x = x_0 + |\partial_x|$.  Those lines, remember, used to be circles of radius $x = x_0$ and $x = x_0 + |\partial_x|$ respectively, so we conclude that the radial component of $\partial_x^\#$ has the same length $|\partial_x|$ as the original.

That is essentially why the map preserves volumes.  It’s more complicated than polar coordinates, where going back from a rectangle to a sector takes squares to rectangles.  In that case you don’t need to parallel shift, but volume is not preserved.

A formalization of this intuition involves a bit of differential geometry and the notion of “exterior powers of tangent spaces” to make the parallelograms into something mathematical.  This topic is discussed somewhat in this previous post on the Cauchy Schwartz inequality.

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## Comments»

1. J - January 15, 2012

hey, just wanted to comment that i appreciate what you’ve posted so far to your blog. your style of explication is pleasantly lucid, and accessible enough for those of us out there with at least a little undergraduate mathematics under our belts. thanks for taking the time to share your thoughts!