Multi-variable integration: the best thing since
sliced bread |

This section is a brief reminder of the principle involved in computing multi-dimensional integrals.

In first-year calculus, you solved integration problems like the following.

Find the area of the region in the first quadrant where the
curve *y = -2x ^{2} + 10x -3* lies above the curve

To do this, you would equate
the heights of the curves to find that they intersect when *x
= 1* and when *x = 3*. Then you would regard the region
between the curves as the union of a large number of narrow
vertical rectangles with width equal to *dx* and with
height equal to the difference of the heights of the curves.
The sum of the areas of the thin rectangles is approximately
the area between the curves, and since the integral is the
limit of such *Riemann sums*,
you would find the exact area to be
*20/3* by integrating * (-2x ^{2} + 10x -3) - (2x^{3} -9x^{2} +
12x)* from

Volumes can be computed by a similar principle. To compute the volume of a loaf of bread, you slice the loaf into thin slices and add their volumes. The volume of a slice is its thickness times the area of its face, and so you get the volume of the loaf by integrating the area of a slice.

For example, to compute the volume of the unit ball (the set in
three-dimensional space where *x ^{2} + y^{2} + z^{2}* is less
than

A typical volume computation is somewhat harder than this,
but only because the slices are not necessarily objects whose
areas you know out of your head. In general, you have to do
an auxiliary integration to compute the area of a slice at a
variable height *z*. In complicated volume problems, you
may need to break the object into pieces to facilitate the
determination of the area of a slice.

**Aside on the Busemann-Petty
problem.** The bread-slicing
principle is the basis of Cavalieri's
principle, which says that if all
slices of one loaf by parallel planes have the same areas as the
corresponding slices of another loaf, then the two loaves have
the same volume. This principle is less obvious than it may
appear.

Indeed, it seems equally obvious that if every slice of one melon by a plane passing through the origin has smaller area than the corresponding slice of a second melon, then the first melon must have smaller volume than the second. Actually, this turns out to be false for badly warped melons, but it is a sensible conjecture for melons that are convex and symmetric with respect to the origin. H. Busemann and C. M. Petty were unable to prove this spherical coordinate version of Cavalieri's principle, and they posed it as a problem in 1956. It was not until 1994 that R. J. Gardner obtained the affirmative answer in three-dimensional Euclidean space.

What about the analogous problem in Euclidean spaces of other
dimensions? The surprising solution was completed in 1997. It
turns out that the melon-slicing problem has a
positive answer in dimensions *2*, *3*, and *4*, but a
negative answer in every higher dimension. Thus, there exist
two convex, centrally symmetric five-dimensional melons with
the property that every four-dimensional slice through the
origin of the first melon has smaller area than the
corresponding slice of the second melon, yet the first melon
has larger volume than the second melon.

(For references, see R. J. Gardner,
A. Koldobsky,
and T. Schlumprecht, An
analytic solution to the Busemann-Petty problem on sections
of convex bodies, *Ann. of
Math. (2)* **149** (1999), no. 2, 691-703; and the
review in *Mathematical Reviews* 2001b:52011.)

The principle that the integral adds up lots of little bits can
be used to compute a variety of geometric quantities. For
example, a little bit of arc length of a
curve in the *xy*-plane is *ds*, which by the Pythagorean
theorem equals the square root of *dx ^{2} + dy^{2}*. So to find
the length of a curve, you integrate

The Math 696 course pages were last modified April 5, 2005.

These pages are copyright © 1995-2005 by Harold P. Boas. All rights reserved.

Multi-variable integration: the best thing since
sliced bread |