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Double & Triple Integrals

A single **[[definite-integral|definite integral]]** chops a line into thin slices and adds them up. But what if the thing you want to total spreads across a flat region — the volume sitting under a curved surface — or fills a solid block, like the mass packed inside a lump of metal? You do the same humble move, only now you tile the region with tiny squares (or pack it with tiny boxes) and add. That is a **[[multiple-integral|multiple integral]]**, and this guide shows you how to actually compute one, one direction at a time.

From slices on a line to tiles on a region

Recall how the single [[definite-integral|definite integral]] was built: you took an interval from a to b, sliced it into thin strips of width dx, measured a height f(x_i) on each, and added up f(x_i) * dx. The answer was the [[limit|limit]] of that [[riemann-sum|Riemann sum]] as the strips grew infinitely thin. Every honest move in this guide is just that same move, lifted up a dimension.

Now picture a surface floating above the floor — say z = f(x, y), a hill whose height changes as you walk around. You want the volume trapped between that surface and a flat region R on the floor below it. A single strip will not do, because R has both length and width. So tile R with a grid of tiny rectangles, each of area dA = dx * dy. Over one tile, treat the hill as flat at height f(x_i, y_j); that little column has volume f(x_i, y_j) * dA. Add up every column and you get an estimate of the whole volume.

single integral:   sum over i      of  f(x_i) * dx          ->  area under a curve

double integral:   sum over i, j   of  f(x_i, y_j) * dA     ->  volume under a surface
                                       where dA = dx * dy

triple integral:   sum over i,j,k  of  f(x_i, y_j, z_k) * dV ->  total stuff in a solid
                                       where dV = dx * dy * dz
Same recipe, more dimensions: chop the region into tiny tiles (dA) or tiny boxes (dV), weight each by the function's value there, add, then take the limit as the pieces shrink to nothing.

Iterated integrals: peel it one direction at a time

Summing over a whole grid at once sounds hopeless. The rescue is wonderfully simple: do it one direction at a time. Hold y fixed and slide along x — that is an ordinary single integral, and it gives you the area of one thin slice of the volume. Then let y vary and add up all those slices. Two single integrals, nested, replace the scary double sum. This nesting is called an iterated integral.

Look closely at the inner integral. While you integrate over x, the variable y is frozen — treated as a constant. That should ring a bell: it is exactly the mindset of a [[partial-derivative|partial derivative]], where you also freeze every variable but one. Differentiation and integration in many dimensions are both done one slot at a time, the others held still.

  1. Set up the outer and inner bounds: the inner integral runs over one variable (say x), the outer over the other (y).
  2. Do the inner integral first, treating the other variable as a frozen constant — just like a partial derivative in reverse.
  3. The result is an expression in the remaining variable. Now do the outer integral over it, with real number bounds.
  4. Out pops a single number — the volume, mass, or total you were after.

One worked example, start to finish

Let's find the volume under the flat-ish surface z = x + y over the unit square, where x runs from 0 to 1 and y runs from 0 to 1. We integrate over x on the inside, then over y on the outside. Watch how y just rides along as a constant during the first step.

Volume = integral (y from 0 to 1) [ integral (x from 0 to 1) of (x + y) dx ] dy

Inner integral, y held constant:
   integral (x from 0 to 1) of (x + y) dx
      = [ x^2/2 + y*x ]  from x = 0 to x = 1
      = (1/2 + y) - 0
      = 1/2 + y

Outer integral, now over y:
   integral (y from 0 to 1) of (1/2 + y) dy
      = [ y/2 + y^2/2 ]  from y = 0 to y = 1
      = (1/2 + 1/2) - 0
      = 1

Volume = 1
Inside-out: integrate over x with y frozen, getting 1/2 + y; then integrate that over y. The nested pair of single integrals delivers the volume, which is exactly 1.

A nice sanity check: the surface z = x + y has average height 1 across the unit square (it runs from 0 at the corner (0,0) up to 2 at (1,1)), and the square has area 1, so the volume ought to be about 1 * 1 = 1. It is. For this surface, because it is a flat tilted plane, you could even have eyeballed it as the volume of a wedge — the integral simply confirms what geometry whispers, and keeps working when the surface stops being flat.

Up to triple integrals, and what they total

Push the same idea one rung higher. A triple integral runs over a solid lump of three-dimensional space. Chop the solid into tiny boxes of volume dV = dx * dy * dz, weight each box by the function's value there, add, and take the limit. As an iterated integral it is three single integrals nested inside one another — peel x, then y, then z, each treating the still-pending variables as constants.

What does a triple integral hand you? It depends entirely on what the function means. If f is density (mass per unit volume), the triple integral is the total mass of the object. If f is charge density, you get total electric charge. If f is simply the constant 1, you get the plain volume of the solid. One machine, many meanings — and that is precisely why multiple integrals show up everywhere from physics to probability to graphics.