Multiple Integrals Preview

A thin metal plate is cut in the shape of the square 0 \le x \le 2, 0 \le y \le 2. The plate is not uniform — near the corner at the origin it is thin and cold, and as you move outward it gets thicker and warmer. The temperature at the point (x, y) on the plate is exactly T(x, y) = x + y degrees above zero.

Here is the question. What is the average temperature of the plate?

You already know how to answer this question for a one-dimensional object. If a thin rod along the interval [0, 2] had temperature T(x) = x at each point, you would compute the average by integrating: take \int_0^2 x \, dx = 2, then divide by the length of the rod, which is 2, to get an average temperature of 1. That works because a single integral adds up the values of a function across a one-dimensional interval.

The plate is two-dimensional. You need a way to add up values of a function across a region — every point (x, y) in the square — and then divide by the area of the region. The old integral cannot do this alone. It sweeps along one variable at a time. You need something bigger.

That something is a double integral. It is what you get when you apply the single-integral idea twice, once for each direction. And once you see how double integrals work, triple integrals — for solids in three dimensions — come almost for free.

Why the old integral is not enough

Remember what a single integral actually does. You have a function f(x) defined on an interval [a, b]. You cut the interval into tiny strips of width \Delta x. On each strip, f(x) is roughly constant, so the value f(x_i) \Delta x is a tiny piece of area — a rectangle of height f(x_i) and width \Delta x. You add all these rectangles. You shrink \Delta x. The sum approaches a number, which you call \int_a^b f(x) \, dx.

The picture is always the same: a shaded region under a curve, with the integral measuring its area.

Now try the same move on a two-variable function. Take f(x, y) = x + y on the square [0, 2] \times [0, 2]. What is the graph of f? It is no longer a curve in a plane — it is a surface in three-dimensional space. Above every point (x, y) in the square, the surface sits at height x + y. At the origin it touches the ground. At the far corner (2, 2) it is 4 units up. The whole thing is a slanted flat roof above the square.

What does it mean to add up the values of f across the square? You want the total of f(x, y) over every single point — but there are infinitely many points, and you cannot add infinitely many numbers directly. The same trick as before is the way out. Cut the square into tiny patches. On each patch, f is roughly constant. Multiply by the patch's area. Add. Shrink.

That sum has a geometric meaning too. Each term f(x_i, y_j) \cdot \Delta A is the volume of a tiny vertical box: a patch of base area \Delta A, standing on the plate, rising to the height f(x_i, y_j). Adding up all the boxes gives the volume of the solid that sits between the square and the slanted roof. So where a single integral measures an area, a double integral measures a volume.

Slicing a volume with parallel cuts

The picture tells you what a double integral is. It does not yet tell you how to compute one. The breakthrough is to realise you already know how to compute volumes, as long as you can slice them into flat slabs.

Go back to the sloped-roof solid over the square [0, 2] \times [0, 2] with roof z = x + y. Imagine slicing it with a knife parallel to the yz-plane. Each cut is made at a fixed value of x, and the knife leaves a flat cross-section behind.

What does a single cross-section look like? Fix x, say x = 1. On this slice, y runs from 0 to 2, and the height of the roof is 1 + y. The cross-section is a two-dimensional region — a trapezoid whose height at each y is 1 + y. Its area is

The same slicing works at any x. Fix x. The cross-section is the region under the curve z = x + y for y \in [0, 2], and its area is

This is a function of x — a number for every slice.

Now you have an infinite stack of slabs. Each slab sits at some value of x, has cross-sectional area A(x), and is \Delta x thick. Its volume is A(x) \Delta x. Add up all the slabs and shrink \Delta x to zero — that is just a single integral in x:

The volume of the solid is 8. You got it by doing a single integral in y, then a single integral in x. Two integrals, one inside the other.

This nested-integral form is what a double integral actually looks like when you sit down to compute one. Written out, it is

The inner integral runs first, with x held fixed. The outer integral runs second, collecting up the results. You do one single integral after another.

The order dy \, dx versus dx \, dy matters for how you set up the integral — which variable you hold fixed first — but on a rectangle it does not matter for the answer. You are measuring the same volume either way.

The formal picture

You now have the mental image. The formal definition just takes this picture and makes it precise.

Reading the definition. dA is a symbol for "tiny patch of area" — you can think of it as dx \, dy or dy \, dx when you are ready to compute. The \iint with two integral signs is notation; it reminds you that there are two directions to sum along. The limit is exactly analogous to the one in a single integral, just with patches instead of strips.

When the region is a rectangle, the double integral can always be rewritten as two nested single integrals. This is called Fubini's theorem, and it is the reason double integrals are computable in practice:

Either order works; pick whichever makes the algebra cleaner.

For the average-temperature question you started with, the answer is now mechanical. The total of the temperature over the plate is \iint_R (x + y) \, dA = 8. The area of the plate is 2 \times 2 = 4. So the average temperature is 8 / 4 = 2 degrees. That is the answer the opening question was asking for.

Computing one end to end

Here is the full mechanical process, applied to a fresh problem.

When the region is not a rectangle

Rectangles are the easy case. Real problems almost always involve regions with curved boundaries — a triangle, a disc, the area between two parabolas. The trick that makes these work is to let one of the integration limits depend on the other variable.

Take the triangle T with corners at (0, 0), (2, 0), and (2, 2). You can describe this region in the following way: x runs from 0 to 2, and for each fixed x, the variable y runs from 0 up to the line y = x. The upper bound for y is no longer a constant — it depends on where you are in x.

The double integral over such a region is

The outer limits are always numbers; the inner limits are allowed to be functions of the outer variable. You still slice the region into vertical slabs, but now the slabs have different heights at different x.

The same region could have been described the other way — for each fixed y \in [0, 2], let x run from y to 2 — giving the alternative integral

Both orders give 2. Learning to switch between them is one of the key skills for harder multiple-integral problems, because some integrands are dramatically easier in one order than the other.

One dimension higher: triple integrals

Go up one more dimension. The reader is now standing in front of a solid cube of metal [0, 2] \times [0, 2] \times [0, 2], and the density varies from point to point — the material is denser in some places than others. The density at the point (x, y, z) is \rho(x, y, z) kilograms per cubic metre.

What is the total mass of the cube?

You already know how to do this. Cut the cube into tiny subcubes, each of volume \Delta V = \Delta x \, \Delta y \, \Delta z. In each subcube, the density is roughly constant, so its mass is roughly \rho(x_i, y_j, z_k) \, \Delta V. Add all the subcube masses. Shrink the grid. The limit of those sums is the triple integral of \rho over the cube.

In practice, a triple integral is three single integrals stacked inside each other. The innermost runs first, with everything else frozen. Its result is then fed into the middle integral, whose result is fed into the outer integral. Nothing new conceptually — just another layer.

A few uses of the triple integral come up again and again:

• Volume. Take f \equiv 1. The triple integral of the constant 1 over a region E is simply \iiint_E 1 \, dV, which counts every subcube once. The answer is the volume of E.
• Mass. Take f = \rho, the density. The triple integral adds up the mass of every subcube, giving the total mass.
• Average of a quantity. The average value of f over a region of volume V is \dfrac{1}{V} \iiint_E f \, dV, exactly analogous to the 1D average.

Volume as a special double integral

There is a small subtlety worth pointing out, because it will come up in exam problems. When you want the volume of a region E in space, there are two ways to compute it.

The triple-integral way: \displaystyle V = \iiint_E 1 \, dV. You treat the region as a 3D blob, integrate the constant 1 over it, and the answer is the volume.

The double-integral way: if the region sits between two surfaces z = g(x, y) (the floor) and z = h(x, y) (the ceiling) over a base region R in the xy-plane, then

These are two views of the same number. The triple integral is the more general one — it works for any solid region, even very complicated ones. The double-integral formula is a shortcut for the common case where the solid has a clean top-and-bottom description.

You can see the connection directly: write the triple integral as a nested single integral with z running innermost.

The innermost z-integral just computes the height of the solid at each (x, y), and then the double integral sweeps that height across the base. That is exactly the picture you built at the start of the article — adding up the heights of a slanted roof over a region in the plane.

Common confusions

A few things students often misread when they first meet multiple integrals.

• "A double integral is just two single integrals." Close, but the order is what it is. When the region is a rectangle and the integrand is nice, yes — you can compute the double integral as a sequence of two single integrals in either order. For non-rectangular regions, one of the two limits has to depend on the other variable, and you cannot just do any two integrals and expect it to work. The nested form encodes the geometry of the region.

• "The double integral of 1 is always 1." No — it is the area of the region you are integrating over. \iint_R 1 \, dA adds up tiny patches of area, so the answer is the total area of R. Similarly, \iiint_E 1 \, dV is the volume of E. The integrand 1 is what counts the region; the result depends entirely on what region you pick.

• "dA and dx \, dy are different things." They are names for the same object. dA means "tiny patch of area"; in rectangular coordinates this patch is a little rectangle of dimensions dx by dy, so dA = dx \, dy. When you later meet polar coordinates, dA takes a different concrete form (r \, dr \, d\theta), but it still means the same thing — a tiny piece of area. The symbol dA is a placeholder that reminds you what you are summing.

• "You integrate in the order shown." The order is innermost first. In \int_0^2 \int_0^x xy \, dy \, dx, the dy is closer to the integrand, so the y-integration runs first, with x held fixed. The dx with its outer limits runs second. People often read the symbols left-to-right and invert the order; do not.

• "Fubini means you can always swap the order." On a rectangle with a well-behaved integrand, yes. On a triangle or more exotic region, swapping the order requires you to re-describe the region in the other direction, not just trade the integral signs. The answer is the same, but the new limits may look nothing like the old ones.

Going deeper

What Fubini's theorem actually says

Fubini's theorem is the reason that \iint_R f \, dA can be computed as a nested integral at all. In full it says: if f is continuous (or even just integrable) on a rectangle R = [a, b] \times [c, d], then the three numbers

all exist and are equal. The double integral on the left — defined as a limit of sums — agrees with the nested single integrals, and the two nested orders agree with each other.

For continuous functions on a closed bounded region, Fubini holds without any fuss. Without continuity, things can go wrong. The classic counterexample uses the function

on the square [0, 1] \times [0, 1]. Compute the two nested integrals in the two orders, and you get different answers: \int_0^1 \int_0^1 f \, dy \, dx = \pi/4 while \int_0^1 \int_0^1 f \, dx \, dy = -\pi/4. The issue is that f blows up near the origin, and the integral \iint_R |f| \, dA is infinite. Fubini needs that absolute-value integral to be finite. For JEE-level problems this almost never matters — you are integrating polynomials and trig functions over clean regions — but it is the first place where "just compute it as two integrals" can actually fail.

Coordinates that fit the region

Cartesian coordinates are not always the right choice. If the region is a disc, or a sector, or anything with circular symmetry, describing it in (x, y) is awkward — the boundary comes out as y = \pm\sqrt{r^2 - x^2} — and the integrand may become ugly too.

The fix is polar coordinates. Every point (x, y) in the plane can be described by a distance r from the origin and an angle \theta from the positive x-axis. The substitution x = r\cos\theta, y = r\sin\theta turns a disc of radius a into a clean rectangle: 0 \le r \le a, 0 \le \theta \le 2\pi.

But the patch of area dA is no longer dr \, d\theta. A small patch in polar coordinates is a tiny ring-piece — it stretches out as r grows. A careful calculation shows that the correct patch is

The extra r is not optional. It comes from the fact that a fixed angular slice d\theta covers a larger arc at larger r. With this adjustment, the double integral in polar coordinates is

where R' is the description of the region in (r, \theta).

For example, the integral of 1 over a disc of radius a in polar form is

That is the area of the disc, as it should be.

The same idea works in three dimensions. Cylindrical coordinates (r, \theta, z) use polar in the xy-plane and leave z alone, giving dV = r \, dr \, d\theta \, dz. Spherical coordinates (\rho, \theta, \varphi) describe points by distance from the origin and two angles, and the volume element is dV = \rho^2 \sin\varphi \, d\rho \, d\theta \, d\varphi. These change-of-variable formulas are the right tool when the solid has spherical or cylindrical symmetry.

How far this idea goes

Everything in this article is a special case of a much bigger machine called the theory of integration on manifolds. The key insight is simple: a single integral adds a function along a 1D curve, a double integral adds it across a 2D region, a triple integral adds it through a 3D solid. In theory, you can keep going — integrals in four, five, or even infinitely many dimensions are perfectly well-defined, and they show up in probability and physics whenever you have more than a few quantities to average over.

The other direction leads to line integrals and surface integrals, where you integrate a function along a curved path or across a curved surface rather than over a flat region. Those are the tools of vector calculus, and the bridge theorems — Green's theorem, Stokes' theorem, the divergence theorem — tie them back to the double and triple integrals of this article. That is a story for later; it all starts here, with the idea that a nested pair of single integrals can measure a volume.

Where this leads next

You now know what double and triple integrals are, how they relate to volume, and how to compute them as nested single integrals. These are the starting points for a surprising amount of mathematics and physics.

• Area Under a Curve — the one-dimensional ancestor of everything in this article. Start here if the single integral still feels new.
• Area Between Curves — the 2D version of "integrate a difference of heights," which is the shortcut formula for volume by double integration.
• Vector Calculus Preview — where line integrals, surface integrals, and the big bridge theorems (Green, Stokes, divergence) come from. Multiple integrals are the baseline that vector calculus builds on.
• Applications of Differential Equations — many real-world setups (heat flow, fluid mass, centre of mass) depend on computing integrals over 2D and 3D regions, exactly the way this article sets up.