A double integral represents the integration of a function of two variables (say, x and y) over a region R in the xy-plane. If the function is f(x,y), a surface can be formed by setting z = f(x,y) in the three dimensional coordinate system. When you evaluate the double integral of this function over the region R, you would get the volume under the surface to the xy-plane over that region. (This should not be confused with a surface integral).

(Yes, this is a The Calculus writeup. You ought to know, generally, about the calculus and integrals to understand the math and the ∫ symbol and the d-symbol. Otherwise, I hope the idea behind a double integral is clear, esp. in the "Physical Meanings" section.)

This writeup will explain the notation of a double integral, explain some of the physical meanings behind a double integral, and finally, show how to calculate double integrals depending on the type of region that must be integrated over.

PART I: Notation

The notation for a regular old double integral over a region, sans function, is this:

         ∫∫  dA
         where dA = dx dy = dy dx

This is said thus: "The double integral over the region R." Next, we have a more generalized double integral.

The a double integral over the region R of a function f(x,y) is written:

         ∫∫  f(x,y) dA
         where dA = dx dy = dy dx

This is spoken thus: "The double integral of f over the region R."

If you're familiar with the typical definite integral, which has two bounds, you might be confused at the moment. What is that R there? What does it mean?

The R-notation is useful because a bounded region in a plane can take on virtually any shape. The R is a generalization. The region contains a whole bunch of points of the form (x,y). The double integral says, "integrate the function over the all points in the region R." The analogue to this for a single/definite integral is, "integrate from a to b." You might also say, "integrate over the interval." The difference between a double integral and your regular old definite single integral is that the way you represent the region in the integral's bounds changes depending on how R is mathematically represented/bounded (more on this later). Yes, it can be confusing, translating R into something mathematically meaningful. Hopefully this writeup will help you understand the notion of R and finding an integral over a region.

For the rest of this writeup, I will explain the meaning of a double integral, why it's useful, and how to evaluate one (if you must).

PART II: Why the Hell Would I Ever Want To Calculate A Double Integral?

Why, you ask? Why, oh why, is it necessary to know? Well, double integrals are great fun... well, maybe. Either way, they do serve some practical purposes in calculating stuff.

The first type of integral above (without the f(x,y)) is relatively simple. The most basic interpretation of this is the (signed) area bounded by R in the xy-plane (By signed, I mean that it can be positive, negative, or even 0, based on the way the curve is oriented). In other words, the area of R. Depending on the type of problem you want to solve, you have to figure out if you want signed area or unsigned area. The same applies for the other interpretations. I'll give a few tips for this, but not much else.

Next, a geometric interpretation (the one with f(x,y)): the double integral can represent a (signed) volume. If you think of f(x,y) as forming a surface in three dimensions (where z = f(x,y)), the value of the double integral can be thought of as the (signed) volume beneath the surface to the xy-plane above R. This is analogous to the idea of a definite single integral finding the area under a curve (or between two curves). For those of you confused about "surface", I explain the idea of f(x,y) as a surface in Part III. If you already understand this idea, move on to Part IV.

Finally, if f(x,y) represents a function for density at the point (x,y) (i.e. mass per unit area), then the value of the integral would be the mass of the region R. However, a density function is typically denoted ρ(x,y) (that's called a "rho"), but the letter used doesn't really matter, because evaluating is all the same. This is a bit of an odd extension, though: the surface integral better generalizes the notion of mass of a surface (R is really a surface, albeit a flat one with only a 2D shape. Don't confuse this surface with the surface in the above paragraph. They're separate).

In my examples later, I'll give an example of each of these.

PART III: f(x,y) As A Surface

f(x,y) is a single-valued function of two variables, x and y. For example, if f(x,y) = 3x2y + cos(y), then f(3, π) = 3 * 32 * π + cos(π) = 27π - 1. The "graph" of a function with two variables is analogous to y = f(x), but kicked up a notch. We're working in three dimensions when we talk of the graph of f(x,y). Imagine the xy-plane, and now imagine the z-axis perpendicular to the plane, running up and down. The graph z = f(x,y) will be a surface: for all coordinates (x,y) in the xy-plane, there will be a corresponding z-value, unless f is undefined at a particular point (where "undefined" can also mean "not real"). Here's an example:

f(x,y) = √(1 - y2)

Even though there's no x in the function, it's still appropriate. The function of y looks like the function for a semi-circle with radius of 1. This function's graph looks like half a cylinder with a radius of 1, the center of which is along the x-axis. It's an infinitely long half-cylinder.

Oh yeah: for every (x,y), f(x,y) should have AT MOST one value. One, period, no more, unless it's undefined. This is just a general property that all functions should have.

For a demonstration, take a cereal bowl, flip it over, and set it on a table. This bowl can be thought of as a surface. Between the bowl and the table, there is volume. The table is like the xy-plane here. Now, if you put the bowl UNDERNEATH the table, with the upper rim touching the underside of the table, then this would be like having a "negative" volume, or a negative double integral (the surface is beneath the xy-plane, meaning z is negative, meaning f(x,y) is negative). Capische? Good.

PART IV: The Mysterious Region R

So, R is a region, a bounded area, whatever you want to call it. It's some shape, named or unnamed, in the xy-plane. R could take the shape of any of the following: rectangle, circle, triangle, hexagon, polygon, region bounded by two functions of x XOR y, cardioid, blob, outline of a flower, silhouette of your mother, &c.

There are various ways to express R mathematically. Ah, there's the rub: You need to figure out HOW to express R mathematically, so you may put bounds on the integrals, giving you two definite integrals, one inside the other. How do I "cover" every point in the region when evaluating the integral, and get every single infinitesimal? Move on to the next section, cowboy.

PART V: Determining How to Evaluate the Buggers

Now I begin the actual math. Here are a couple of precautions.

  1. For all (x,y) in region R, the function f(x,y) should be defined (I'm unsure about handling improper integrals here). For example: if f(x,y) = 1/(xy), and (0,0) is in the region R, then you can evaluate the double integral because f(x,y) isn't defined at any coordinate where x = 0 or y = 0 (can't divide by 0, foo).
  2. Pay attention to things like signed areas, "orientation" of R (important in polar and parametric regions), whether R is composed of other, distinct regions (you can split up R), that kind of thing. I leave these things up to you and your instructor, colleagues, business, whatever. Just know when and how to use these integrals correctly. (When I say, "split up R," think of what you have to do when you're evaluating a definite integral involving |x| (absolute value), when the bounds cross 0).

A. The Region R is a Rectangle or Square

This is perhaps the simplest region you will encounter, since the only bounds on the integrals are numbers.

Region R is a rectangle in the xy-plane. x-coordinates of the corners are at a and b and the y-coordinates of the corners are c and d. THEN the double integral can be found:

                        d  b
       ∫∫  f(x,y) dA = ∫  ∫  f(x,y) dx dy
         R              c  a
                        b  d
                     = ∫  ∫  f(x,y) dy dx
                        a  c

Notice the order of the bounds and the differentials dx and dy: be sure to match up the order of the bounds to the order of your differentials (for all integrals). Also, for something like this, it doesn't matter WHICH order you integrate in, an idea that comes from Fubini's Theorem. However, be VERY CAREFUL when using the theorem, for it doesn't ALWAYS apply (as the writeup in Fubini's Theorem explains). So be careful here, is all.

EXAMPLE: f(x,y) = cos(x) + 2xy + 1, and R is the region in the xy-plane of the rectangle, whose corners are at (0,0), (0,3), (π,3), (π,0). Find the volume under the surface formed by f(x,y) over the region R.

The x-bounds a and b are 0 and π; the y-bounds c and d are 0, 3.

                         3  π
        ∫∫  f(x,y) dA = ∫  ∫  (cos(x) + 2xy + 1) dx dy
          R              0  0

First, we calculate the inside integral.

         π                                          |x=π
        ∫  (cos(x) + 2xy + 1) dx = sin(x) + x2y + x |    = (0 - 0) + π2y + π = π2y + π
         0                                          |x=0

Now we take that and evaluate the outside integral.

         3                            |y=3
        ∫ (π2y + π) dy = π2y2/2 + π y |    = π232/2 – 0 + 3π = 9π2/2 + 3π (units3)
         0                            |y=0

For most double-integrals (which are integrals inside an integral), you apply the same basic technique: ignore the other variable, treating it as a constant; evaluate the definite integral on the inside; then evaluate the definite integral on the outside.

B. Region R is Bounded by Two Functions of EITHER x or y (not both), and Endpoints in the xy-plane.

Defined by good ol' functions. I like functions. Perfect example is this: region bounded two curves y = x2 and y = 1 and two x-bounds x = -1 and x = 1.

NOTA BENE: The order in which you evaluate integrals becomes very important here.

CASE 1: Functions of x

R is bounded by the graph of the curves y = h(x), y = g(x), and the lines x = a and x = b. h(x) ≥ g(x) for all x on (a,b) (interval notation), and a ≤ b. (If R isn't like this, then you'll have to split up the double integral). THEN:

                         b  h(x)
        ∫∫  f(x,y) dA = ∫  ∫  f(x,y) dy dx
          R              a  g(x)

It is Very Important that you evaluate the integral IN THAT ORDER. If you don't, then when you finish evaluating the integral, you'll have some exes leftover. This is a Bad Thing.


R is bounded by the graph of the curves x = h(y), x = g(y), and the lines y = c and y = d. h(y) ≥ g(y) for all y on (c,d), and cd. (Same note as above: you ought to split up the double integral if these aren't true). THEN:
                         d  h(y)
        ∫∫  f(x,y) dA = ∫  ∫  f(x,y) dx dy
          R              c  g(y)

Again, integrate in that order or else you die.

EXAMPLE (uses Case 1):
Region R is bounded by the curves y = 1 and y = x2, between x = -1 and x = 1. The density for each point in the region is:
ρ(x,y) = 3 x2 + y (unit: grams/meters2)
Find the mass of region R.

Since 1 ≥ x2 on this interval, treat h(x) = 1 is the upper bound for y and g(x) = x2 is the lower bound for y.

                         1  1
        ∫∫  ρ(x,y) dA = ∫  ∫  (3 x2 + y) dy dx
          R             -1  x2

First do the inside integral:

         1                              |y=1
        ∫  (3 x2 + y) dy = 3 x2y + y2/2 |    = 3 x2 - 3 x5 + 1/2 -  x6/ 2
         x2                             |y=x2

Now the outside:

         1                                                            |x=1
        ∫  (3 x2 - 3 x5 + 1/2 – x6/2) dx = x3 - 3 x6 / 6 + x/2 – x7/14 |
        -1                                                            |x=-1
                  = (1 - 1/2 + 1/2 - 1/14) - (-1 - 1/2 - 1/2 + 1/14)
                  = 20/7

So the mass of R is 20/7 grams. Yay, yet another (exciting) double integral solved.

C. Region Bounded by the Graph of a Polar Function

Deciding when to use this CAN be tricky, but need not be. For example, if region R is a circle centered at (0,0) in the xy-plane, you have two options: use two semi-circle functions and integrate using method B above, or define the circle polarly, and (maybe) make the integral easier... Sometimes. Anyway, I haven't used this one too frequently myself, but I imagine it can be done.

(Note 1: I'm assuming you know about polar functions and coordinates. If you need a review... see this pipelink.)

(Note 2: There is a distinction between r and r(θ). The former represents a variable, r. The latter represents the polar function r(θ), a generalized polar function. Say it with me: r is a variable, r of theta is a function. r is a variable, r of theta is a function. Got it? Good. Don't confuse either of these with each other or the region R, or else You will get defenestrated.)

(Note 3: There is no note 3.)

The region R is bounded in the xy-plane by the polar function r = r(θ), between the angles θ1 and θ2. The double integral of f(x,y) over the region R is:
                         θ2  r(θ)
        ∫∫  f(x,y) dA = ∫   ∫  f(r cos(θ), r sin(θ)) r dr dθ
          R              θ1  0
  1. r(θ) ≥ 0 for all θ in (θ1, θ2) for the above to work.
  2. If r(θ) ≤ 0 for all θ in (θ1, θ2), then multiply the integral by -1.
  3. You have to split up the double integral into two double integrals if r(θ) changes sign on (θ1, θ2) (It may need to be split up more than once if it changes sign more than once).
  4. θ1θ2 (i.e. positively oriented) THIS MUST BE TRUE. ... I think, anyway.

The f(x,y) changes noticeably. This is because of a relationship between polar and cartesian coordinates, which says that:

      x = r cos(θ)
      y = r sin(θ)

Likewise, dA changed into r dr dθ. WHY this happens, I am unsure myself (I'd like to see a "proof" for the area element being that. /msg, anyone?), but the good people at MathWorld have confirmed this, and I have seen it used in an actual multivariable calculus class I once sat in on.

Find the area bounded an ellipse with a minor axis length of .6 and major axis length of .8.

First off, we need to find two things: semi-latus rectum length (stop snickering) (denoted L), and eccentricity (denoted e). They're related to an equation for the polar graph of an ellipse I found at Wikipedia (search: Ellipse). You'll have to trust me that the polar function graphs the appropriate ellipse. I can't really explain it without making a whole other node.

      L = (.6)2/.8 = .45
      e = √(1 – (.6)2/(.8)2) = √(7)/4

According to Wikipedia dot com on ellipses, the polar function for this ellipse should be:

      r(θ) = L / (1 – e cos(θ)) = .45 / (1 - √(7)/4 cos(θ))

Since we're only looking for area, we can say that f(x,y) = 1, and just evaluate the first type of double integral. To trace out the full ellipse, θ runs from 0 to 2π. So now we have:

        ∫∫  dA = ∫   ∫  r dr dθ
          R       0   0

Inside integral:

         r(θ)          |r=r(θ)
        ∫  r dr = r2/2 |    = r2(θ) / 2
         0             |r=0
                            = (.45 / (1 - √(7)/4 cos(θ)))2 / 2
                            = .10125 / (1 - √(7)/4 cos(θ))2

Now we integrate the last integral:

        ∫  .10125 / (1 - √(7)/4 cos(θ))2 dθ 

I use a TI-83 Plus to numerically evaluate. The input I gave was: fnInt(.10125 / (1 - √(7)/4 cos(θ))2, θ, 0, 2π). It yielt

        ∫  .10125 / (1 - √(7)/4 cos(θ))2 dθ = .48π 

We can check this one, actually. Again, according to Wikipedia, the area of the length is π a b, where a is the major axis length and b is the minor axis length. So the area should be π(.6)(.8) = .48π. Oh happy days!

D. Region Whose Boundary is Traced by the Graph of a Parametric Function

The region R is defined by the parametric function (x(t), y(t)), where t is the parameter and it runs from a to b. There are two cases for this:

CASE 1: f(x,y) is constant

This one is actually fairly easy, and really cool. We can turn this sucker into a single integral. All we really have to do is evaluate the simple double integral over the region, and multiply that by the constant at the end. So this case really comes down to the double integral over the region. Here's what we have:

                  b                  a
        ∫∫  dA = ∫  y(t) x'(t) dt = ∫  x(t) y'(t) dt
          R       a                  b

That's it. Easy enough to express, anyway. For specific functions, it can at least be evaluated on a TI-83 now, if you can't actually solve the integral symbolically. There's a really neat reason for this, and it involves the way you write differentials, and the way you find an area under the function f(x) with a regular integral. dx/dt = x'(t), so dx = x'(t) dt; y = f(x), so ∫ y dx = ∫ y(t) x'(t) dt. Do notice in the second equation that the bounds are flipped. Not an error. I forget why at the moment, but you need to toss in a negative sign. (If you followed that explanation, you get cookies)

CASE 2: f(x,y) isn't constant

This is one is tricky. It involves using something called Green's Theorem. There's an odd application of it I learned in my Calc II online class. It was weird, but kind of neat. What this basically comes down to is turning a double integral into a single integral (like above, only more complex). Here's what Green's Theorem says:

        ∫∫  (∂g/∂x - ∂h/∂y) dx dy = ∫  (h(x,y) dx + g(x,y) dy)
SOURCE: http://mathworld.wolfram.com/GreensTheorem.html
(Don't worry if you don't fully understand what it means. I don't either. I don't know why it works, why it's true, or any other general usage for it. I just know it is without a doubt true.)

If we transform this a little, taking into account that x = x(t), y = y(t), dx = x'(t) dt and dy = y'(t) dt:

        ∫∫  (∂g/∂x - ∂h/∂y) dx dy = ∫  (h(x(t),y(t)) x'(t) dt + g(x(t),y(t)) y'(t) dt)
          R                          a

On the left side of the equation, all that's going on is taking a double integral over a region R, but the integrand doesn't look like f(x,y). Now, using this equation.

Make arbitrary functions, g(x,y) and h(x,y) such that ∂g/∂x - ∂h/∂y = f(x,y).
The simplest way to do this is to define them thus:
        h(x,y) = 0

        g(x,y) = ∫  f(s, y) ds

By the Fundamental Theorem of Calculus, ∂g/∂x = f(x,y) and ∂h/∂y = 0. Thus, ∂g/∂x - ∂h/∂y = f(x,y). Here's the final restatement of this method to calculate a double integral (THIS IS IMPORTANT):

                         b   x(t)
        ∫∫  f(x,y) dA = ∫  (∫  f(s, y(t)) ds) y'(t) dt
          R              a   0
  1. (x(t), y(t)) is traced out counterclockwise as t runs from a to b (positive orientation). If it's not, then reverse the sign of the integral's final value.
  2. If on the interval (a,b) the curve intersects and passes through itself, you'll have to split up the double integral. If it only forms a closed curve that doesn't intersect itself, you're safe. No worries.
  3. If the curve isn't closed, region R will be bound by the x-axis, itself, and any parts of the curve which curve over or under itself. This is kind of confusing to demonstrate sans pictures. All I'm saying is this: BE CAREFUL YOU FOOL!
  4. The parametric function traces the boundary of the region ONCE from a to b.

Well, that about wraps things up for double integrals. Now go on. Go. Get out of here. Class is over. Really. I mean it. Unless you want to read the pseudo-footnotes (which might be important). Well?


  • Pictures (ASCII or links)
  • More examples
  • More techniques, meanings, information (if you know of any, /msg me about them)


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