Chapter 16

Double Integrals

We now begin a new chapter---after studying in detail various means of studying change via multivariate differentiation, we will switch to study accumulation via multivariate integration. As Calculus I and II focused on defining the integral of a single variable over a 1-dimensional region (the closed interval $[𝑎, 𝑏]$ ), we will continue in Calculus III to define the integral of multivariate functions over two and three dimensional regions.

16.1From Riemann Sums to Double Integrals

In one dimension, an integral measures the area under a graph by breaking it into slices, and adding up approximate areas of each slice, via a Riemann sum, before taking a limit. We will begin with a similar process here: we define the double integral of a function $𝑓 (𝑥, 𝑦)$ over a region $𝑅$ in the plane by a two dimensional Riemann sum.

This two dimensional Riemann sum works by breaking the region $𝑅$ into small rectangular regions which we will denote $Δ 𝐴$ , choosing a point $(𝑥 𝑖, 𝑦 𝑗)$ in each such region, and then summing

𝑁 \sum 𝑖 = 1 𝑁 \sum 𝑗 = 1 𝑓 (𝑥 𝑖, 𝑦 𝑗) Δ 𝐴 .

As the number of regions goes to infinity, and the size of each rectangle $Δ 𝐴$ goes to zero, this becomes an integral, with $Σ$ becoming $\int$ and $Δ$ becoming $𝑑$ .

Definition 16.1 (Double Integral). If the Riemann sums approach a limit as the rectangular pieces become arbitrarily small, we define the double integral of $𝑓$ over $𝑅$ to be

\iint 𝑅 𝑓 (𝑥, 𝑦) 𝑑 𝐴 = l i m m a x (Δ 𝐴) \to 0 \sum 𝑖 \sum 𝑗 𝑓 (𝑥 𝑖, 𝑦 𝑗) Δ 𝐴 .

Figure 16.1 Cut the rectangle into cells, and over each one stand a column as tall as

𝑓

is at its middle. One such column is picked out: its base is

Δ 𝐴 = Δ 𝑥 Δ 𝑦

and its height is

𝑓 (𝑥 𝑖, 𝑦 𝑗)

, so its volume is a single term of the sum. Add up all of them and you have an approximation to the solid; add up more of them and the blocks close on the surface floating above. That limit is what

\iint 𝑅 𝑓 𝑑 𝐴

means---how to compute it is the next question, not this one. Take

𝑓 = 1

and every column is one unit tall, so the volume is nothing but the area of

𝑅

. Take the ripple and the columns below the plane are counted negative, which is why a signed integrand gives net volume rather than volume.

If $𝑓 \geq 0$ , this measures the volume under the graph of $𝑓$ and above the region $𝑅$ , instead of the area under a curve. For a function that takes both positive and negative values, it measures signed, or net, volume: volume below the $𝑥 𝑦$ -plane is counted negatively.

But how do we evaluate this thing? We can first hold $𝑥$ constant and add along the $𝑦$ -direction, giving a function of $𝑥$ , and then add these results over $𝑥$ . Or we can do the opposite: hold $𝑦$ constant, add along the $𝑥$ -direction, and then add those results over $𝑦$ . Either way, we add up all the little volumes, and this gives the total volume under the surface.

You can see this in the animation below: one of the side bar graphs gives the result of summing along rows first, the other along columns, and these two side graphs have the same total area under their curves.

Figure 16.2 Nothing here is a new approximation---these are the same columns as above, added up in a different order. Take a strip running in the

𝑦

-direction and add just those columns: one number,

𝐴 (𝑥 𝑖)

, which stands out in front as a single bar. Do it for every strip and the row of bars is a one-dimensional picture whose own area is the whole double sum. Now add the strips running the other way instead and the same array of numbers collapses onto the other axis, giving a completely different row of bars---with exactly the same area. That is Fubini's theorem: the order you sum in is free, which is what lets a double integral be computed as two ordinary integrals, one after the other.

The Riemann sum is the definition of the double integral. The following theorem is what lets us compute it using the single-variable integration we already know.

Theorem 16.2 (Fubini's Theorem for a Rectangle). If $𝑓$ is continuous on the closed rectangle

𝑅 = [𝑎, 𝑏] \times [𝑐, 𝑑],

then both iterated integrals exist and

\iint 𝑅 𝑓 (𝑥, 𝑦) 𝑑 𝐴 = \int 𝑏 𝑎 (\int 𝑑 𝑐 𝑓 (𝑥, 𝑦) 𝑑 𝑦) 𝑑 𝑥 = \int 𝑑 𝑐 (\int 𝑏 𝑎 𝑓 (𝑥, 𝑦) 𝑑 𝑥) 𝑑 𝑦 .

The same idea can be seen without the approximating columns. One ordinary integral finds the area of a slice, and a second ordinary integral adds up those slice areas.

Figure 16.3 The same idea again with the steps taken out. Cut the solid at

𝑥

and the face you expose is an ordinary plane region; its area is

𝐴 (𝑥) = \int 𝑓 (𝑥, 𝑦) 𝑑 𝑦

, a single-variable integral in which

𝑥

is just a constant. Sweep the cut across and watch that area rise and fall---and watch it drawn out, as it goes, as a curve standing in front of the solid. The area under that curve is the volume. A double integral is two ordinary integrals: one to find the area of a slice, another to add the slices up.

Volume is only one interpretation of this accumulated quantity. Taking $𝑓 = 1$ counts area, taking $𝑓 = 𝜌$ accumulates a density to find mass, and dividing an integral by the area of its region gives the average value:

A r e a (𝑅) = \iint 𝑅 1 𝑑 𝐴, 𝑀 = \iint 𝑅 𝜌 𝑑 𝐴, 𝑓 a v g = 1 A r e a (𝑅) \iint 𝑅 𝑓 𝑑 𝐴 .

16.2Rectangular Regions

Let $𝑅$ be the region $𝑎 \leq 𝑥 \leq 𝑏$ and $𝑐 \leq 𝑦 \leq 𝑑$ . Say we want to compute the integral $\iint 𝑅 𝑓 (𝑥, 𝑦) 𝑑 𝐴$ . By the observation above (Fubini's theorem) we can compute this by integrating all the $𝑥$ 's first then integrating $𝑦$ , or vice versa:

\iint 𝑅 𝑓 (𝑥, 𝑦) 𝑑 𝐴 = \int 𝑏 𝑎 (\int 𝑑 𝑐 𝑓 (𝑥, 𝑦) 𝑑 𝑦) 𝑑 𝑥 = \int 𝑑 𝑐 (\int 𝑏 𝑎 𝑓 (𝑥, 𝑦) 𝑑 𝑥) 𝑑 𝑦 .

This is a massive simplification: it means that we can compute two dimensional integrals by just doing two one dimensional integrals, one after the other!

Example 16.3. Evaluate $\iint 𝑅 𝑥 2 𝑦 𝑑 𝐴$ for $𝑅 = [1, 2] \times [3, 4]$ .

Example 16.4. Evaluate $\iint 𝑅 𝑥 (3 - 𝑦 2) 𝑑 𝐴$ for $𝑅 = [0, 2] \times [1, 2]$ .

Oftentimes, the order one performs the integrals in does not matter---both are equally straightforward. But this is not always the case!

Example 16.5. Integrate $𝑦 s i n (𝑥 𝑦)$ over the region $𝑅 = {0 \leq 𝑥 \leq 1, 0 \leq 𝑦 \leq 𝜋}$ .

Try both orders, see which is easier!

Sometimes, when the function you are integrating is a product of a function of $𝑥$ and a separate function of $𝑦$ , things can simplify even further! If $𝑓 (𝑥, 𝑦) = 𝑔 (𝑥) ℎ (𝑦)$ then we may write

\iint 𝑅 𝑔 (𝑥) ℎ (𝑦) 𝑑 𝐴 = \int 𝑏 𝑎 \int 𝑑 𝑐 𝑔 (𝑥) ℎ (𝑦) 𝑑 𝑦 𝑑 𝑥 = (\int 𝑏 𝑎 𝑔 (𝑥) 𝑑 𝑥) (\int 𝑑 𝑐 ℎ (𝑦) 𝑑 𝑦) .

We get this by realizing that $𝑔 (𝑥)$ is a constant with respect to the $𝑦$ integral so we can pull it out: but then once we have done the $𝑦$ integral the result is a constant, so we can pull it out of the $𝑥$ integral!

Example 16.6. Compute

\iint 𝑅 𝑒 𝑥 s i n (𝑦) 𝑑 𝐴

on the region $𝑅 = [0, 𝜋 / 2] \times [0, 𝜋 / 2]$ .

This is essentially all there is to the theory of double integrals when the domain is a rectangle, where both variables are bounded by constants. But the regions of the plane over which we might wish to integrate are much more varied.

16.3Variable Bounds and Choice of Order

In one variable calculus, the only sort of region over which you could perform an integral is a single interval. But in two variables, the regions of the plane over which you could wish to integrate are much more varied!

We have learned how to deal with rectangular regions by slicing---and this same technique will serve us well in many other cases. To start, we won't focus on completely general regions, but rather on regions where the top and bottom are bounded by functions of $𝑥$ .

Here, if we integrate with respect to $𝑦$ first, our vertical slices will each be of a different length---but they will still always be intervals. The top and bottom are functions, meaning they pass the vertical line test, so each boundary intersects a vertical strip exactly once.

At the fixed value $𝑥$ , what is the interval we are integrating over? Well, it runs from the bottom function $𝑐 (𝑥)$ to the top function $𝑑 (𝑥)$ , and so the integral along this slice is

\int 𝑑 (𝑥) 𝑐 (𝑥) 𝑓 (𝑥, 𝑦) 𝑑 𝑦 .

Then all that remains is to integrate these slices along the $𝑥$ -direction:

\iint 𝑅 𝑓 (𝑥, 𝑦) 𝑑 𝐴 = \int 𝑏 𝑎 \int 𝑑 (𝑥) 𝑐 (𝑥) 𝑓 (𝑥, 𝑦) 𝑑 𝑦 𝑑 𝑥 .

The inner bounds $𝑐 (𝑥)$ and $𝑑 (𝑥)$ describe movement along one slice. The outer bounds $𝑎$ and $𝑏$ describe the full range of slices. The next animation separates those two jobs.

Figure 16.4 The same slicing, over a region that is not a rectangle. Now the slice does not run the full width every time: it begins on one boundary curve and ends on the other, so the inner bounds are the functions

𝑔 (𝑥)

and

ℎ (𝑥)

rather than two numbers. Watch the slice shorten toward each end and vanish---the

𝑥

where it vanishes is where the outer bounds come from. Inner bounds describe one slice; outer bounds describe the range of slices.

This suggests a consistent process for setting up a double integral:

Sketch the region.
Choose whether to use vertical or horizontal slices.
Find the inner bounds by describing one slice.
Find the outer bounds by finding the full range of those slices.

Exercise 16.7. Find the integral of $𝑥 + 2 𝑦$ over the region $𝑅$

𝑅 = {(𝑥, 𝑦) ∣ 0 \leq 𝑥 \leq 2, 𝑥 2 - 2 \leq 𝑦 \leq 𝑥} .

Sometimes the $𝑥$ bounds don't even need to be given explicitly---they are just the region between where the curves intersect:

Exercise 16.8. Find the volume above the $𝑥 𝑦$ -plane under the graph of $𝑧 = 𝑥 2 + 𝑦 2$ , within the region $𝑅$ bounded by $𝑦 = 2 𝑥$ and $𝑦 = 𝑥 2$ .

There's nothing special about using vertical slices; we can also do integrals with horizontal slices, integrating $𝑑 𝑥$ first. If a region runs from a left boundary $𝑥 = 𝑝 (𝑦)$ to a right boundary $𝑥 = 𝑞 (𝑦)$ while $𝑐 \leq 𝑦 \leq 𝑑$ , then

\iint 𝑅 𝑓 (𝑥, 𝑦) 𝑑 𝐴 = \int 𝑑 𝑐 \int 𝑞 (𝑦) 𝑝 (𝑦) 𝑓 (𝑥, 𝑦) 𝑑 𝑥 𝑑 𝑦 .

Indeed, the above example can be redone this way no problem!

Exercise 16.9. Find the volume above the $𝑥 𝑦$ -plane under the graph of $𝑧 = 𝑥 2 + 𝑦 2$ , within the region $𝑅$ bounded by $𝑦 = 2 𝑥$ and $𝑦 = 𝑥 2$ , this time first slicing horizontally, with constant $𝑦$ .

However, not every example is just as easy both ways. For example the following integral is easy to write down using horizontal slices, but harder when using vertical slices:

Exercise 16.10. Integrate $𝑥 + 𝑦$ on the region $𝑅$ determined by

𝑅 = {(𝑥, 𝑦) ∣ - 2 \leq 𝑦 \leq 2, 𝑦 2 - 1 \leq 𝑥 \leq 3} .

16.3.1Changing the Order of Integration

To do the last integral instead with vertical slices, we would need to solve for the $𝑦$ bounds as functions of $𝑥$ . One vertical description does not work across the whole region, so the region must be split into simpler pieces.

Exercise 16.11. Set up the integral of $𝑥 + 𝑦$ on the region

𝑅 = {(𝑥, 𝑦) ∣ - 2 \leq 𝑦 \leq 2, 𝑦 2 - 1 \leq 𝑥 \leq 3}

using vertical slices.

One of the most important things about setting up a double integral correctly is thinking through which order of integration will be more useful, and why. Sometimes, one way of slicing will lead to an impossible integral, but the other way will be easy!

Example 16.12. Compute the integral

\int 10 \int 1 𝑥 s i n (𝑦 2) 𝑑 𝑦 𝑑 𝑥

by switching the order of integration first.

Here the geometry of the region stays fixed while the description of it changes. The animation displays both descriptions at once and shows why one order succeeds where the other does not.

Figure 16.5 One triangle, described twice. Slice it vertically and each slice runs from

𝑦 = 𝑥

up to

𝑦 = 1

; slice it horizontally and each runs from

𝑥 = 0

across to

𝑥 = 𝑦

. Both are correct and both give the same number---but with

s i n (𝑦 2)

as the integrand only one of them can be finished, because

\int s i n (𝑦 2) 𝑑 𝑦

has no elementary antiderivative while

\int 𝑦 0 s i n (𝑦 2) 𝑑 𝑥

is just

𝑦 s i n (𝑦 2)

. Changing the order is not cosmetic; sometimes it is the whole problem.

Just like there is a subdivision rule for one dimensional integrals,

\int 𝑏 𝑎 𝑓 𝑑 𝑥 = \int 𝑐 𝑎 𝑓 𝑑 𝑥 + \int 𝑏 𝑐 𝑓 𝑑 𝑥,

there is a similar rule for double integrals. If a region $𝑅$ is broken into subregions $𝑅 1$ and $𝑅 2$ with disjoint interiors (they may meet along their boundaries), then

\iint 𝑅 𝑓 𝑑 𝐴 = \iint 𝑅 1 𝑓 𝑑 𝐴 + \iint 𝑅 2 𝑓 𝑑 𝐴 .

This lets us perform integrals we otherwise could not, by breaking the domain down into simpler pieces, which we can then slice with respect to either $𝑥$ or $𝑦$ .