5.1 Double Integrals over Rectangular Regions

Big idea

A single integral adds up values of a function along an interval. The interval is one dimensional, so the pieces you add are widths, and the sum measures area under a curve. This section runs the same construction one dimension higher. The domain becomes a rectangle in the plane, the pieces become little rectangles of area, and the sum measures volume under a surface.

The machinery is deliberately parallel. Cut the rectangle $R$ with a grid, pick a sample point in each cell, multiply the height of the surface at that point by the area of the cell, and add. That double Riemann sum is a stack of thin boxes approximating the solid between $R$ and the graph of $f$. Refine the grid and the approximation improves; the limit, when it exists, is the double integral. For a continuous function on a closed rectangle the limit always exists and does not depend on which sample points you chose.

None of that computes anything. The computational content of the section is a different statement: a double integral over a rectangle can be evaluated as two ordinary integrals performed one after the other. Hold $x$ fixed and integrate in $y$; you get a function of $x$ alone, which you then integrate. Each step is a single-variable problem, so every technique you already own carries over unchanged.

Two facts make this practical. First, the order is yours to choose, because integrating in $y$ first and integrating in $x$ first produce the same number. Second, the limits on a rectangle are four constants, which is why rectangles come before anything else. When the region has curved sides the limits start depending on the outer variable, and that complication is the whole subject of the next section.

Decoder

If $f$ is continuous on the rectangle $R = [a,b] \times [c,d]$, then the double integral of $f$ over $R$ equals the iterated integral in either order, and both iterated integrals give the same number.

The phrase carrying the content is “either order”. The definition of the double integral says nothing about $x$ or $y$ individually; it is a limit of sums over cells. The theorem promises that two quite different sequences of one-variable computations both land on that limit. So when one order leads to an antiderivative you cannot find, you are free to switch, and you are not approximating anything when you do.

Read the notation carefully, because it is where mistakes start. In

$$ \int_a^b \int_c^d f(x,y)\, dy\, dx $$

the inner differential $dy$ pairs with the inner limits $c$ and $d$, and the outer $dx$ pairs with $a$ and $b$. Differentials and limits nest like parentheses. The inner integral treats $x$ as a constant, exactly as partial differentiation treats one variable as a constant, and it must come out as a function of $x$ with no $y$ left in it. If $y$ survives the inner step, you substituted the limits into the wrong variable.

The classic error is swapping the limits when you swap the order. On a rectangle the limits are attached to the variables, not to the positions, so reversing the order moves $c$ and $d$ out with $dy$ and brings $a$ and $b$ in with $dx$. Writing $\int_c^d \int_a^b f\, dy\, dx$ silently integrates $y$ from $a$ to $b$, which is a different problem.

Definitions and results

Rectangular region. The set $R = [a,b] \times [c,d]$ consists of all points with $a \le x \le b$ and $c \le y \le d$. Its area is $(b-a)(d-c)$.

Double Riemann sum. Split $[a,b]$ into $m$ equal pieces and $[c,d]$ into $n$ equal pieces, producing $mn$ cells each of area $\Delta A = \Delta x\, \Delta y$. Choosing a sample point $(x_{ij}^{*}, y_{ij}^{*})$ in each cell gives

$$ S_{mn} = \sum_{i=1}^{m} \sum_{j=1}^{n} f\big(x_{ij}^{*}, y_{ij}^{*}\big)\, \Delta A $$

The double integral. If $S_{mn}$ approaches a single number as $m$ and $n$ both grow without bound, regardless of the sample points, that number is $\iint_R f(x,y)\, dA$. Every function continuous on a closed rectangle is integrable there.

Volume reading. When $f \ge 0$ on $R$, the integral is the volume of the solid lying above $R$ and below the surface $z = f(x,y)$. When $f$ takes both signs, the integral is a signed volume: regions below the plane $z = 0$ count negatively.

Iterated integral. The inner integral $\int_c^d f(x,y)\, dy$ is computed with $x$ held constant and is called a partial integral; it produces a function of $x$. Integrating that function from $a$ to $b$ gives the iterated integral.

Equality of the two orders. For $f$ continuous on $R$,

$$ \iint_R f(x,y)\, dA = \int_a^b \!\! \int_c^d f(x,y)\, dy\, dx = \int_c^d \!\! \int_a^b f(x,y)\, dx\, dy $$

Separable integrands. If the integrand factors as $f(x,y) = g(x)h(y)$ and the region is a rectangle with constant limits, the double integral splits into a product:

$$ \int_a^b \!\! \int_c^d g(x) h(y)\, dy\, dx = \left( \int_a^b g(x)\, dx \right) \left( \int_c^d h(y)\, dy \right) $$

Both conditions are required. A factored integrand over a non-rectangular region does not split, because the inner limits then depend on the outer variable and cannot be pulled out of the outer integral.

Linearity and average value. Double integrals respect sums and constant multiples, and if $R$ is cut into two non-overlapping pieces the integrals over the pieces add. The average value of $f$ on $R$ is the integral divided by the area of $R$.

Worked examples

A plane over a rectangle, both ways

Find the volume under $z = 4 - x - y$ over $R = [0,1] \times [0,2]$. The surface stays positive on $R$, with a maximum of $4$ at the origin and a minimum of $1$ at the far corner, so expect a volume between $1 \cdot 2 = 2$ and $4 \cdot 2 = 8$.

Integrate in $y$ first, treating $x$ as a constant:

$$ \int_0^2 (4 - x - y)\, dy = \left[ 4y - xy - \frac{y^2}{2} \right]_0^2 = 8 - 2x - 2 = 6 - 2x $$

$$ \int_0^1 (6 - 2x)\, dx = \left[ 6x - x^2 \right]_0^1 = 5 $$

Now reverse the order. Integrating in $x$ first gives $\int_0^1 (4 - x - y)\, dx = 4 - \tfrac{1}{2} - y = \tfrac{7}{2} - y$, and then

$$ \int_0^2 \left( \frac{7}{2} - y \right) dy = 7 - 2 = 5 $$

The same $5$, as promised, and it sits inside the predicted range.

Checking against a Riemann sum

Estimate the same volume with four cells and midpoint samples. Cutting $[0,1]$ in half and $[0,2]$ in half gives cells of area $\tfrac{1}{2}$ with midpoints $(0.25, 0.5)$, $(0.25, 1.5)$, $(0.75, 0.5)$ and $(0.75, 1.5)$. The heights are $3.25$, $2.25$, $2.75$ and $1.75$, summing to $10$, so the estimate is $10 \cdot \tfrac{1}{2} = 5$.

The estimate is exact here, and the reason is worth knowing: for a linear integrand the midpoint sample point on each cell reads the average height of the surface over that cell, so the error cancels cell by cell. For a curved surface the midpoint rule is only an approximation.

A separable integrand

Evaluate $\iint_R x^2 \cos y \, dA$ over $R = [0,3] \times \left[0, \tfrac{\pi}{2}\right]$. The integrand is a function of $x$ times a function of $y$, and the region is a rectangle, so both conditions for splitting hold:

$$ \left( \int_0^3 x^2\, dx \right) \left( \int_0^{\pi/2} \cos y\, dy \right) = \left[ \frac{x^3}{3} \right]_0^3 \cdot \big[ \sin y \big]_0^{\pi/2} = 9 \cdot 1 = 9 $$

Doing it the long way gives the same thing: the inner integral in $y$ is $x^2 \sin y$ evaluated from $0$ to $\tfrac{\pi}{2}$, which is $x^2$, and $\int_0^3 x^2\, dx = 9$.

When the order decides whether you finish

Evaluate $\iint_R x e^{xy}\, dA$ over $R = [0,1] \times [0,2]$. Integrating in $x$ first requires integration by parts. Integrating in $y$ first does not, because $x$ is constant there and $x e^{xy}$ is exactly the derivative of $e^{xy}$ with respect to $y$:

$$ \int_0^2 x e^{xy}\, dy = \big[ e^{xy} \big]_0^2 = e^{2x} - 1 $$

$$ \int_0^1 \left( e^{2x} - 1 \right) dx = \left[ \frac{e^{2x}}{2} - x \right]_0^1 = \frac{e^2}{2} - 1 - \frac{1}{2} = \frac{e^2 - 3}{2} $$

Numerically that is about $2.19$. Sanity check the sign and size: the integrand is nonnegative on $R$, it is $0$ along the edge $x = 0$ and reaches $2e^2 \approx 14.8$ at the corner $(1,2)$, so a positive answer well below the area times the maximum is what you want.

Practice

These drills stay on rectangles, so every limit is a constant. Set up the iterated integral, do the inner integral with the outer variable frozen, then finish. Watch for integrands that factor, and use the shortcut when they do.

Quiz

Five items on double Riemann sums, evaluating iterated integrals in both orders, volumes over rectangles, and the separable case.