Double Integrals over Rectangles

Q: What does a double integral measure?

When f(x, y) ≥ 0, the double integral of f over a region is the volume of the solid between the region and the surface z = f(x, y). In general it is a signed volume: parts below the xy-plane count as negative.

Q: What is an iterated integral?

Two single integrals, one inside the other. The inner integral is taken with the other variable held constant; its result is a function of the outer variable, which the outer integral then integrates.

Q: What does Fubini's theorem say?

For a continuous function on a rectangle, the double integral equals the iterated integral in either order: integrate in y first and then x, or in x first and then y.

Q: How do you find the average value of f over a rectangle?

Divide the double integral of f by the area of the rectangle.

Q: When does a double integral split into a product?

When the region is a rectangle and f(x, y) = g(x)h(y). Then the double integral is the integral of g times the integral of h.

Quick answer

The double integral of f(x, y) over a rectangle adds up f times area, the way a single integral adds up f times length. Cutting the rectangle into small pieces and standing a box on each gives a Riemann sum, and the limit is the volume under the surface. Fubini's theorem turns the double integral into two single integrals, one inside the other, taken in either order. Dividing by the area gives the average value of f.

What you'll learn

Estimate a double integral with a Riemann sum of boxes
Interpret a double integral as a signed volume
Evaluate a double integral as an iterated integral in either order
Find the average value of a function over a rectangle

Volume under a surface

A single integral $\int_a^b f(x)\,dx$ gives the area under a curve. One dimension up, a function $f(x, y) \ge 0$ defined on a rectangle $R = [a, b] \times [c, d]$ has a graph $z = f(x, y)$ , a surface. The solid between $R$ and the surface has a volume, and the double integral $\iint_R f(x, y)\,dA$ is that volume.

The solid under z = 16 − x² − 2y² over a square

Riemann sums with boxes

To estimate the volume, cut $R$ into small rectangles, pick a sample point in each, and stand a box on each rectangle, as tall as $f$ at the sample point. The volume of a box is $f(x_i^*, y_j^*)\,\Delta A$ , so the solid’s volume is about

\sum_{i}\sum_{j} f(x_i^*, y_j^*)\,\Delta A

a double Riemann sum. As the rectangles shrink, the sums approach the double integral:

\iint_R f(x, y)\,dA = \lim_{m, n \to \infty} \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_i^*, y_j^*)\,\Delta A

The limit exists for every continuous $f$ . When $f$ takes negative values, boxes below the $xy$ -plane count as negative, and the integral is a signed volume.

Four boxes estimating the volume

Example 1 — Estimating a volume

Estimate the volume under $f(x, y) = 16 - x^2 - 2y^2$ over $R = [0, 2] \times [0, 2]$ with four unit squares, first using upper right corners, then midpoints.

Each square has area $\Delta A = 1$ . The upper right corners are $(1, 1)$ , $(2, 1)$ , $(1, 2)$ and $(2, 2)$ , where $f$ equals $13$ , $10$ , $7$ and $4$ . The estimate is $13 + 10 + 7 + 4 = 34$ : the boxes in the figure.

The midpoints are $(0.5, 0.5)$ , $(1.5, 0.5)$ , $(0.5, 1.5)$ and $(1.5, 1.5)$ , where $f$ equals $15.25$ , $13.25$ , $11.25$ and $9.25$ , for a total of $49$ . The exact volume, found in Example 2, is $48$ . Here $f$ decreases in both directions, so the upper right corners give the lowest points of each square and underestimate badly; midpoints land close.

Why two single integrals are enough

Slice the solid with a plane $x = \text{constant}$ . The slice is a region under the curve $z = f(x, y)$ for $c \le y \le d$ , and its area is a single integral in $y$ , with $x$ held fixed:

A(x) = \int_c^d f(x, y)\,dy

A thin slab of thickness $\Delta x$ has volume about $A(x)\,\Delta x$ , and adding the slabs is the slicing idea behind the disk and washer methods: the volume is $\int_a^b A(x)\,dx$ . Slicing the other way, with planes $y = \text{constant}$ , gives the same volume. Fubini’s theorem says this holds for every continuous $f$ on a rectangle:

\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

The right-hand sides are iterated integrals. Work from the inside out: the inner integral treats the outer variable as a constant, the same way a partial derivative does. The order of $dy\,dx$ tells which variable goes first, and each pair of limits belongs to its own variable.

Worked examples

Average value

The average value of a function of one variable is its integral divided by the length of the interval. With two variables, divide by area:

f_{\text{avg}} = \frac{1}{\text{area}(R)}\iint_R f(x, y)\,dA

A box on $R$ as tall as $f_{\text{avg}}$ has the same volume as the solid under the surface.

Common mistakes

Practice problems

Estimate $\iint_R (x + y)\,dA$ over $R = [0, 2] \times [0, 2]$ with four unit squares and midpoints. Then find the exact value.

Answer

Estimate $8$ ; exact value $8$

Full solution

At the midpoints $(0.5, 0.5)$ , $(1.5, 0.5)$ , $(0.5, 1.5)$ and $(1.5, 1.5)$ , $x + y$ equals $1$ , $2$ , $2$ and $3$ , for a total of $8$ . Exactly, $\int_0^2 (2x + 2)\,dx = 4 + 4 = 8$ . The midpoint rule is exact for linear functions: on each square, the surface rises above the box on one side as much as it dips below on the other.
Evaluate $\displaystyle\int_0^1 \int_0^2 (x + y)\,dy\,dx$ .

Answer

$3$

Full solution

The inner integral is $\Big[xy + \tfrac{y^2}{2}\Big]_0^2 = 2x + 2$ , and $\int_0^1 (2x + 2)\,dx = 1 + 2 = 3$ .
Evaluate $\displaystyle\int_0^3 \int_0^1 6xy^2\,dy\,dx$ .

Answer

$9$

Full solution

The inner integral is $6x \cdot \tfrac{1}{3} = 2x$ , and $\int_0^3 2x\,dx = 9$ .
Evaluate $\iint_R e^{x + y}\,dA$ over $R = [0, 1] \times [0, 1]$ . Give the exact value and a decimal.

Hint

$e^{x + y} = e^x e^y$ .

Answer

$(e - 1)^2 \approx 2.9525$

Full solution

The integrand is a product, so the integral is $\left(\int_0^1 e^x\,dx\right)\left(\int_0^1 e^y\,dy\right) = (e - 1)(e - 1)$ .
Find the volume of the solid under the plane $z = 4 - x - y$ and over $R = [0, 1] \times [0, 2]$ .

Answer

$5$

Full solution

The plane stays above $z = 1$ on $R$ , so the integral is a true volume. The inner integral is $\int_0^2 (4 - x - y)\,dy = 8 - 2x - 2 = 6 - 2x$ , and $\int_0^1 (6 - 2x)\,dx = 6 - 1 = 5$ .
Find the average value of $f(x, y) = xy$ on $R = [0, 2] \times [0, 4]$ .

Answer

$2$

Full solution

The integral is $\left(\int_0^2 x\,dx\right)\left(\int_0^4 y\,dy\right) = 2 \cdot 8 = 16$ , and the area is $8$ , so the average is $2$ .
Evaluate $\displaystyle\int_1^2 \int_0^1 (2x + y)\,dx\,dy$ .

Answer

$\tfrac{5}{2}$

Full solution

The inner integral, in $x$ , is $\Big[x^2 + xy\Big]_0^1 = 1 + y$ , and $\int_1^2 (1 + y)\,dy = 1 + \tfrac{3}{2} = \tfrac{5}{2}$ .
Without integrating, decide whether $\iint_R (y - 1)\,dA$ over $R = [0, 3] \times [0, 2]$ is positive, negative or zero.

Answer

Zero

Full solution

On the half of $R$ with $y > 1$ the surface lies above the $xy$ -plane, and on the half with $y < 1$ it lies below by the same amounts, mirrored across $y = 1$ . The signed volumes cancel. Directly, $\int_0^2 (y - 1)\,dy = 0$ .
A student evaluates $\displaystyle\int_0^2 \int_1^2 x^2 y\,dx\,dy$ and expects the answer $4$ from Example 3. They get $\tfrac{14}{3}$ . Who is right?

Hint

Which variable runs from $1$ to $2$ here?

Answer

The student computed correctly, but this integral is over a different rectangle, so $4$ does not apply.

Full solution

With $dx$ inside, the limits $1$ to $2$ belong to $x$ and $0$ to $2$ belong to $y$ : the rectangle is $[1, 2] \times [0, 2]$ . The value is $\left(\int_1^2 x^2\,dx\right)\left(\int_0^2 y\,dy\right) = \tfrac{7}{3} \cdot 2 = \tfrac{14}{3}$ . Switching the order of integration means moving the limits along with their differentials.

Frequently asked questions

What does a double integral measure?