Difficulty: Intermediate
Prerequisites: Single-variable integration (definite integrals, substitution, integration by parts). Familiarity with functions of two variables and rectangular regions in the xy-plane.
Double integrals extend the idea of a single integral to functions of two variables. Where a single integral gives the area under a curve, a double integral gives the volume under a surface over a flat region. This section focuses on the simplest case: integrating over a rectangle in the xy-plane, which lets you build intuition before tackling more complex regions. If you are comfortable evaluating definite integrals of one variable, you already have the core skill; the new idea is doing it twice, once for each variable.
A double integral over a rectangle computes the "signed volume" between a surface z = f(x, y) and the xy-plane. Fubini's Theorem tells you that you can evaluate it by doing two single-variable integrals in sequence (an iterated integral), and the order you integrate in does not matter, as long as f is continuous on the rectangle.
Double integral
The integral of a function f(x, y) over a two-dimensional region R, written as the double integral of f(x, y) dA over R. It represents the signed volume between the surface z = f(x, y) and the xy-plane.
In simple terms, it is the 2D version of a definite integral: instead of adding up thin slices along one axis, you are adding up tiny rectangular patches across a flat region.
Rectangular region (R)
A region defined by R = [a, b] x [c, d], meaning a <= x <= b and c <= y <= d. Think of it as a box on the xy-plane whose sides are parallel to the axes.
Iterated integral
A double integral broken into two nested single-variable integrals. You integrate with respect to one variable first (treating the other as a constant), then integrate the result with respect to the second variable.
In simple terms, it means "do one integral inside another," the same way you would evaluate nested parentheses from the inside out.
Fubini's Theorem
If f(x, y) is continuous on a rectangle R = [a, b] x [c, d], then the double integral of f over R equals the iterated integral in either order: integrate x first then y, or y first then x. The result is the same.
Think of it as permission to choose whichever order of integration makes the calculation easier.
Average value of f over R
The average value of a function f(x, y) over a rectangle R is the double integral of f over R divided by the area of R. It generalises the single-variable average-value formula to two dimensions.
The region R is always R = [a, b] x [c, d].
The double integral of f(x, y) dA over R means "add up the values of f over every tiny patch in R."
dA represents a small area element; on a rectangle it equals dx dy (or dy dx).
If f(x, y) is continuous on R = [a, b] x [c, d], then:
Integrate x first: integral from c to d of (integral from a to b of f(x, y) dx) dy
Integrate y first: integral from a to b of (integral from c to d of f(x, y) dy) dx
Both orders give the same answer.
The inner integral is evaluated first, treating the other variable as a constant. The result is a function of the remaining variable, which you then integrate.
Integrating x first (inner integral over x, outer over y):
Inner: integral from 0 to 2 of x e^y dx = (x^2 / 2) e^y evaluated from 0 to 2 = 2e^y
Outer: integral from 0 to 1 of 2e^y dy = 2e^y evaluated from 0 to 1 = 2e - 2
Integrating y first (inner integral over y, outer over x):
Inner: integral from 0 to 1 of x e^y dy = x e^y evaluated from 0 to 1 = x(e - 1)
Outer: integral from 0 to 2 of x(e - 1) dx = (e - 1)(x^2 / 2) evaluated from 0 to 2 = 2e - 2
Both orders yield 2e - 2, confirming Fubini's Theorem.
Example 1: Integral of e^(x - y) over R, where R has x from 0 to some bound, y from 1 to 3.
Inner integral over x gives e^(x - y) terms.
After evaluating, the answer involves differences of exponentials: -1/e + e^3 + 1 - 1/e^3.
Example 2: Integral of (x^2 + y^2) over a rectangle with x from -5 to 5, y from 1 to 5.
Inner integral over x: (x^3/3 + xy^2) evaluated from -5 to 5.
Result after full evaluation: 520/3.
Example 3: Integral of 2 dA over R = {(x, y): 2 <= x <= 5, 3 <= y <= 5}.
This computes a volume: integral from 3 to 5 of (integral from 2 to 5 of 2 dx) dy.
Inner: 2 times (5 - 2) = 6. Outer: 6 times (5 - 3) = 12.
Geometrically, this is the volume of a box with base 3 x 2 and height 2.
Example 4: Integral of (1 + 2xy) over R = [0, 3] x [0, 1].
Inner integral over x: x + x^2 y evaluated from 0 to 3 = 3 + 9y.
Outer: integral from 0 to 1 of (3 + 9y - (0)) ... the final result is 6.
Formula: Avg = (1 / Area(R)) times the double integral of f over R.
Area(R) = (b - a)(d - c).
Example: f(x, y) = x^2 y + 2x over R = [-2, 1] x [-2, 2].
Area(R) = 3 times 4 = 12.
After evaluating the double integral, the result is -12.
Average value = -12 / 12 = -1.
Example: Integral of 1/(x + 2y)^2 over a rectangle with x from 0 to 1, y from 1 to 2.
For the inner integral over x, use substitution u = x + 2y, du = dx.
This converts the inner integral into integral of 1/u^2 du, which gives -1/u.
After evaluating and integrating the remaining expression over y, the answer is (1/2) ln(6/5).
Problem 1: Double integral of 2x e^(-xy) over R = [-1, 1] x [0, 1].
Integrate y first: inner integral gives (-2e^(-xy)) evaluated from 0 to 1 = -2e^(-x) + 2.
Outer: integral from -1 to 1 of (-2e^(-x) + 2) dx = 2(1/e - e + 2).
Problem 2: Volume under z = 3x^2 + 2y^2 over R = [-1, 1] x [-1, 2].
Inner integral over x: (x^3 + 2y^2 x) evaluated from -1 to 1 = 2 + 4y^2 - (-1 - 2y^2) = 2 + 4y^2.
Outer: integral from -1 to 2 of (2 + 4y^2) dy = 2y + (4/3)y^3 evaluated from -1 to 2 = 88/3.
\iint_R f(x,y)\,dA = \int_c^d \int_a^b f(x,y)\,dx\,dy = \int_a^b \int_c^d f(x,y)\,dy\,dxwhere R = [a, b] x [c, d] and f is continuous on R.
\text{Avg}(f) = \frac{1}{\text{Area}(R)} \iint_R f(x,y)\,dAwhere Area(R) = (b - a)(d - c).
Double integrals over rectangles are used in engineering and physics to compute quantities like mass (when density varies across a flat plate), total charge on a rectangular conductor, and average temperature over a flat surface. If you know how temperature varies as a function T(x, y) across a rectangular heating element, the average value formula gives you the mean temperature directly.
Students often think the order of integration changes the answer. It does not, provided f is continuous on the rectangle. Fubini's Theorem guarantees both orders yield the same result.
Students sometimes forget to treat the "other" variable as a constant when evaluating the inner integral. If you are integrating with respect to x first, y is a constant in that step, not a variable.
Confusing the limits: the inner integral's limits correspond to the variable you are integrating, and the outer integral's limits correspond to the remaining variable. Mixing them up is one of the most common setup errors.
Thinking that the double integral of a constant, say c, over R equals just c. It actually equals c times the area of R.
Fubini's Theorem is almost always tested. You need to state the continuity condition and show you can evaluate the integral in both orders.
Average value problems appear frequently. Remember to divide by the area of R, not by the lengths of the sides separately.
Exam questions often ask you to "find the volume" under a surface over a rectangle. This is simply the double integral of f(x, y) dA.
Substitution within the inner integral (as in the 1/(x + 2y)^2 example) is a common exam technique. Be ready to substitute and adjust limits for the inner variable only.
True or False: The double integral of f(x, y) over R = [a, b] x [c, d] can only be evaluated by integrating x first.
False. Fubini's Theorem allows either order.
Fill in the blank: The average value of f over R equals the double integral of f over R divided by ______.
The area of R.
True or False: If f(x, y) = 5 and R is a 3 x 4 rectangle, then the double integral of f over R is 5.
False. It is 5 times 12 = 60.
Fill in the blank: When evaluating the inner integral with respect to x, the variable y is treated as a ______.
Constant.
True or False: Fubini's Theorem requires f to be continuous on R.
True.
Q: Evaluate the double integral of (1 + 2xy) dA over R = [0, 3] x [0, 1].
A: Integrate x first. Inner: integral from 0 to 3 of (1 + 2xy) dx = (x + x^2 y) from 0 to 3 = 3 + 9y. Outer: integral from 0 to 1 of (3 + 9y) dy = (3y + 9y^2/2) from 0 to 1 = 3 + 9/2 = 15/2. The answer is 15/2. (Note: the lecture notes record a slightly different simplification path arriving at 6; check your bounds carefully.)
Q: Find the volume under z = 2 over the rectangle R = {(x, y): 2 <= x <= 5, 3 <= y <= 5}.
A: The double integral of 2 dA = 2 times Area(R) = 2 times (3)(2) = 12.
Q: Compute the double integral of x e^y dA over R = [0, 2] x [0, 1]. Verify using both orders of integration.
A: Integrating x first: inner = (x^2/2)e^y from 0 to 2 = 2e^y. Outer = 2e^y from 0 to 1 = 2e - 2. Integrating y first: inner = x e^y from 0 to 1 = x(e - 1). Outer = (e - 1)(x^2/2) from 0 to 2 = 2(e - 1) = 2e - 2. Both give 2e - 2.
Q: Find the average value of f(x, y) = x^2 y + 2x over R = [-2, 1] x [-2, 2].
A: Area(R) = 3 times 4 = 12. The double integral evaluates to -12. Average = -12/12 = -1.
Q: Evaluate the double integral of 2x e^(-xy) dA over R = [-1, 1] x [0, 1].
A: Integrate y first. Inner: integral from 0 to 1 of 2x e^(-xy) dy = -2e^(-xy) from 0 to 1 = -2e^(-x) + 2. Outer: integral from -1 to 1 of (-2e^(-x) + 2) dx = (2e^(-x) + 2x) from -1 to 1 = (2e^(-1) + 2) - (2e^(1) - 2) = 2(1/e - e + 2).
This material connects directly to double integrals over general (non-rectangular) regions, which you will study next. There, the limits of the inner integral become functions of the outer variable rather than constants. The same Fubini's Theorem idea applies, but choosing the right order of integration becomes more important because one order may be much simpler than the other.
Double integrals also connect to triple integrals (adding a third variable for volume in 3D), and later to surface integrals and vector calculus topics like flux. The average-value formula generalises further in those settings.
Double integral, iterated integral, Fubini's Theorem, Fubini theorem, rectangular region, double integral over a rectangle, volume under a surface, average value of a function of two variables, integrating in either order, switching order of integration, dA, dx dy, dy dx, multivariable calculus, integral calculus, MATH 203, double integration examples, double integral practice problems