Difficulty: Intermediate | Prerequisites: Single-variable integration techniques, sketching regions in the xy-plane
Double integrals extend single-variable integration to functions of two variables, letting you compute volumes, averages, and accumulated quantities over 2D regions. The practical skill is evaluating them as iterated integrals, where the order of integration (dx dy or dy dx) can make the difference between a straightforward calculation and an impossible one. This section also introduces Riemann sums in two dimensions and techniques for bounding integrals without computing them exactly.
Double integral
The integral of a function f(x, y) over a two-dimensional region D, written as the double integral of f(x, y) dA over D. It gives the signed volume between the surface z = f(x, y) and the xy-plane. Think of it as adding up f's values over every tiny patch of D.
Iterated integral
A double integral evaluated as two nested single-variable integrals, one inside the other. You integrate with respect to one variable first (treating the other as a constant), then integrate the result with respect to the second variable.
Fubini's theorem
The theorem that guarantees a double integral can be computed as an iterated integral (in either order) provided f is continuous on the region. In simple terms, you can choose whether to integrate x first or y first, and you will get the same answer.
Order of integration
Whether you integrate with respect to x first (dx dy) or y first (dy dx). Changing the order requires rewriting the limits of integration by re-describing the region D.
Riemann sum (2D)
An approximation of a double integral formed by dividing the region into small rectangles, evaluating f at a sample point in each rectangle, and summing f times the area of each rectangle. The 2D trapezoid rule averages the Riemann sums using all four corners of each sub-rectangle.
Integral bounds / estimation
Techniques for showing a double integral lies between two values without computing it exactly. If m <= f(x, y) <= M on D, then m times Area(D) <= double integral of f over D <= M times Area(D).
Some iterated integrals are impossible (or very hard) to evaluate in the given order because the inner antiderivative has no closed form. The classic signal: the inner integral's variable appears inside a function you cannot antidifferentiate (e.g. e to the power x cubed, or cos of x cubed minus 1).
The fix: reverse the order of integration. This requires re-describing the region D by swapping which variable has the constant outer limits and which has the variable inner limits.
Step-by-step for reversing order:
Sketch the region D from the original limits.
Identify the boundary curves.
Re-express D with the other variable as the outer integral: find the range of that variable, and for each value of it, find the range of the inner variable.
Write the new iterated integral with the updated limits.
Problem 5a: integral from 0 to 1 of integral from y to 1 of y times e to the x cubed, dx dy. The inner integral (with respect to x) of e to the x cubed has no elementary antiderivative. Reverse the order: the region is 0 <= y <= x, 0 <= x <= 1. The new integral is integral from 0 to 1 of integral from 0 to x of y times e to the x cubed, dy dx. Now the inner integral with respect to y is straightforward.
Problem 5b follows the same logic with y cos(x cubed minus 1).
Sometimes a region D is described by a piecewise set of limits, giving a sum of two or more iterated integrals in one order.
To reverse: sketch D from both sets of limits, identify it as a single region, then write one iterated integral in the reversed order.
The key is getting the sketch right. Plot the boundary curves, find where they intersect, and read off the new limits.
Use the property: if m <= f(x, y) <= M on D, then m times Area(D) <= double integral of f dA <= M times Area(D).
For sums of functions, bound each piece separately and add.
Useful tricks for bounding:
Drop positive terms from a denominator to make the fraction larger.
Add positive terms to a denominator to make the fraction smaller.
Use symmetry: if f is odd in x and D is symmetric about the y-axis, the integral of f over D is zero.
Problem 7 splits the integrand into two pieces: xe to the 10y squared over (1 + x to the 8 + y squared) plus 1 over (1 + x squared + y to the 4). The first piece is odd in x over the symmetric rectangle, so its integral is zero. The second piece is bounded between computable constants.
Divide the region into sub-rectangles. Each sub-rectangle has area delta x times delta y.
For each sub-rectangle, evaluate f at a chosen corner (top-left, top-right, bottom-left, or bottom-right).
Sum f(sample point) times delta x times delta y over all sub-rectangles to get one Riemann sum.
The 2D trapezoid rule: average the four Riemann sums (one per corner choice). This is equivalent to weighting each data point by how many sub-rectangles share it: corner points get weight 1, edge points get weight 2, interior points get weight 4, then multiply by delta x times delta y and divide by 4.
Fubini's theorem (Type I region)
\iint_D f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dxwhere D is the region a <= x <= b, g1(x) <= y <= g2(x).
Fubini's theorem (Type II region)
\iint_D f(x,y)\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dywhere D is the region c <= y <= d, h1(y) <= x <= h2(y).
Integral bounds
m \cdot \text{Area}(D) \;\leq\; \iint_D f(x,y)\,dA \;\leq\; M \cdot \text{Area}(D)where m and M are the minimum and maximum values of f on D.
2D Riemann sum
\sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i^*, y_j^*)\,\Delta x\,\Delta yThe 2D trapezoid rule averages the four sums obtained by choosing each corner of each sub-rectangle as the sample point.
"I can always integrate in the given order." You cannot. Some iterated integrals require reversing the order because the inner antiderivative does not exist in closed form. If the inner integrand involves e to the x cubed, cos(x cubed), or similar, try the other order.
"Reversing the order just means swapping dx and dy." Swapping the differentials is necessary but not sufficient. You must also rewrite the limits of integration by re-describing the region. Sketching is essential.
"The integral of an odd function is always zero." Only when the domain is symmetric about the relevant axis. The integral of an odd-in-x function over a rectangle symmetric in x is zero, but not over an arbitrary region.
"The 2D trapezoid rule uses the midpoint." It does not. The 2D trapezoid rule averages the four corner-based Riemann sums. The midpoint rule is a different (and often better) approximation, but it is not what this problem asks for.
Double integrals are the foundation for everything that follows in the second half of Calculus III: triple integrals, surface integrals, and the big theorems (Green's, Stokes', Divergence). Getting comfortable with setting up limits and reversing order now saves a great deal of pain later.
Reversing the order of integration is one of the most commonly tested skills on MATH 215 exams. Practise sketching the region from the original limits and reading off the new ones.
Bounding integrals without evaluating them (the "show that" style of Problem 7) appears regularly as a proof-style question. Know the bounding property and how to exploit symmetry.
Riemann sum estimation with tabulated data (Problem 8) tests whether you understand the definition of the double integral, not just the mechanics of antidifferentiation.
Always sketch the region. On an exam, a quick sketch catches limit errors before they propagate.
True or false: Fubini's theorem says you can always swap the order of integration without changing the limits. False. You can swap the order, but you must rewrite the limits to describe the same region in the new order.
Fill in the blank: If f(x, y) is odd in x and D is symmetric about the y-axis, then the double integral of f over D equals ___. Answer: zero.
True or false: The integral from 0 to 1 of the integral from y to 1 of e to the x cubed dx dy cannot be evaluated in the given order because e to the x cubed has no elementary antiderivative. True.
Fill in the blank: To approximate a double integral using the 2D trapezoid rule, average the Riemann sums using ___ different corners. Answer: four.
True or false: If 0 <= f(x, y) <= 3 on a region D with area 5, then the double integral of f over D is at most 15. True.
Q: Evaluate the integral from 0 to 1 of the integral from y to 1 of y times e to the x cubed dx dy by reversing the order of integration.
A: The region is 0 <= y <= x, 0 <= x <= 1. Reversed: integral from 0 to 1 of integral from 0 to x of y times e to the x cubed dy dx. The inner integral gives (x squared / 2) times e to the x cubed. Then integrate with respect to x using the substitution u = x cubed, du = 3x squared dx. The result is (1/6)(e minus 1).
Q: A double integral over D is given as a sum of two iterated integrals in the order dx dy. Describe the general strategy for expressing it as a single iterated integral in the order dy dx.
A: Sketch the region D from both sets of limits. Identify the boundary curves and their intersections. Then describe D as a Type I region (x on the outside): find the overall x-range, and for each x, express the y-range in terms of x. Write the single iterated integral with these new limits.
Q: Let D = [-2, 2] times [-1, 1]. Show that the double integral of xe to the 10y squared over (1 + x to the 8 + y squared) over D is zero.
A: The integrand is an odd function of x (replacing x with -x negates the function). The domain D is symmetric about x = 0. By the odd-function symmetry property, the integral is zero.
Q: A swimming pool has a 12-by-12 foot square base. Depth measurements are given at a grid of points spaced 3 feet apart (a 5-by-5 grid of values). How do you estimate the volume using the average of four corner-based Riemann sums?
A: Divide the base into 16 squares, each 3 by 3 feet (area 9 square feet). Compute four Riemann sums, each using a different corner of each sub-square as the sample point (top-left, top-right, bottom-left, bottom-right). For each sum, multiply each sampled depth by 9 and add. Average the four sums. Equivalently, weight each grid point: corner points (appearing in 1 sub-square) by 1, edge points (appearing in 2) by 2, interior points (appearing in 4) by 4, sum all weighted depths, multiply by 9, and divide by 4.
Double integrals build on single-variable definite integrals and the Fundamental Theorem of Calculus. The region-sketching skills here are the same ones used in Lagrange multiplier problems from section 14.8, where you needed to identify feasible regions.
Looking ahead, triple integrals (section 15.6) extend the same ideas to three dimensions. Changing the order of integration becomes even more important there, as does switching to polar, cylindrical, or spherical coordinates (sections 15.3, 15.7, 15.8). The bounding techniques from Problem 7 reappear whenever you need to prove convergence of an improper integral.
In applications, double integrals compute mass (when f is a density), centre of mass, moments of inertia, probability over 2D distributions, and work done by a force over a region.
Double integral, iterated integral, Fubini's theorem, order of integration, reversing order of integration, change of order, Type I region, Type II region, Riemann sum 2D, trapezoid rule two dimensions, bounding integrals, integral estimation, odd function symmetry, MATH 215, Calculus III, sections 15.1 15.2, University of Michigan