Double Integrals

Evaluate iterated double integrals of a polynomial over a rectangular region.

A single integral adds up infinitely many thin slices to get the area under a curve. A double integral does the same thing one dimension up: it adds up infinitely many thin vertical slabs to get the volume under a surface z = f(x, y).

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Formally, it's the limit of a sum over a fine grid partition of the rectangle into m x n small patches, each of area ΔA:

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

Double integral as a limit of Riemann sums

$\iint_R$ Double integral over the region R — add up contributions across the whole 2D region.
$f(x,y)$ The integrand: the function whose values are being summed.
$dA$ An infinitesimal patch of area — the 2D equivalent of dx.
$\lim_{m,n\to\infty}$ Let the grid get finer in both directions at once: more columns (m) and more rows (n).
$\sum_{i=1}^{m}$ Sum over the m columns of the grid — the x-direction.
$\sum_{j=1}^{n}$ Sum over the n rows of the grid — the y-direction.
$f(x_{ij}^*, y_{ij}^*)$ The function's value at a sample point inside grid cell (i, j).
$\Delta A$ The area of each grid cell — the same for every cell, since the grid is uniform. $= \Delta x\,\Delta y$
$\Delta x$ The width of each grid column. $= \dfrac{b-a}{m}$
$\Delta y$ The height of each grid row. $= \dfrac{d-c}{n}$

A general region doesn't tile so neatly, so its Riemann sum is usually written as a single sum over irregular patches. A rectangle tiles into a uniform grid, so the same idea splits cleanly into the double sum above — one index for columns, one for rows.

That definition works over any region, but it isn't something you evaluate directly. Over a rectangle, it reduces to two ordinary single-variable integrals done back to back:

This is an iterated integral — the exact form used in the practice problems below:

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

Iterated integral (rectangular region)

$\int_c^d$ Outer integral: sweep y from c to d.
$\int_a^b$ Inner integral: sweep x from a to b, with y momentarily fixed.
$f(x,y)$ The integrand, evaluated at each (x, y) point as the sweep proceeds.
$dx$ Integrate with respect to x first, holding y constant.
$dy$ Then integrate that result with respect to y.

Over a rectangle [a, b] x [c, d], the order of integration doesn't change the result:

\iint_{[a,b]\times[c,d]} 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\,dx

Fubini's theorem (rectangular region)

$\iint_{[a,b]\times[c,d]}$ Double integral over the rectangle formed by x in [a,b] and y in [c,d].
$\times$ Cartesian product: pairs every x in [a,b] with every y in [c,d] to form the rectangle.
$f(x,y)$ The same integrand throughout — Fubini's point is that its integral doesn't depend on which variable you integrate first.
$dA$ The area element from the double-integral definition, which the iterated integral reproduces exactly.
$\int_c^d$ Outer integral: sweep y from c to d.
$\int_a^b$ Inner integral: sweep x from a to b, with y momentarily fixed.
$dx$ Integrate with respect to x, holding y constant.
$dy$ Integrate with respect to y, holding x constant.

Fubini's theorem guarantees both orders agree when the region is a rectangle (or more generally, when neither pair of bounds depends on the other variable) and f is continuous on it. That's also what the visualization below is demonstrating: sweep along x, then along y, and the running sum converges to the same exact total either way. Once a region has curved or variable boundaries, swapping the order of integration means swapping the limits too, not just relabeling dx and dy.

Worked example

\iint_R 6x^2y\,dA, \quad R = [0,2]\times[1,3]

$\int_1^3 6x^2y\,dy = 6x^2\cdot\dfrac{y^2}{2}\Big|_1^3 = 3x^2(9-1) = 24x^2$
Inner integral first, with respect to y, holding x frozen — the 6x^2 factor is constant here, so it just comes along for the ride. Once evaluated, the y is gone: the inner integral eats one variable and hands back a function of x alone.
$\int_0^2 24x^2\,dx = 8x^3\Big|_0^2 = 64$
Outer integral, with respect to x, finishes the job.
$\int_0^2 6x^2y\,dx = 2yx^3\Big|_0^2 = 16y$
Now do it in the other order, because Fubini says you may — inner integral with respect to x this time, holding y fixed.
$\int_1^3 16y\,dy = 8y^2\Big|_1^3 = 64$
Then the outer integral with respect to y. Same answer — that's not a coincidence, it's the theorem.
$\left(\int_0^2 6x^2\,dx\right)\left(\int_1^3 y\,dy\right) = (16)(4) = 64$
A third way: 6x^2y separates into g(x)h(y), so the double integral is just the product of two single integrals.

Answer:

64

How to solve one of these

1.
Identify the region. Read off $a, b$ (the $x$ -bounds) and $c, d$ (the $y$ -bounds) from $R = [a,b] \times [c,d]$ .
2.
Choose the order. Fubini says either works. Pick the one that makes the inner integral easier — or if the integrand factors as $g(x)h(y)$ , skip ahead to step 6.
3.
Write the iterated integral with matched bounds. Inner differential pairs with inner bounds — $dy$ inner means the $y$ -bounds go on the inner sign. This is where most errors happen.
4.
Evaluate the inner integral. Treat the other variable as a constant — it factors out exactly like a number would. Substitute the limits. The inner variable disappears.
5.
Evaluate the outer integral. You now have a single-variable integral. Do it. Substitute the limits. You have a number.
6.
(Shortcut) If $f(x,y) = g(x)h(y)$ and $R$ is a rectangle, compute $\left(\int_a^b g\right)\left(\int_c^d h\right)$ instead. Rectangle only.
7.
Check. Redo it in the opposite order. Fubini guarantees agreement, so a mismatch means you made an arithmetic error — not that the theorem failed. Step 7 is the one no textbook lists and every student needs: it turns Fubini from a fact you memorize into a tool you use.

\int_{c}^{d}\int_{a}^{b} f(x,y)\,dx\,dy

Slice the region into tiny rectangles, multiply each by the function's value there, and add every last one up.

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