Double integrals

A definite integral $\int_a^b f(x)\,dx$ sums a quantity along a line. A double integral does the same job over a two-dimensional region: it sums the values of a function $f(x,y)$ over every point of a flat region $R$ , and the result reads as the volume trapped under the surface $z = f(x,y)$ and above $R$ . Double integrals open the "multiple integrals" half of MA10002's Integral Calculus unit, right after single-variable techniques such as reduction formulae and Beta/Gamma integrals. The core skill is entirely mechanical once the picture is right: sketch the region $R$ , describe it as a stack of thin strips, and turn the double integral into two ordinary single integrals done one after another — an iterated integral. The one genuinely conceptual skill is reversing the order of that stacking, which occasionally turns an unsolvable integral into an easy one. This node builds both: setting up iterated integrals over curved regions, and switching the order when the region — or the integrand — demands it.

The core insight

A double integral $\iint_R f(x,y)\,dA$ is evaluated by slicing $R$ into thin strips and integrating twice in succession — first across a strip holding the outer variable fixed, then along the strips — and choosing which variable to hold fixed first (the order of integration) can make an otherwise intractable integral solvable.

The core idea

What the double integral means. For $f(x,y)$ continuous on a region $R$ , $\iint_R f(x,y)\,dA$ is built the same way a single integral is: chop $R$ into tiny patches of area $dA$ , multiply each by the local value of $f$ , and sum. Unlike a single integral, there is no single natural order to sum in — you can sweep the patches column by column or row by row, and both give the same total (Fubini's idea). That freedom is exactly what makes reversing the order legal.

Vertically simple regions. If $R$ sits between $x = a$ and $x = b$ , with $y$ running from a lower curve $g_1(x)$ to an upper curve $g_2(x)$ , the double integral collapses into an iterated integral:

\iint_R f(x,y)\,dA = \int_a^b \left(\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\right)dx.

where: the inner integral runs over $y$ with $x$ held fixed at a chosen strip, producing a function of $x$ alone; the outer integral then sweeps that function across $x$ from $a$ to $b$ . Geometrically, the inner integral is the "area" of one vertical strip of the solid, and the outer integral adds up all the strips.

Horizontally simple regions. The same region, described the other way, has $y$ from $c$ to $d$ and $x$ running between two curves $h_1(y) \le x \le h_2(y)$ , giving

\iint_R f(x,y)\,dA = \int_c^d \left(\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\right)dy.

Changing the order of integration. Both formulas describe the same region and give the same number, but the algebra of actually carrying out the integration can differ enormously — one order may lead to an elementary antiderivative and the other to none at all. Switching orders means re-reading the same sketched region with $x$ and $y$ 's roles swapped and re-deriving the new limits from the boundary curves; it is never a matter of just swapping the integral signs symbolically.

Special case: area. Setting $f(x,y) = 1$ turns the double integral into a pure area calculator, $\text{Area}(R) = \iint_R dA$ — a fact that resurfaces constantly once you reach polar coordinates and change-of-variables problems later in the unit.

Figure

Region between the curves y equals x squared and y equals x from zero to one with a vertical strip shown running from the lower parabola up to the line

Derivation ASetting up an iterated integral from a sketch

The goal is to convert a double integral over a curved region into two ordinary single integrals by reading the strip limits directly off a sketch, using $\displaystyle\iint_R xy\,dA$ over the region bounded by $y = x^2$ and $y = x$ for $0 \le x \le 1$ as the running example, since this is the exact region the worked example below evaluates.

Step 1. Sketch $R$ and decide which curve bounds it above and which below. For $x \in [0,1]$ , compare $x$ and $x^2$ : since $0 \le x \le 1$ , $x \ge x^2$ , so the line $y = x$ lies above the parabola $y = x^2$ throughout the interval. This ordering is the single most error-prone step in the whole process — get it backwards and every limit that follows is wrong.

Step 2. Choose vertical strips because the region is naturally described by $x$ running over a fixed interval with $y$ bounded between two functions of $x$ . A strip at a fixed $x$ -value runs from the lower curve $y = x^2$ up to the upper curve $y = x$ , so the strip is described by $x^2 \le y \le x$ , with $x$ itself ranging over $0 \le x \le 1$ .

Step 3. Write the inner integral for one strip. Holding $x$ fixed, integrate over $y$ from $x^2$ to $x$ : $\displaystyle\int_{x^2}^{x} xy\,dy$ . Here $x$ is treated purely as a constant multiplier and as the (fixed) limits of $y$ ; only $y$ is the variable of integration.

Step 4. Wrap the outer integral around the strips. Since the strips themselves range over $x$ from $0$ to $1$ , the double integral becomes the iterated integral $\displaystyle\int_0^1\left(\int_{x^2}^{x} xy\,dy\right)dx$ , which is now two ordinary single-variable integrals done one after the other — exactly the setup evaluated step by step in the worked example.

Worked example

Problem. Evaluate $\displaystyle\iint_R xy\,dA$ where $R$ is the region bounded by $y = x^2$ and $y = x$ for $0 \le x \le 1$ .

Step 1. Using the sketch from the derivation above, $R$ is vertically simple: $0 \le x \le 1$ , $x^2 \le y \le x$ . The iterated integral is $\displaystyle\int_0^1\int_{x^2}^{x} xy\,dy\,dx$ .

Step 2. Evaluate the inner integral, treating $x$ as constant: $\displaystyle\int_{x^2}^{x} xy\,dy = x\left[\frac{y^2}{2}\right]_{x^2}^{x} = x\left(\frac{x^2}{2} - \frac{x^4}{2}\right) = \frac{x^3}{2} - \frac{x^5}{2}$ .

Step 3. Evaluate the outer integral: $\displaystyle\int_0^1\left(\frac{x^3}{2} - \frac{x^5}{2}\right)dx = \left[\frac{x^4}{8} - \frac{x^6}{12}\right]_0^1 = \frac18 - \frac1{12} = \frac{3}{24} - \frac{2}{24} = \frac{1}{24}$ . So $\displaystyle\iint_R xy\,dA = \boldsymbol{\dfrac{1}{24}}$ .

Competitive stretch

Problem. Change the order of integration in $\displaystyle\int_0^1\int_{y}^{1} e^{x^2}\,dx\,dy$ and evaluate it.

Step 1. As given, $R$ is $0 \le y \le 1$ , $y \le x \le 1$ — the triangle bounded by $y = 0$ , $x = 1$ , and $y = x$ . Re-describing the same triangle with $x$ as the outer variable gives $0 \le x \le 1$ , $0 \le y \le x$ , so $\displaystyle\int_0^1\int_y^1 e^{x^2}\,dx\,dy = \int_0^1\int_0^x e^{x^2}\,dy\,dx$ .

Step 2. This reversal is what makes the integral tractable: $\int e^{x^2}\,dx$ has no elementary antiderivative, but with $y$ as the inner variable, $e^{x^2}$ is a constant with respect to $y$ , giving $\displaystyle\int_0^1 e^{x^2}\Big[y\Big]_0^x dx = \int_0^1 x e^{x^2}\,dx$ .

Step 3. Substitute $u = x^2$ , $du = 2x\,dx$ : $\displaystyle\int_0^1 xe^{x^2}\,dx = \frac12\int_0^1 e^u\,du = \frac12\left[e^u\right]_0^1 = \frac{e-1}{2}$ . The original integral equals $\boldsymbol{\dfrac{e-1}{2} \approx 0.8591}$ — the standard reason to reverse order: the given order was analytically stuck.

Common traps

Sketching the region wrong, or skipping the sketch entirely — nearly every setup or reversal error traces back to a bad picture of $R$ .
Letting the "constant" variable creep into the inner integral — inside $\int \cdots dy$ , $x$ must stay frozen throughout.
Writing reversed limits as plain constants instead of re-deriving them as functions of the new outer variable from the sketch.
Assuming both orders are equally easy to integrate — some problems are set precisely because only one order has an elementary antiderivative.

Ask the tutor: if sketching a curved region or re-deriving limits after reversing order feels shaky, ask the AI tutor to walk through two or three more region-sketching examples with you step by step.

Why does the order of integration ever matter if both give the same number?

Both orders give the same numeric value for the integral — that is Fubini's theorem — but "same value" does not mean "equally easy to compute." The inner antiderivative you must find is different in each order, and sometimes one direction has no elementary antiderivative at all, as with $e^{x^2}$ . Practically, you only reverse order when the given order is stuck or the limits are awkward to write; if the original order integrates cleanly, there is no need to touch it.

How do I know whether to use vertical strips or horizontal strips?

Look at which description of the region needs fewer pieces. If a vertical line at a fixed $x$ enters and leaves $R$ through exactly two curves for the whole interval of $x$ , the region is vertically simple and you integrate $y$ first. If instead a horizontal line at fixed $y$ gives the cleaner two-curve description, use horizontal strips and integrate $x$ first. Some regions need to be split into pieces to be simple in either direction — always sketch first before deciding.

In the worked example, why is x treated as a constant in the inner integral but not in the outer one?

The inner integral $\int_{x^2}^{x} xy\,dy$ computes the "area" of one vertical strip at a specific, fixed value of $x$ — inside that single strip, $x$ genuinely does not change, only $y$ sweeps from the lower curve to the upper curve. Once that strip's contribution is found as a function of $x$ , the outer integral then lets $x$ vary, sweeping across every strip from $x=0$ to $x=1$ and adding them all up. The two integrals play different roles: one measures a single slice, the other assembles the slices.

Double integrals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads