What a double integral is
A single integral \(\int_a^b f(x)\,dx\) adds up \(f\) along an interval. A double integral adds up a function \(f(x, y)\) over a region \(R\) of the plane:
\[\iint_R f(x, y)\,dA\]
If \(f(x, y)\) is a height, the double integral is the volume under the surface \(z = f(x, y)\) above \(R\). If \(f = 1\), it is simply the area of \(R\).
Vertical strips (integrate \(y\) first)
\[\iint_R f\,dA = \int_{x = a}^{b}\left[\int_{y = g_1(x)}^{g_2(x)} f(x, y)\,dy\right]dx\]
Horizontal strips (integrate \(x\) first)
\[\iint_R f\,dA = \int_{y = c}^{d}\left[\int_{x = h_1(y)}^{h_2(y)} f(x, y)\,dx\right]dy\]
Work from the inside out: do the inner integral treating the other variable as a constant, then integrate the result.
Reading the limits from a picture
- Sketch the region and find where the boundary curves meet.
- Draw a typical strip. For a vertical strip, the inner limits are the lower and upper curves, as functions of \(x\).
- Outer limits are the extreme values of \(x\) that the strip sweeps over. They must be numbers.
Over a rectangle \(a \le x \le b\), \(c \le y \le d\), all four limits are constants and the order doesn't matter.
Solved examples
Example 1: Over a rectangle
EasyEvaluate \(\displaystyle\int_0^1\!\!\int_0^2 (x + y)\,dy\,dx\).
- Inner: \(\displaystyle\int_0^2 (x + y)\,dy = \Big[xy + \tfrac{y^2}{2}\Big]_0^2 = 2x + 2\).
- Outer: \(\displaystyle\int_0^1 (2x + 2)\,dx = 1 + 2\).
Answer\(3\)
Example 2: Variable inner limit
EasyEvaluate \(\displaystyle\int_0^1\!\!\int_0^x xy\,dy\,dx\).
- Inner: \(\displaystyle\int_0^x xy\,dy = x\cdot\frac{x^2}{2} = \frac{x^3}{2}\).
- Outer: \(\displaystyle\int_0^1 \frac{x^3}{2}\,dx = \frac{1}{8}\).
Answer\(\dfrac{1}{8}\)
Example 3: Between two curves
MediumEvaluate \(\displaystyle\iint_R xy\,dA\), where \(R\) is the region between \(y = x^2\) and \(y = x\).
- The curves meet at \(x = 0\) and \(x = 1\). For a vertical strip, \(y\) runs from \(x^2\) to \(x\) (see the picture above).
- Inner: \(\displaystyle\int_{x^2}^{x} xy\,dy = \frac{x}{2}\big(x^2 - x^4\big)\).
- Outer: \(\displaystyle\frac{1}{2}\int_0^1 (x^3 - x^5)\,dx = \frac{1}{2}\left(\frac{1}{4} - \frac{1}{6}\right)\).
Answer\(\dfrac{1}{24}\)
Example 4: Over a triangle
MediumEvaluate \(\displaystyle\iint_R (x + y)\,dA\) over the triangle \(x \ge 0,\ y \ge 0,\ x + y \le 1\).
- For a vertical strip, \(y\) runs from 0 to \(1 - x\), and \(x\) from 0 to 1.
- Inner: \(\displaystyle\int_0^{1-x}(x + y)\,dy = x(1 - x) + \frac{(1 - x)^2}{2} = \frac{1 - x^2}{2}\).
- Outer: \(\displaystyle\frac{1}{2}\int_0^1 (1 - x^2)\,dx = \frac{1}{2}\cdot\frac{2}{3}\).
Answer\(\dfrac{1}{3}\)
Example 5: Inner integral in x
Exam levelEvaluate \(\displaystyle\int_1^2\!\!\int_0^{\log y} e^x\,dx\,dy\).
- The inner variable is \(x\): \(\displaystyle\int_0^{\log y} e^x\,dx = e^{\log y} - 1 = y - 1\).
- Outer: \(\displaystyle\int_1^2 (y - 1)\,dy = \frac{1}{2}\).
Answer\(\dfrac{1}{2}\)
Read \(dy\,dx\) from the inside out: \(dy\) is next to the inner integral, so the inner limits belong to \(y\). Always check that the outer limits are plain numbers. If an outer limit contains a variable, the order is wrong.
Common mistakes
1. Putting the limits on the wrong variable
In \(\displaystyle\int_0^1\!\!\int_0^x \dots\,dy\,dx\), the inner limits \(0\) to \(x\) are for \(y\).
2. Choosing the wrong curve as the lower limit
Between \(y = x^2\) and \(y = x\) on \([0, 1]\), \(x^2\) is the lower curve. Test a value: at \(x = \tfrac{1}{2}\), \(x^2 = \tfrac{1}{4} < \tfrac{1}{2}\).
3. Not drawing the region
Almost every wrong limit comes from skipping the sketch.
Practice questions
Q1\(\displaystyle\int_0^2\!\!\int_0^1 (x^2 + y^2)\,dx\,dy\)
Inner: \(\tfrac{1}{3} + y^2\). Outer: \(\tfrac{2}{3} + \tfrac{8}{3} = \dfrac{10}{3}\).
Q2\(\displaystyle\int_0^1\!\!\int_0^x e^{y/x}\,dy\,dx\)
Inner: \(x(e - 1)\). Outer: \(\dfrac{e - 1}{2}\).
Q3\(\displaystyle\iint_R x^2y\,dA\) over the triangle with vertices \((0, 0), (1, 0), (1, 1)\)
\(\displaystyle\int_0^1\!\!\int_0^x x^2y\,dy\,dx = \int_0^1\frac{x^4}{2}\,dx = \frac{1}{10}\).
Q4\(\displaystyle\int_0^\pi\!\!\int_0^{\sin x} y\,dy\,dx\)
\(\displaystyle\int_0^\pi\frac{\sin^2 x}{2}\,dx = \frac{\pi}{4}\).
Q5\(\displaystyle\int_0^1\!\!\int_1^2\frac{dx\,dy}{(x + y)^2}\)
Inner: \(\dfrac{1}{1 + y} - \dfrac{1}{2 + y}\). Outer: \(\log\dfrac{2\cdot 2}{3} = \log\dfrac{4}{3}\).
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Frequently asked questions
What does a double integral represent?
The volume under the surface \(z = f(x, y)\) above the region \(R\). With \(f = 1\), it gives the area of \(R\).
Does the order of integration matter?
For a rectangle with constant limits, no. For other regions, the limits change with the order; see change of order of integration.
Where double integrals lead next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.