What a triple integral is
Before you start: triple integrals extend double integrals by one more variable.
A triple integral \(\displaystyle\iiint_V f(x, y, z)\,dV\) adds up \(f\) over a solid region \(V\). If \(f\) is a density, it gives the mass. With \(f = 1\), it gives the volume.
\[\iiint_V f\,dV = \int_{x = a}^{b}\int_{y = g_1(x)}^{g_2(x)}\int_{z = h_1(x, y)}^{h_2(x, y)} f(x, y, z)\,dz\,dy\,dx\]
The innermost limits may depend on both outer variables; the middle limits on the outer one only; the outer limits are numbers.
Setting up the limits
- Inner (\(z\)): from the bottom surface to the top surface of the solid.
- Project the solid onto the \(xy\)-plane to get a region \(R\).
- Middle and outer: set up \(R\) exactly as for a double integral.
For spheres, cones and cylinders, switch to spherical or cylindrical coordinates.
Solved examples
Example 1: Over a box
EasyEvaluate \(\displaystyle\int_0^1\!\!\int_0^2\!\!\int_0^3 xyz\,dz\,dy\,dx\).
- All limits are constant and the integrand is a product, so the integral splits: \(\displaystyle\int_0^1 x\,dx\cdot\int_0^2 y\,dy\cdot\int_0^3 z\,dz = \frac{1}{2}\cdot 2\cdot\frac{9}{2}\).
Answer\(\dfrac{9}{2}\)
Example 2: Volume of a tetrahedron
MediumEvaluate \(\displaystyle\int_0^1\!\!\int_0^{1-x}\!\!\int_0^{1-x-y} dz\,dy\,dx\).
- Inner: \(1 - x - y\). Middle: \(\displaystyle\int_0^{1-x}(1 - x - y)\,dy = \frac{(1 - x)^2}{2}\).
- Outer: \(\displaystyle\int_0^1\frac{(1 - x)^2}{2}\,dx = \frac{1}{6}\).
Answer\(\dfrac{1}{6}\), the volume of the tetrahedron \(x + y + z \le 1\) in the first octant
Example 3: Over the same tetrahedron
MediumEvaluate \(\displaystyle\int_0^1\!\!\int_0^{1-x}\!\!\int_0^{1-x-y} x\,dz\,dy\,dx\).
- Inner: \(x(1 - x - y)\). Middle: \(\dfrac{x(1 - x)^2}{2}\).
- Outer: \(\displaystyle\frac{1}{2}\int_0^1 x(1 - x)^2\,dx = \frac{1}{2}\cdot\frac{1}{12}\).
Answer\(\dfrac{1}{24}\)
Example 4: Exponential integrand
Exam levelEvaluate \(\displaystyle\int_0^a\!\!\int_0^x\!\!\int_0^{x + y} e^{x + y + z}\,dz\,dy\,dx\).
- Inner: \(\Big[e^{x + y + z}\Big]_0^{x + y} = e^{2(x + y)} - e^{x + y}\).
- Middle: \(\Big[\tfrac{1}{2}e^{2x + 2y} - e^{x + y}\Big]_0^x = \tfrac{1}{2}e^{4x} - \tfrac{3}{2}e^{2x} + e^x\).
- Outer: \(\displaystyle\int_0^a\Big(\tfrac{1}{2}e^{4x} - \tfrac{3}{2}e^{2x} + e^x\Big)dx = \frac{e^{4a}}{8} - \frac{3e^{2a}}{4} + e^a - \left(\frac{1}{8} - \frac{3}{4} + 1\right)\).
Answer\(\dfrac{e^{4a}}{8} - \dfrac{3e^{2a}}{4} + e^a - \dfrac{3}{8}\)
Do one integral at a time and write the intermediate result on its own line. Triple integrals are rarely hard, but three layers of substitution give three chances for a slip.
Common mistakes
1. Outer limits that contain variables
Only the inner limits may depend on other variables. The outermost limits must be numbers.
2. Order of \(dz\,dy\,dx\) not matching the limits
The innermost differential matches the innermost limits.
3. Using Cartesian coordinates for a sphere
It works but is very messy. Use spherical coordinates.
Practice questions
Q1\(\displaystyle\int_0^1\!\!\int_0^1\!\!\int_0^1 (x + y + z)\,dx\,dy\,dz\)
Each term contributes \(\tfrac{1}{2}\): total \(\dfrac{3}{2}\).
Q2\(\displaystyle\int_0^2\!\!\int_0^x\!\!\int_0^y dz\,dy\,dx\)
\(\displaystyle\int_0^2\!\!\int_0^x y\,dy\,dx = \int_0^2\frac{x^2}{2}\,dx = \frac{4}{3}\).
Q3\(\displaystyle\int_0^1\!\!\int_0^{1-x}\!\!\int_0^{1-x-y} z\,dz\,dy\,dx\)
By symmetry with Example 3: \(\dfrac{1}{24}\).
Q4\(\displaystyle\iiint x^2\,dV\) over the box \(0 \le x \le 1\), \(0 \le y \le 2\), \(0 \le z \le 3\)
\(\dfrac{1}{3}\cdot 2\cdot 3 = 2\).
Q5\(\displaystyle\int_0^{\pi/2}\!\!\int_0^{a\sin\theta}\!\!\int_0^{(a^2 - r^2)/a} r\,dz\,dr\,d\theta\)
Inner: \(\dfrac{r(a^2 - r^2)}{a}\). Middle: \(a^3\left(\dfrac{\sin^2\theta}{2} - \dfrac{\sin^4\theta}{4}\right)\). Outer: \(a^3\left(\dfrac{\pi}{8} - \dfrac{3\pi}{64}\right) = \dfrac{5\pi a^3}{64}\).
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Frequently asked questions
What does a triple integral represent?
The total of \(f\) over a solid. With \(f = 1\) it is the volume; with \(f\) a density it is the mass.
Which order should I integrate in?
Whichever makes the limits simplest, usually \(z\) first (from the bottom surface to the top surface).
Where triple integrals lead next
← All 33 Calculus topics
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.