Triple Integral of xyz Over a Box: Worked Example
The problem: evaluate ∭ xyz dV over the box 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3.
Students find triple integrals intimidating mostly because of the notation: three integral signs, three differentials, six limits. The anxiety is understandable but misplaced here, because this particular integral has a beautiful shortcut. When the integrand is a product of single-variable pieces and the region is a rectangular box, the triple integral splits into a product of three ordinary one-variable integrals. This example shows both the shortcut and the long way, so you trust the shortcut.
Worked solution
We want ∭ xyz dV over the box [0, 1] × [0, 2] × [0, 3].
Step 1. Check that separation is allowed. Two conditions, both satisfied here. First, the integrand factors: xyz = (x)·(y)·(z), a product of a function of x alone, a function of y alone, and a function of z alone. Second, the region is a box: every limit is a constant, none of them depends on another variable. When both hold, the triple integral equals the product of three single integrals.
Step 2. Write the separated form.
∭ xyz dV = (∫₀¹ x dx) · (∫₀² y dy) · (∫₀³ z dz)
Step 3. Evaluate the x integral. ∫₀¹ x dx = [x²/2]₀¹ = 1/2 − 0 = 1/2. ✓
Step 4. Evaluate the y integral. ∫₀² y dy = [y²/2]₀² = 4/2 − 0 = 2. ✓
Step 5. Evaluate the z integral. ∫₀³ z dz = [z²/2]₀³ = 9/2 − 0 = 9/2. ✓
Step 6. Multiply. (1/2) · (2) · (9/2) = (1 · 2 · 9) / (2 · 2) = 18/4 = 9/2.
So ∭ xyz dV = 9/2.
The long way: iterated integration. To see that the shortcut is honest, integrate one variable at a time, inside out:
∫₀³ ∫₀² ∫₀¹ xyz dx dy dz
Inner integral (x first, treating y and z as constants): ∫₀¹ xyz dx = yz · ∫₀¹ x dx = yz · (1/2) = (yz)/2.
Middle integral (y next, z constant): ∫₀² (yz)/2 dy = (z/2) · ∫₀² y dy = (z/2) · 2 = z.
Outer integral: ∫₀³ z dz = [z²/2]₀³ = 9/2. ✓
Same answer, 9/2, as it must be. The shortcut just does the three one-variable integrations independently instead of nesting them.
Why this method works
Integration over a box with constant limits means each variable’s integration never interferes with the others: the x limits do not care what y and z are doing. And because xyz is a product, the integrand never mixes the variables either. With no mixing in the integrand and no mixing in the limits, the triple integral is genuinely three independent one-dimensional problems multiplied together. This is a special case of Fubini’s theorem, which guarantees that iterated integrals over a box can be computed in any order.
Common mistakes
- Splitting when the region is not a box. If a limit depends on another variable, for example 0 ≤ y ≤ x, the y integral cannot be separated from x. Splitting is only legal when every limit is constant.
- Splitting when the integrand does not factor. For x + y + z or sin(xy), there is no product structure, so you must integrate term by term or iterate properly. Check the factorization before you split.
- Mixing up which limit belongs to which variable. The x integral uses the x limits (0 to 1), the y integral uses the y limits (0 to 2), the z integral uses the z limits (0 to 3). Pairing ∫₀² x dx by accident gives a wrong factor of 2 instead of 1/2.
Check (never skip)
Two independent checks agree. First, the iterated integration above produced 9/2, matching the separated product exactly. ✓ Second, an average-value sanity check: the box has volume 1 · 2 · 3 = 6, so the average value of xyz over the box is (9/2)/6 = 3/4. But the average of a product of independent factors is the product of the averages: average of x on [0, 1] is 1/2, average of y on [0, 2] is 1, average of z on [0, 3] is 3/2, and (1/2)(1)(3/2) = 3/4. ✓ Both roads lead to the same place, so 9/2 is solid.
Frequently Asked Questions
Q: When can I split a triple integral into a product of single integrals?
A: Exactly when two things hold: the integrand factors as f(x)·g(y)·h(z), and the region is a rectangular box with constant limits. If either fails, iterate the integrals in an order the limits allow instead.
Q: Does the order of integration matter here?
A: No. For a continuous integrand over a box, Fubini’s theorem says every order gives the same answer. We did dx dy dz; dz dy dx or any other order also yields 9/2.
Q: What if the integrand were x + y + z instead of xyz?
A: Then it does not factor, so integrate term by term: ∭x dV = (1/2)(2)(3) = 3, ∭y dV = (1)(2)(3) = 6, ∭z dV = (1)(2)(9/2) = 9, for a total of 18. Sums split; products factor.
Try it with steps on our triple integral calculator.