Triple Integral Calculator

Enter the constant k and the box bounds to evaluate the triple integral.

How it is calculated

Because the integrand is the same constant everywhere, the three nested integrals reduce to the constant times the box dimensions: ∭ k dV = k · (x1 − x0) · (y1 − y0) · (z1 − z0). Each factor in brackets is the edge length of the box along one axis, and their product is the volume V. All seven fields are plain numbers; if x, y and z are lengths in metres, the volume is in cubic metres and the integral carries the unit of k times m³. Each upper bound must be strictly greater than its lower bound, otherwise there is no result. k may be negative or zero, which gives a negative or zero integral. Results are shown with four decimals. The tool does not integrate functions of x, y or z.

Worked example

Let k = 2.5 over the box x from 0 to 2, y from −1 to 3, z from 1 to 4. The edge lengths are 2 − 0 = 2, 3 − (−1) = 4 and 4 − 1 = 3. The volume is 2 × 4 × 3 = 24.0000. Multiply by the constant: 2.5 × 24 = 60.0000. If k were a density in kg/m³ and the edges in metres, this would be a mass of 60 kg.

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Common mistakes

  • Entering the bounds in the wrong order, for example x0 = 2 and x1 = 0. The tool gives no result, because each upper limit must exceed its lower limit.
  • Treating a negative lower bound as a length. The edge is x1 − x0, so from −1 to 3 it is 4, not 3 or 2.
  • Typing a function such as x²y instead of a constant in k. Only a constant integrand works; a variable integrand needs real integration, not k times volume.
  • Mixing units between axes, such as x in centimetres and z in metres. The volume then has no meaning and k times it is wrong.
  • Forgetting that the answer carries the unit of k times volume; it is not a volume itself unless k = 1.

About this calculator

A triple integral extends the idea of an integral to three dimensions, summing a function’s values over a solid region — commonly a rectangular box defined by bounds [x0, x1] × [y0, y1] × [z0, z1]. When the integrand is a constant k rather than a function of x, y, and z, the integral simplifies dramatically: it equals k times the volume of the box, i.e. k × (x1−x0) × (y1−y0) × (z1−z0).

This special case shows up constantly in multivariable calculus courses as a stepping stone before tackling variable integrands, and in physics and engineering it directly computes quantities like the mass of a uniform-density block (density × volume) or a constant charge density’s total charge.

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