Triple Integrals

A triple integral adds up the values of f(x, y, z) over a solid region E in space. You evaluate triple integrals as iterated integrals, one variable at a time from the inside out, usually z first. With f = 1, the triple integral gives the volume of E. With a density f, it gives the mass.

Before this lesson: Double Integrals over General Regions, Applications of Double Integrals and 3D Coordinate System

From double integrals to triple integrals

A double integral adds up f over a flat region. A triple integral adds up f over a solid. Cut the solid into tiny boxes of volume ΔV, multiply each by the value of f there, and add. As the boxes shrink, the sum approaches the triple integral.

To compute it, stand at a point (x, y) of the floor region D and look straight up. A vertical column rises from a bottom surface z = u₁(x, y) to a top surface z = u₂(x, y). Integrate f up that column in z. Then do a double integral of the result over D.

Triple integral formula

\iiint_E f(x, y, z)\,dV = \iint_D \Big[\int_{u_1(x, y)}^{u_2(x, y)} f(x, y, z)\,dz\Big]\,dA

\text{Volume of } E = \iiint_E 1\,dV

1. Sketch the solid. Find the bottom surface z = u₁(x, y) and the top surface z = u₂(x, y).
2. Find D, the shadow of the solid on the xy-plane. Describe it as a Type I or Type II region.
3. Integrate in z, then do the double integral over D from the inside out.
4. For a box [a, b] × [c, d] × [p, q], all six limits are numbers, and any order works.

Triple integrals examples

Example 1: A box

Evaluate \iiint_B xyz\,dV, where B = [0, 1] × [0, 2] × [0, 3].

\int_0^3 xyz\,dz = xy\Big[\dfrac{z^2}{2}\Big]_0^3

= \dfrac{9xy}{2}

\int_0^2 \dfrac{9xy}{2}\,dy = \dfrac{9x}{2}\Big[\dfrac{y^2}{2}\Big]_0^2

= 9x

\int_0^1 9x\,dx = \Big[\dfrac{9x^2}{2}\Big]_0^1

= \dfrac{9}{2}

Answer: 9/2.

Example 2: Volume of a tetrahedron

Find the volume of the solid in the first octant under the plane 2x + y + z = 4.

z = 4 - 2x - y

0 \le x \le 2, \quad 0 \le y \le 4 - 2x

V = \int_0^2 \int_0^{4 - 2x} \int_0^{4 - 2x - y} 1\,dz\,dy\,dx

\int_0^{4 - 2x - y} 1\,dz = 4 - 2x - y

\int_0^{4 - 2x} (4 - 2x - y)\,dy = \dfrac{(4 - 2x)^2}{2}

= 8 - 8x + 2x^2

\int_0^2 (8 - 8x + 2x^2)\,dx = \Big[8x - 4x^2 + \dfrac{2x^3}{3}\Big]_0^2

= 16 - 16 + \dfrac{16}{3}

= \dfrac{16}{3}

Answer: the volume is 16/3. Check with the volume of a pyramid: (1/3)(base area 4)(height 4) = 16/3.

Example 3: A function over a tetrahedron

Evaluate \iiint_E z\,dV, where E is the tetrahedron in the first octant under x + y + z = 1.

\int_0^{1 - x - y} z\,dz = \dfrac{(1 - x - y)^2}{2}

\int_0^{1 - x} \dfrac{(1 - x - y)^2}{2}\,dy = \Big[-\dfrac{(1 - x - y)^3}{6}\Big]_0^{1 - x}

= \dfrac{(1 - x)^3}{6}

\int_0^1 \dfrac{(1 - x)^3}{6}\,dx = \Big[-\dfrac{(1 - x)^4}{24}\Big]_0^1

= \dfrac{1}{24}

Answer: 1/24.

Example 4: A solid with a curved side

Find the volume of the solid under the plane z = y and above the region D in the xy-plane between y = x² and y = 1.

x^2 = 1 \Rightarrow x = \pm 1

V = \int_{-1}^{1} \int_{x^2}^{1} \int_0^{y} 1\,dz\,dy\,dx

\int_0^{y} 1\,dz = y

\int_{x^2}^{1} y\,dy = \dfrac{1 - x^4}{2}

\int_{-1}^{1} \dfrac{1 - x^4}{2}\,dx = \dfrac{1}{2}\Big[x - \dfrac{x^5}{5}\Big]_{-1}^{1}

= \dfrac{1}{2}\Big(2 - \dfrac{2}{5}\Big)

= \dfrac{4}{5}

Answer: the volume is 4/5.

Example 5: Mass and balance height of a block

A cube of material fills [0, 2] × [0, 2] × [0, 2], in meters. It gets lighter toward the top: its density is ρ = 3 − z kilograms per cubic meter. Find its mass and the height of its center of mass.

m = \int_0^2 \int_0^2 \int_0^2 (3 - z)\,dz\,dy\,dx

\int_0^2 (3 - z)\,dz = \Big[3z - \dfrac{z^2}{2}\Big]_0^2

= 6 - 2

= 4

m = 4 \cdot 4

= 16

M_{xy} = \int_0^2 \int_0^2 \int_0^2 z(3 - z)\,dz\,dy\,dx

\int_0^2 (3z - z^2)\,dz = \Big[\dfrac{3z^2}{2} - \dfrac{z^3}{3}\Big]_0^2

= 6 - \dfrac{8}{3}

= \dfrac{10}{3}

M_{xy} = 4 \cdot \dfrac{10}{3}

= \dfrac{40}{3}

\bar{z} = \dfrac{40/3}{16}

= \dfrac{5}{6}

Answer: the mass is 16 kg, and the center of mass is 5/6 ≈ 0.83 m above the floor. That is below the middle height of 1 m, because the bottom is heavier.

Common mistakes

Related methods

Other orders of integration

There are six orders, such as dz dy dx or dx dz dy. Choose the one where the inner limits are easy to read from the solid. Project onto the plane that gives the simplest shadow.

Round solids

For cylinders, cones and spheres, the limits are simpler in cylindrical or spherical coordinates. Start with triple integrals in cylindrical coordinates; spherical coordinates come right after.

Triple integrals as double integrals

When f = 1, the inner integral is top − bottom, so the volume is ∬_D (u₂ − u₁) dA. That is the volume under a surface from double integrals.

Practice problems

1. Evaluate \int_0^1 \int_0^1 \int_0^1 (x + y + z)\,dz\,dy\,dx.
2. Evaluate \int_0^2 \int_0^1 \int_0^3 x y^2 z\,dz\,dy\,dx.
3. Evaluate \int_0^1 \int_0^{x} \int_0^{x + y} 1\,dz\,dy\,dx.
4. Find the volume of the tetrahedron in the first octant under x + y + z = 3.
5. Evaluate \iiint_E y\,dV, where E is given by 0 ≤ x ≤ 2, 0 ≤ y ≤ x and 0 ≤ z ≤ y.
6. Find the volume of the solid between the paraboloid z = x² + y² and the plane z = 4.
7. Evaluate \iiint_B e^{x + y + z}\,dV over the cube B = [0, 1] × [0, 1] × [0, 1].
8. Find the volume of the solid under z = 4 − y² and above the rectangle 0 ≤ x ≤ 3, −2 ≤ y ≤ 2.
9. Find the average value of f(x, y, z) = x + y + z over the cube [0, 2] × [0, 2] × [0, 2].
10. A wedge-shaped block fills the tetrahedron in the first octant under x + y + z = 1, in meters. Its density is 2400z kilograms per cubic meter. Find its mass.

\int_0^1 (x + y + z)\,dz = x + y + \dfrac{1}{2}

\int_0^1 \Big(x + y + \dfrac{1}{2}\Big)dy = x + 1

\int_0^1 (x + 1)\,dx = \dfrac{3}{2}

\Big(\int_0^2 x\,dx\Big)\Big(\int_0^1 y^2\,dy\Big)\Big(\int_0^3 z\,dz\Big)

= 2 \cdot \dfrac{1}{3} \cdot \dfrac{9}{2}

= 3

\int_0^{x + y} 1\,dz = x + y

\int_0^{x} (x + y)\,dy = x^2 + \dfrac{x^2}{2}

= \dfrac{3x^2}{2}

\int_0^1 \dfrac{3x^2}{2}\,dx = \dfrac{1}{2}

V = \int_0^3 \int_0^{3 - x} \int_0^{3 - x - y} 1\,dz\,dy\,dx

\int_0^{3 - x - y} 1\,dz = 3 - x - y

\int_0^{3 - x} (3 - x - y)\,dy = \dfrac{(3 - x)^2}{2}

\int_0^3 \dfrac{(3 - x)^2}{2}\,dx = \Big[-\dfrac{(3 - x)^3}{6}\Big]_0^3

= \dfrac{27}{6}

= \dfrac{9}{2}

\int_0^{y} y\,dz = y^2

\int_0^{x} y^2\,dy = \dfrac{x^3}{3}

\int_0^2 \dfrac{x^3}{3}\,dx = \Big[\dfrac{x^4}{12}\Big]_0^2

= \dfrac{16}{12}

= \dfrac{4}{3}

V = \iint_D (4 - x^2 - y^2)\,dA

x^2 + y^2 \le 4

V = \int_0^{2\pi} \int_0^2 (4 - r^2)\,r\,dr\,d\theta

\int_0^2 (4r - r^3)\,dr = \Big[2r^2 - \dfrac{r^4}{4}\Big]_0^2

= 8 - 4

= 4

V = 2\pi \cdot 4 = 8\pi

(e - 1)(e - 1)(e - 1)

(e - 1)^3 \approx 5.073

V = \int_0^3 \int_{-2}^{2} \int_0^{4 - y^2} 1\,dz\,dy\,dx

\int_0^{4 - y^2} 1\,dz = 4 - y^2

\int_{-2}^{2} (4 - y^2)\,dy = \Big[4y - \dfrac{y^3}{3}\Big]_{-2}^{2}

= \Big(8 - \dfrac{8}{3}\Big) - \Big(-8 + \dfrac{8}{3}\Big)

= \dfrac{32}{3}

V = 3 \cdot \dfrac{32}{3} = 32

\iiint_B (x + y + z)\,dV = 3\iiint_B x\,dV

\iiint_B x\,dV = \Big(\int_0^2 x\,dx\Big)(2)(2)

= 2 \cdot 4

= 8

3 \cdot 8 = 24

f_{\text{avg}} = \dfrac{24}{8} = 3

m = 2400\iiint_E z\,dV

\iiint_E z\,dV = \dfrac{1}{24}

m = 2400 \cdot \dfrac{1}{24} = 100

Triple integrals FAQ

What does a triple integral represent?

It adds up f(x, y, z) over a solid. With f = 1 it gives volume. With a density it gives mass, and with other functions it gives totals such as charge or heat.

How do you find the limits of a triple integral?

Find the bottom and top surfaces for the z limits. Then find the shadow D on the xy-plane and give it limits as in a double integral. The outer limits must be numbers.

How do you find volume with a triple integral?

Integrate 1 over the solid: V = ∭_E 1 dV. The inner z integral gives the height of each column, top minus bottom.

Does the order of integration matter in a triple integral?

For a continuous function, every correct order gives the same answer. The work can be much easier in one order, so pick the order with the simplest limits.

Summary

• A triple integral adds f(x, y, z) dV over a solid E.
• Integrate in z from the bottom surface to the top surface, then over the shadow D.
• Inner limits can contain outer variables; the outermost limits are numbers.
• Volume: integrate 1. Mass: integrate the density.
• Center of mass: divide each moment by the mass.

Previous lesson: Applications of Double Integrals

Next lesson: Triple Integrals in Cylindrical Coordinates

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