Engineering Math Practice

Triple Integrals Over a Box

Evaluate an iterated triple integral of a polynomial over a rectangular box.

The idea. A double integral adds up over a 2D region; a triple integral adds up over a 3D solid. Same move, one dimension up:

∭Ef(x,y,z) dV\iiint_E f(x,y,z)\, dV

Triple integral

  • ∭E\iiint_Etriple integral over the solid E.

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Over a rectangular box:

B=[a,b]×[c,d]×[p,q]B = [a,b] \times [c,d] \times [p,q]

Rectangular box

  • [a,b],[c,d],[p,q][a,b],[c,d],[p,q]the constant bounds of a rectangular box.

it's three nested integrals, innermost first:

∭Bf dV=∫ab∫cd∫pqf(x,y,z) dz dy dx\iiint_B f\, dV = \int_a^b \int_c^d \int_p^q f(x,y,z)\, dz\, dy\, dx

Iterated triple integral over a box

Read the differentials inside-out: dz is innermost, so the innermost integral runs over z with x and y frozen. Do it, and z disappears. Then y. Then x. Three integrations, three variables killed, one number out.

dV is the little bit of volume:

dV=dz dy dxdV = dz\,dy\,dx

Volume element over a box

  • dV=dz dy dxdV = dz\,dy\,dxa little bit of volume, over a box.

Fubini still holds over a box — all six orderings give the same answer. Over a box, swapping is free; over a general solid, swapping means re-deriving the bounds.

Volume of a solid falls out with f = 1:

V(E)=∭E1 dVV(E) = \iiint_E 1\, dV

Volume as a triple integral

General solids. The bounds nest by dependency:

∭Ef dV=∫ab∫g1(x)g2(x)∫h1(x,y)h2(x,y)f dz dy dx\iiint_E f\, dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{h_1(x,y)}^{h_2(x,y)} f\, dz\, dy\, dx

Triple integral over a general solid

  • g1(x),g2(x)g_1(x), g_2(x)middle bounds, may depend on x.
  • h1(x,y),h2(x,y)h_1(x,y), h_2(x,y)innermost bounds, may depend on x and y.

The ladder: innermost z-bounds can depend on both x and y; middle y-bounds can depend on x; outermost x-bounds must be constants. Each variable's bounds may only depend on variables further out — outermost bounds are always numbers.

Worked example

∭Bxyz dV,B:[0,1]×[0,2]×[0,3]\iiint_B xyz\, dV, \qquad B: [0,1] \times [0,2] \times [0,3]
  1. ∫03xyz dz=xy⋅z22∣03=xy⋅92=9xy2\int_0^3 xyz\, dz = xy \cdot \frac{z^2}{2}\bigg|_0^3 = xy \cdot \frac{9}{2} = \frac{9xy}{2}

    Innermost, over z (hold x, y).

  2. ∫029xy2 dy=9x2⋅y22∣02=9x2⋅2=9x\int_0^2 \frac{9xy}{2}\, dy = \frac{9x}{2} \cdot \frac{y^2}{2}\bigg|_0^2 = \frac{9x}{2} \cdot 2 = 9x

    Middle, over y (hold x).

  3. ∫019x dx=9⋅x22∣01=92\int_0^1 9x\, dx = 9 \cdot \frac{x^2}{2}\bigg|_0^1 = \frac{9}{2}

    Outermost, over x.

  4. (∫01x dx)(∫02y dy)(∫03z dz)=12⋅2⋅92=92\left(\int_0^1 x\,dx\right)\left(\int_0^2 y\,dy\right)\left(\int_0^3 z\,dz\right) = \frac{1}{2} \cdot 2 \cdot \frac{9}{2} = \frac{9}{2}

    Since xyz separates over a box, check it as the product of three single integrals. Same number.

Answer:

92\frac{9}{2}

How to solve one of these

  1. 1.

    Set up the order of integration, innermost differential first. For a box, any order works; pick the easiest.

  2. 2.

    Integrate the innermost variable, treating the other two as constants. Substitute its bounds. That variable disappears.

  3. 3.

    Integrate the middle variable, treating the outer one as constant. Substitute. Gone.

  4. 4.

    Integrate the outermost variable. Its bounds must be constants. You get a number.

  5. 5.

    For a general solid, build bounds inside-out. Innermost may depend on the two outer variables, middle on the one outer variable, outermost are constants. A variable in a bound outside itself means the setup is wrong.

  6. 6.

    Check separable cases. If ff factors and the region is a box, the triple integral is the product of three single integrals.

∭Bf dV=∫ab∫cd∫pqf(x,y,z) dz dy dx\iiint_B f\, dV = \int_a^b \int_c^d \int_p^q f(x,y,z)\, dz\, dy\, dx

Integrate innermost to outermost — z first holding x and y constant, then y holding x constant, then x — and dV becomes dz dy dx.

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