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Triple Integrals and Solid Regions · Tutorial 612 of 999

Splitting Solids for Valid Iterated Bounds

Learn to split a solid into pieces with valid iterated bounds and evaluate the resulting sum of triple integrals.

Intermediate 10 min read

What You'll Learn

  • Recognize when a projection or a set of fiber bounds changes across a solid
  • Locate where competing boundary functions meet to choose a partition
  • Write a triple integral over each piece and add the results
  • Evaluate split integrals in a consistent order
  • Identify and correct bounds that include points outside the solid

Why a Solid May Need to Be Split

A triple integral is easiest to evaluate when one set of iterated bounds describes the solid throughout the entire integration region. Sometimes the geometry does not cooperate: a projected region may have a boundary that changes formula, or the lower and upper surfaces of a fiber may switch roles. A single set of simple bounds then fails to describe the solid correctly.

The remedy is to partition the solid into pieces. Each piece is chosen so its projection and bounds have one consistent description. The integral over the original solid is then the sum of the integrals over those pieces. This is not an approximation: the pieces cover the original solid, and their interiors do not overlap.

Key idea: Split a solid where the formula for a boundary changes or where competing bounds meet. Write valid bounds on each resulting piece, integrate each piece, and add the results.

Partitions and Additivity

Suppose a solid \(S\) is the union of finitely many solids \(S_1,\ldots,S_n\), with disjoint interiors. If \(f\) is integrable over \(S\), then the triple integral is additive across the pieces:

$$ \iiint_S f\,dV = \sum_{i=1}^{n}\iiint_{S_i} f\,dV. $$

The pieces may share boundary surfaces. Those shared boundaries do not contribute volume, so counting them as the boundary of both pieces does not change the integral. In the usual setting of solids with piecewise-smooth boundaries, such surfaces have three-dimensional volume zero.

To find a useful partition, inspect the boundary functions in the direction of the proposed iterated bounds. If a projection has an upper edge given by the smaller of two functions, determine where those functions are equal. If a fiber lies between two surfaces, find where the surfaces meet; on opposite sides of that meeting, their lower and upper roles may reverse.

1
Choose a direction for the fibers.
Decide which variable will have the innermost bounds and identify the projection onto the other two coordinates.
2
Find where the description changes.
Equate competing boundary functions, or locate where the projected region changes its edge formula.
3
Partition the projection or the solid.
Use the change locations to create pieces on which each bound has a consistent formula and order.
4
Integrate and add.
Write one iterated integral per piece, evaluate them, and sum the results.

Worked Example: Splitting a Projection with a Changing Edge

Worked Example: Splitting a Projection with a Changing Edge

Let \(S\) be the solid with \(0\le x\le 2\), \(0\le z\le 1\), and \(0\le y\le \min(x,2-x)\). Find its volume by evaluating \(\iiint_S 1\,dV\).

The projection onto the \(xy\)-plane has upper \(y\)-boundary \(\min(x,2-x)\). The two expressions meet when \(x=2-x\), so \(2x=2\) and \(x=1\). For \(0\le x\le 1\), the smaller expression is \(x\); for \(1\le x\le 2\), it is \(2-x\). Thus split the solid at \(x=1\):

$$ \iiint_S 1\,dV = \int_0^1\int_0^x\int_0^1 1\,dz\,dy\,dx + \int_1^2\int_0^{2-x}\int_0^1 1\,dz\,dy\,dx. $$

For the first piece, integrate from the inside outward:

$$ \int_0^1\int_0^x\int_0^1 1\,dz\,dy\,dx = \int_0^1\int_0^x [z]_0^1\,dy\,dx = \int_0^1\int_0^x 1\,dy\,dx = \int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac12. $$

For the second piece:

$$ \int_1^2\int_0^{2-x}\int_0^1 1\,dz\,dy\,dx = \int_1^2\int_0^{2-x} 1\,dy\,dx = \int_1^2 (2-x)\,dx = \left[2x-\frac{x^2}{2}\right]_1^2 = 2-\frac32 = \frac12. $$

Adding the two contributions gives the volume \(1\). As a check, the projected region is a triangle with base \(2\) and height \(1\), so its area is \(1\); the solid has constant height \(1\), giving volume \(1\) as well.

Worked Example: Splitting Where a Fiber's Upper Bound Changes

Worked Example: Splitting Where a Fiber's Upper Bound Changes

Consider the solid \(S\) defined by \(0\le x\le 2\), \(0\le y\le 2\), and \(0\le z\le \min(x,2-y)\). Find its volume.

The projection is the full square \(0\le x,y\le 2\), but the upper \(z\)-bound changes where \(x=2-y\), or equivalently \(y=2-x\). For a fixed \(x\), the bound is \(z\le x\) when \(0\le y\le 2-x\), and \(z\le 2-y\) when \(2-x\le y\le 2\). Therefore,

$$ \iiint_S 1\,dV = \int_0^2\int_0^{2-x}\int_0^x 1\,dz\,dy\,dx + \int_0^2\int_{2-x}^{2}\int_0^{2-y} 1\,dz\,dy\,dx. $$

Evaluate the first integral:

$$ \int_0^2\int_0^{2-x}\int_0^x 1\,dz\,dy\,dx = \int_0^2\int_0^{2-x}x\,dy\,dx = \int_0^2 x(2-x)\,dx = \left[x^2-\frac{x^3}{3}\right]_0^2 = 4-\frac83 = \frac43. $$

For the second integral, integrate first with respect to \(z\), then \(y\):

$$ \begin{aligned} \int_0^2\int_{2-x}^{2}\int_0^{2-y}1\,dz\,dy\,dx &=\int_0^2\int_{2-x}^{2}(2-y)\,dy\,dx\\ &=\int_0^2\left[2y-\frac{y^2}{2}\right]_{2-x}^{2}\,dx\\ &=\int_0^2\left(2-\left(2-\frac{x^2}{2}\right)\right)\,dx\\ &=\int_0^2\frac{x^2}{2}\,dx =\left[\frac{x^3}{6}\right]_0^2 =\frac43. \end{aligned} $$

The volume is \(\frac43+\frac43=\frac83\). As a check on the split, the changeover line \(y=2-x\) divides the square into two triangular regions. The substitution \(u=2-y\) turns the second contribution into the same integral as the first, consistent with the two equal values.

Worked Example: Splitting Where the Lower and Upper Surfaces Switch

Worked Example: Splitting Where the Lower and Upper Surfaces Switch

Let \(S\) consist of points with \(0\le x\le 2\), \(0\le y\le 1\), and \(z\) between the two surfaces \(z=x\) and \(z=2-x\). Find \(\iiint_S z\,dV\).

The surfaces meet where \(x=2-x\), which gives \(x=1\). On \(0\le x\le 1\), \(x\le 2-x\), so \(z\) ranges from \(x\) to \(2-x\). On \(1\le x\le 2\), \(2-x\le x\), so the lower and upper bounds reverse. The correct integral is

$$ \iiint_S z\,dV = \int_0^1\int_0^1\int_x^{2-x}z\,dz\,dy\,dx + \int_1^2\int_0^1\int_{2-x}^{x}z\,dz\,dy\,dx. $$

For the first piece:

$$ \begin{aligned} \int_0^1\int_0^1\int_x^{2-x}z\,dz\,dy\,dx &=\int_0^1\int_0^1\left[\frac{z^2}{2}\right]_x^{2-x}\,dy\,dx\\ &=\int_0^1\int_0^1\frac{(2-x)^2-x^2}{2}\,dy\,dx\\ &=\int_0^1(2-2x)\,dx =\left[2x-x^2\right]_0^1 =1. \end{aligned} $$

For the second piece:

$$ \begin{aligned} \int_1^2\int_0^1\int_{2-x}^{x}z\,dz\,dy\,dx &=\int_1^2\int_0^1\left[\frac{z^2}{2}\right]_{2-x}^{x}\,dy\,dx\\ &=\int_1^2\int_0^1\frac{x^2-(2-x)^2}{2}\,dy\,dx\\ &=\int_1^2(2x-2)\,dx =\left[x^2-2x\right]_1^2 =0-(-1) =1. \end{aligned} $$

The total is \(1+1=2\). The bounds cover the full interval between the two surfaces on each side of their intersection, and the two computed contributions add to the integral over the whole solid.

Why the Split Is Necessary

In the examples, the changeover locations were found by equating boundary functions. That step is essential: a split chosen elsewhere may leave a piece whose bounds still change inside it. Once a changeover is found, test the order of the functions on each side. For example, in the third example, the inequality \(x\le 2-x\) holds only when \(x\le 1\). Keeping the same lower and upper bounds on both sides would make the second integral invalid.

A minimum or maximum can sometimes be written directly in an iterated bound. But replacing a changing bound with a single expression is useful only if it describes the correct fiber everywhere and can be evaluated as written. Splitting makes the regions of validity explicit and often gives simpler integrals.

Common Mistake: Using One Bound Across the Entire Projection

For the solid in the second example, it is incorrect to use \(0\le z\le x\) over the whole square. That would give

$$ \int_0^2\int_0^2\int_0^x 1\,dz\,dy\,dx = \int_0^2\int_0^2 x\,dy\,dx = \int_0^2 2x\,dx = 4. $$

This integral counts points outside the solid. At \(x=2,y=2\), for instance, the actual bound is \(z\le\min(2,0)=0\), whereas the incorrect bound allows \(z\) to reach \(2\). The correction is to split along \(y=2-x\) and use \(z\le x\) on one side and \(z\le 2-y\) on the other. The correct volume is \(\frac83\), not \(4\).

Practical check: Before integrating, choose a point from each proposed piece and verify that its bounds agree with the original inequalities. A bound that works on one side of a changeover may fail on the other.

Why Partitioning Matters

Splitting a solid is a way to match the description of the integral to the actual geometry. It allows one iterated integral to handle each region where the bounds remain consistent, without omitting parts of the solid or including points that do not belong to it. The same principle applies whether the change is in the projected boundary, the upper surface of a fiber, or the order of two surfaces that bound the fiber.

Before evaluating a triple integral with changing bounds, check that the pieces cover the entire solid, that their interiors do not overlap, and that each set of limits describes its piece throughout. Then add the resulting integrals. This geometric check is as important as the algebraic evaluation.

Check Your Understanding

Use the geometry of the bounds to answer each question.

  1. For \(0\le x\le 3\), \(0\le y\le \min(x,3-x)\), where should the projection be split, and what is the upper \(y\)-bound on each side?
  2. In the solid \(0\le x,y\le 1\), \(0\le z\le\min(y,1-x)\), what equation determines where the upper \(z\)-bound changes?
  3. Why may two pieces of a partition share a boundary surface without changing the sum of their triple integrals?
  4. For a solid between \(z=x\) and \(z=4-x\), \(0\le x\le 4\), at what value of \(x\) do the lower and upper surfaces switch roles?
  5. What geometric error can occur if a bound that is valid on one side of a changeover is used over the entire projection?

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