Cross-Section Slicing for Triple Integrals

Why Slice a Solid?

A three-dimensional solid can be difficult to describe all at once. A useful alternative is to examine it one plane at a time. Fixing one coordinate gives a planar cross-section; as that coordinate varies, the cross-section changes and traces out the solid.

This viewpoint connects the geometry of a solid to the bounds of a triple integral. First describe the region in each slice. Then integrate over that planar region to obtain its area, and accumulate those areas along the slicing direction.

Central idea: A solid can be described by its cross-sections perpendicular to a coordinate axis. Integrating the area of each cross-section over the range of that coordinate gives the triple integral of \(1\) over the solid.

Cross-Sections and Their Areas

Let \(E\) be a bounded solid, and suppose its \(z\)-coordinates range from \(a\) to \(b\). For a fixed value \(z=t\), the horizontal cross-section is the planar region consisting of points \((x,y)\) for which \((x,y,t)\) lies in \(E\). Denote this region by \(E_t\).

Definition: The horizontal cross-section of \(E\) at height \(t\) is \(E_t=\{(x,y):(x,y,t)\in E\}\). Its area is denoted by \(A(t)\).

If the cross-section has Cartesian bounds \(g_1(t)\le x\le g_2(t)\) and \(h_1(x,t)\le y\le h_2(x,t)\), its area can be written as a double integral of \(1\):

A(t)=\int_{g_1(t)}^{g_2(t)} \int_{h_1(x,t)}^{h_2(x,t)}1\,dy\,dx.

The bounds are allowed to depend on \(t\), because the planar region can change as the slice moves. Combining this area integral with an integral over the full range of \(z\) gives the corresponding triple integral.

Cross-section formula: For a bounded solid whose cross-sections are measurable and whose boundary is sufficiently regular for the usual iterated-integral results to apply,

\iiint_E 1\,dV = \int_a^b A(t)\,dt = \int_a^b\iint_{E_t}1\,dA\,dt.

Here \(dA\) means integration over the slice’s two coordinates. For a horizontal slice, \(dA=dy\,dx\) or \(dx\,dy\), according to the chosen planar bounds. The notation \(dV\) denotes integration over all three coordinates.

Why the Cross-Section Formula Works

The formula follows from treating the solid as a collection of thin slabs. Imagine dividing the interval from \(a\) to \(b\) into short subintervals. Over a sufficiently thin subinterval near \(t\), the solid is approximated by a slab whose thickness is \(\Delta z\) and whose face has area approximately \(A(t)\). Its volume is therefore approximately \(A(t)\Delta z\). Adding the slab volumes gives a Riemann sum for the integral of \(A(t)\).

There is also a direct integral justification. Let \(\mathbf{1}_E(x,y,z)\) equal \(1\) when \((x,y,z)\in E\) and \(0\) otherwise. Integrating this indicator over a rectangular region containing \(E\) gives the triple integral over \(E\). Fubini’s theorem allows the integrations to be grouped by fixing \(z\) first. At a fixed height \(t\), the indicator is \(1\) precisely on \(E_t\), so its double integral is the area \(A(t)\).

\iiint_E 1\,dV = \int_a^b \left(\iint_{\mathbb{R}^2}\mathbf{1}_E(x,y,t)\,dA\right)dt = \int_a^b A(t)\,dt.

The integral over \(\mathbb{R}^2\) can equivalently be taken over any fixed planar region large enough to contain every cross-section. Outside \(E_t\), the indicator is zero. Thus the inner integral contributes exactly the area of the slice, and the outer integral accumulates those areas through the solid.

Deriving Bounds from the Slices

A practical construction follows a consistent sequence. Choose the slicing coordinate, identify the full interval for that coordinate, describe the planar region at a fixed value, and then put the slice bounds into the corresponding double integral. If the slices are horizontal, the resulting order has \(z\) as the outer variable when the area is integrated first.

Choose a slicing direction.
For horizontal slices, fix \(z=t\); for slices perpendicular to the \(x\)-axis, fix \(x=t\).

Find the coordinate interval.
Determine the minimum and maximum values of the coordinate being fixed.

Describe a typical slice.
Translate the solid’s inequalities into planar bounds, allowing those bounds to depend on the fixed coordinate.

Integrate the slice, then accumulate.
Write the area as a double integral and integrate it over the coordinate interval.

Worked Examples

Worked Example: A Square Pyramid with Shrinking Slices

Consider the solid \(E\) described by \(0\le z\le2\), \(-(2-z)\le x\le2-z\), and \(-(2-z)\le y\le2-z\). Find the triple integral of \(1\) over \(E\) by slicing horizontally.

Fix \(z=t\). Both \(x\) and \(y\) range from \(-(2-t)\) to \(2-t\). The cross-section is a square with side length \(2(2-t)\). Its area, calculated from the Cartesian bounds, is

A(t)=\int_{-(2-t)}^{2-t}\int_{-(2-t)}^{2-t}1\,dy\,dx =\int_{-(2-t)}^{2-t}2(2-t)\,dx =4(2-t)^2.

Accumulate these areas from the bottom at \(z=0\) to the top at \(z=2\):

\iiint_E1\,dV =\int_0^2 4(2-z)^2\,dz =\left[-\frac{4}{3}(2-z)^3\right]_0^2 =0-\left(-\frac{32}{3}\right) =\frac{32}{3}.

The bounds give the same triple integral directly:

\iiint_E1\,dV = \int_0^2 \int_{-(2-z)}^{2-z} \int_{-(2-z)}^{2-z}1\,dy\,dx\,dz =\int_0^2 4(2-z)^2\,dz =\frac{32}{3}.

As a check, the area formula \(4(2-z)^2\) is zero at the top \(z=2\) and is \(16\) at the base \(z=0\), as the bounds require. Differentiating the antiderivative \(-\frac{4}{3}(2-z)^3\) gives \(4(2-z)^2\), confirming the integration.

Worked Example: Triangular Slices That Shrink with Height

Let \(E\) be defined by \(0\le z\le1\), \(x\ge0\), \(y\ge0\), and \(x+y\le1-z\). Describe its horizontal slices and derive the triple integral of \(1\) over \(E\).

At height \(z=t\), the inequalities become \(x\ge0\), \(y\ge0\), and \(x+y\le1-t\). Thus \(x\) ranges from \(0\) to \(1-t\), and for each \(x\), \(y\) ranges from \(0\) to \(1-t-x\). The slice area is

A(t)=\int_0^{1-t}\int_0^{1-t-x}1\,dy\,dx =\int_0^{1-t}(1-t-x)\,dx =\left[(1-t)x-\frac{x^2}{2}\right]_0^{1-t} =\frac{(1-t)^2}{2}.

Now integrate the slice area over \(0\le z\le1\):

\iiint_E1\,dV =\int_0^1\frac{(1-z)^2}{2}\,dz =\left[-\frac{(1-z)^3}{6}\right]_0^1 =0-\left(-\frac16\right) =\frac16.

The full bounds express the same construction: \(z\) runs from \(0\) to \(1\), \(x\) runs from \(0\) to \(1-z\), and \(y\) runs from \(0\) to \(1-z-x\). Therefore the triple integral has the value \(\frac16\). At \(z=0\), the slice has area \(\frac12\); at \(z=1\), it collapses to a point with area \(0\). Differentiating \(-\frac{(1-z)^3}{6}\) gives \(\frac{(1-z)^2}{2}\), verifying the outer integration.

Worked Example: Slicing Perpendicular to the \(x\)-Axis

Consider \(E\) given by \(0\le x\le1\), \(0\le y\le x\), and \(0\le z\le2-x\). Use slices perpendicular to the \(x\)-axis to derive the triple integral of \(1\) over \(E\).

Fix \(x=t\), where \(0\le t\le1\). The slice is a rectangle: \(y\) ranges from \(0\) to \(t\), while \(z\) ranges from \(0\) to \(2-t\). Its area is

A(t)=\int_0^t\int_0^{2-t}1\,dz\,dy =\int_0^t(2-t)\,dy =t(2-t).

Accumulating the areas along the \(x\)-axis gives

\iiint_E1\,dV =\int_0^1t(2-t)\,dt =\left[t^2-\frac{t^3}{3}\right]_0^1 =1-\frac13 =\frac23.

In iterated-integral form, this is

\int_0^1\int_0^t\int_0^{2-t}1\,dz\,dy\,dt =\int_0^1t(2-t)\,dt =\frac23.

The cross-section has area \(0\) at \(t=0\), where its \(y\)-width is zero, and area \(1\) at \(t=1\). The antiderivative \(t^2-\frac{t^3}{3}\) differentiates to \(2t-t^2=t(2-t)\), confirming the result. This example shows that the fixed coordinate need not be \(z\): the same slicing principle works along another coordinate axis.

A Common Pitfall: Treating Every Slice as the Same

For the square pyramid in the first example, the base has area \(16\) and the height is \(2\). A tempting but incorrect calculation is to multiply those values and claim the triple integral is \(32\). That calculation assumes the cross-sectional area stays equal to \(16\) through the full height.

But the bounds show that the side length at height \(z\) is \(2(2-z)\), so the area is \(4(2-z)^2\), not a constant. It shrinks from \(16\) at the base to \(0\) at the top. The correct accumulation is

\int_0^2 4(2-z)^2\,dz=\frac{32}{3},

not \(16\cdot2=32\). The correction is to derive the cross-section at a general coordinate value before integrating. Checking the bounds at the ends of the interval is a useful way to catch an incorrectly assumed constant slice.

When This Viewpoint Is Useful

Cross-section slicing is especially useful when the solid’s planar sections have simple descriptions, even if the full three-dimensional boundary is less convenient to visualize. A section may be a rectangle, triangle, or another region with straightforward Cartesian bounds. Its bounds then reveal the changing inner limits of a triple integral.

The method also makes the geometric meaning of the bounds explicit. The outer interval records how far the slices travel; the inner two bounds describe the shape of a typical slice. If the shape changes as the slice moves, that change must appear in the bounds and therefore in the area being accumulated.

Takeaway: Fix one coordinate, describe the resulting planar region, express its area as a double integral, and integrate that area across the full coordinate interval. The slice bounds are what make the triple integral match the solid.

Check Your Understanding

Use the cross-section viewpoint to answer the following questions.

For a horizontal slice \(E_t\), what does the area \(A(t)\) measure?
If the cross-section has bounds \(0\le x\le t\) and \(0\le y\le 3-t\), write an integral for its area.
Why do the inner bounds in the triangular-slice example depend on \(z\)?
What does the outer integral do after the area of each slice has been found?
For the solid in the third worked example, what is the cross-section area at \(x=0\), and why?