Deriving Volume as an Integral of Cross-Sectional Area

From Cross-Sections to Volume

A cross-section of a three-dimensional solid is the two-dimensional shape obtained by cutting the solid with a plane. If the cross-sections change as we move through the solid, their areas change too. The basic idea of the volume formula is to accumulate these areas along an interval: a cross-section with area \(A(x)\), extending through a very small thickness \(\Delta x\), suggests a small volume of about \(A(x)\Delta x\).

The word “suggests” matters. A product \(A(x)\Delta x\) is the volume of a prism with constant cross-sectional area \(A(x)\), but an actual thin part of the solid may have different cross-sections at different positions. We therefore need to be precise about what justifies the volume formula. The rigorous connection uses the slicing theorem for volume; the Riemann sums then describe the integral of the area function.

The Interval, Thickness, and Area Function

Choose an axis along which to measure position through the solid. For slices perpendicular to the \(x\)-axis, suppose the solid extends from \(x=a\) to \(x=b\), where \(a<b\). At each position \(x\), let \(A(x)\) be the area of the cross-section in the plane perpendicular to that axis.

The interval \([a,b]\) records the full range of positions. A partition divides it into smaller intervals. If the partition points are \(a=x_0<x_1<\cdots<x_n=b\), the thickness of the \(i\)th interval is \(\Delta x_i=x_i-x_{i-1}\). Choose a sample position \(x_i^*\) within that interval. The corresponding cross-sectional area is \(A(x_i^*)\).

For example, if position is measured in centimetres and cross-sectional area in square centimetres, then \(A(x_i^*)\Delta x_i\) has units \(\text{cm}^2\cdot\text{cm}=\text{cm}^3\). This product is a volume estimate, not automatically the exact volume of the part of the solid in that interval.

The Slicing Theorem and the Riemann-Sum Formula

Here is the independent geometric result that permits the connection between cross-sections and volume. It is often called the slicing theorem, or the volume form of Cavalieri’s principle. In a rigorous treatment it follows from the theorem that volume can be computed by integrating the indicator function of a solid in three-dimensional space and applying Fubini’s theorem. In particular, it is not a consequence of simply asserting that a variable-area slab lies between two constant-area prisms.

The theorem accounts for the full three-dimensional solid, including the way its cross-sections vary. The Riemann-sum limit below follows from this theorem together with the definition of the definite integral. It is important to keep these two roles separate: the slicing theorem relates volume to cross-sectional area, while Riemann sums approximate the integral of that area.

For a partition \(P\) of \([a,b]\), the Riemann sum for \(A\) is $$ \sum_{i=1}^{n} A(x_i^*)\Delta x_i. $$

If \(A\) is continuous on \([a,b]\), it is Riemann integrable. As the largest subinterval width tends to zero, every choice of sample points gives sums converging to the same integral. Writing \(\lVert P\rVert=\max_i\Delta x_i\) for the mesh of the partition, the result is $$ \lim_{\lVert P\rVert\to 0}\sum_{i=1}^{n}A(x_i^*)\Delta x_i = \int_a^b A(x)\,dx = V(E). $$

Here is why the Riemann sums converge when \(A\) is continuous. Continuity on a closed, bounded interval implies uniform continuity. Given any \(\varepsilon>0\), choose the partition fine enough that the values of \(A\) within each subinterval differ by less than \(\varepsilon/(b-a)\). Let \(m_i\) and \(M_i\) be the minimum and maximum values of \(A\) on the \(i\)th subinterval. Then \(M_i-m_i<\varepsilon/(b-a)\). The Riemann sum and the integral both lie between the lower and upper sums, so their difference is at most $$ \sum_{i=1}^{n}(M_i-m_i)\Delta x_i < \frac{\varepsilon}{b-a}\sum_{i=1}^{n}\Delta x_i = \varepsilon. $$ Thus the difference tends to zero as the mesh tends to zero. This establishes the Riemann-sum limit for the integral; the slicing theorem supplies its equality with volume.

Worked Example: A Linear Cross-Sectional Area

Worked Example: Slicing Along the \(y\)-Axis

Worked Example: A Quadratic Area Function

A Common Mistake: Treating a Sample Slice as the Whole Solid

A tempting shortcut is to use the area at one endpoint and multiply it by the full interval length. For the area function \(A(y)=4-y^2\) on \([0,2]\), this shortcut using the left endpoint gives $$ A(0)(2-0)=4\cdot 2=8\text{ cm}^3. $$

This is not the volume: it treats every cross-section as if it had the area at \(y=0\), even though the area decreases to \(A(2)=0\). The correct calculation accumulates the varying area: $$ V=\int_0^2(4-y^2)\,dy=\frac{16}{3}\text{ cm}^3, $$ which is not \(8\) cm\(^3\). A single product \(A(y)\Delta y\) can be useful as an estimate for a thin interval, but the integral—or equivalently the limit of the Riemann sums—is needed to account for the whole solid.

A related error is to claim that the minimum and maximum cross-sectional areas automatically bound the volume of a slab by constant-area prisms. Such a comparison needs a separate volume theorem; it does not follow merely from comparing the areas. The slicing theorem provides the sound foundation for the integral formula without relying on that unsupported step.

Why the Formula Matters

The formula $$ V=\int_a^b A(x)\,dx $$ turns a geometric problem into an accumulation problem. Once the interval and the area function are known, the definite integral combines the contributions from all positions. The same reasoning works with another axis: if slices perpendicular to the \(y\)-axis run from \(y=c\) to \(y=d\), and their area is \(B(y)\), then the volume is \(\int_c^d B(y)\,dy\), under the same slicing-theorem and integrability conditions.

For a practical setup, check three items before evaluating: the interval must span the solid in the slicing direction; the area function must describe the entire cross-section at each position; and the differential, such as \(dx\), must correspond to thickness along that direction. If \(A\) is continuous, its integrability is guaranteed. More generally, the slicing theorem requires an integrable cross-sectional area function.