Volumes by Slicing: Known Cross-Sections
If you know the area \(A(x)\) of every cross-section of a solid perpendicular to an axis, the volume is just \(V = \int_a^b A(x)\,dx\) — no revolution required. The cross-sections can be squares, semicircles, triangles, anything.
By the end you'll be able to build \(A(x)\) from a base region and a cross-section shape (square, semicircular, …) and integrate it to get an exact volume.
Predict: as \(x\) grows, does the square cross-section get bigger or smaller? Drag the slice (or the x slider) and read A(x) to check.
The base is the region between \(y = \sqrt{x}\) and the x-axis on \([0, 4]\). At each \(x\), a cross-section stands up perpendicular to the axis with side \(s = \sqrt{x}\) — a square by default, or toggle to a semicircle with that segment as its diameter. Add slabs to stack the slices toward the solid and watch the slab sum close in on \(\int_0^4 A(x)\,dx\).
s = √x = 1.414 · A(x) = x = 2.000 · slab sum over 8 slabs ≈ 7.000 → V = ∫₀⁴ A(x) dx = 8
Slice a solid into thin slabs perpendicular to an axis; each slab is nearly a prism of volume \(A(x)\,dx\), and adding them up gives \(V = \int_a^b A(x)\,dx\).
If a solid extends along the x-axis from \(x = a\) to \(x = b\) and its cross-section perpendicular to the axis at position \(x\) has area \(A(x)\), then \[V = \int_a^b A(x)\,dx.\] Cut the solid into \(n\) slabs of thickness \(\Delta x\): each slab's volume is approximately \(A(x_i)\,\Delta x\), and the Riemann sum \(\sum A(x_i)\,\Delta x\) converges to the integral as the slabs get thin — exactly the accumulation you watched the slab slider perform. Evaluate the integral with the FTC as usual. The only real work is building \(A(x)\): for a fixed shape whose size is set by a side (or diameter) \(s\), the standard areas are square \(A = s^2\), semicircle of diameter \(s\): \(A = \pi s^2/8\), and equilateral triangle \(A = (\sqrt{3}/4)s^2\).
Picture a loaf of bread. Every slice has a face — some shape with an area — and each slice's volume is roughly (face area) × (thickness). The whole loaf is the sum of its slices. In the demo above, the green segment from the axis up to \(y = \sqrt{x}\) is the side \(s\) of the face at that \(x\); the face itself stands up out of the page, square or semicircular, and its area A(x) shrinks to 0 at the left tip and grows as \(x\) increases. In the previous lesson you revolved a region and every face came out a circle — here the base region stays put and you choose the face shape.
A solid has base the region between \(y = \sqrt{x}\) and the x-axis on \([0, 4]\); cross-sections perpendicular to the x-axis are squares whose side runs from the axis up to the curve, so \(s = \sqrt{x}\). Build the area first: \(A(x) = s^2 = (\sqrt{x})^2 = x\). Then integrate: \[V = \int_0^4 x\,dx = \left[\frac{x^2}{2}\right]_0^4 = \frac{16}{2} = \textbf{8}.\] That's the exact value the slab sum in the demo approaches as you add slabs.
Same base, different face: the cross-sections are now semicircles with diameter \(s = \sqrt{x}\). The radius is \(s/2\), so \(A(x) = \frac{1}{2}\pi \left(\frac{\sqrt{x}}{2}\right)^2 = \frac{\pi x}{8}\). Now finish it: \[V = \int_0^4 \frac{\pi x}{8}\,dx = \frac{\pi}{8}\int_0^4 x\,dx = \frac{\pi}{8}\cdot \underline{\;\;?\;\;} = \underline{\;\;?\;\;}\]
Reveal the answer
\(\int_0^4 x\,dx = 8\) (computed in the worked example above), so \(V = \frac{\pi}{8}\cdot 8 = \boldsymbol{\pi} \approx 3.142\). Flip the demo's toggle to semicircle and push the slab slider right — the sum closes in on \(\pi\).
More info — disks and washers are slicing in disguise
The disk method from the previous lesson is not a separate idea: revolving a region around an axis just guarantees every cross-section is a circle of radius \(R(x)\), so the general formula \(V = \int A(x)\,dx\) specializes to \(V = \int \pi[R(x)]^2\,dx\). A washer subtracts an inner circle: \(A(x) = \pi[R(x)]^2 - \pi[r(x)]^2\). Once you see that, there's only ONE volume formula to remember — integrate the cross-sectional area — and the OpenStax section in Dive deeper below develops it exactly this way. Two pitfalls to keep off your back: if \(s\) is a diameter, the radius is \(s/2\) (so a semicircle gives \(\pi s^2/8\), not \(\pi s^2/2\)); and the cross-sections must be perpendicular to the variable you integrate — slices perpendicular to the y-axis mean \(A(y)\,dy\) with y-limits.
Check your understanding
A solid has base the region between \(y = x\) and the x-axis on \([0, 3]\); cross-sections perpendicular to the x-axis are squares of side \(s = x\). What is the volume?
Cross-sections perpendicular to the x-axis are semicircles whose diameter is \(s(x)\). Which area function is correct?
How does the disk method for volumes of revolution relate to the general slicing formula \(V = \int_a^b A(x)\,dx\)?
A solid's cross-sections are squares perpendicular to the y-axis, with side set by a bounding curve. How do you set up the volume integral?
Recap
- Known cross-sectional area \(A(x)\) perpendicular to the x-axis on \([a,b]\) gives \(V = \int_a^b A(x)\,dx\) — each thin slab contributes \(A(x)\,dx\).
- Build \(A(x)\) from the shape and the region: square of side \(s\): \(A = s^2\); semicircle of diameter \(s\): \(A = \pi s^2/8\); equilateral triangle of side \(s\): \(A = (\sqrt{3}/4)s^2\) — usually \(s = f(x)\), the height of the bounding curve.
- Watch diameter vs. radius: a semicircle on diameter \(s\) has radius \(s/2\), so \(A = \pi s^2/8\), not \(\pi s^2/2\).
- Cross-sections must be perpendicular to the variable of integration — slices perpendicular to the y-axis need \(A(y)\) and \(dy\).
- Disks and washers are the special case where the cross-section is a circle: \(A(x) = \pi R^2\) (minus \(\pi r^2\) for a washer).
Dive deeper
- OpenStax Calculus Volume 1 — 6.2 Determining Volumes by Slicing Establish V = the integral of A(x) dx for known cross sections
Sources
- Volumes by Slicing — Known Cross Sections