Cavalieri's principle says that two solids with equal cross-section areas at every height have the same volume. Calculus turns that idea into a formula: if you know the area $A(x)$ of every cross-section, you know the volume. This builds on Area Between Curves, where you added thin strips; now you add thin slabs.
Why it works
Slice the solid perpendicular to the $x$-axis into slabs of thickness $\Delta x$. A slab at $x$ is close to a prism (a "cylinder" in the general sense) whose base is the cross-section, so its volume is about $A(x)\,\Delta x$. Adding the slabs gives a Riemann sum:
$$V \approx \sum_{i=1}^{n} A(x_i)\,\Delta x \;\longrightarrow\; \int_a^b A(x)\,dx.$$
Cavalieri's principle follows at once: equal $A(x)$ gives equal integrals.
Key idea
$$V = \int_a^b A(x)\,dx,$$ where $A(x)$ is the area of the cross-section at $x$. All the work is in writing $A(x)$ as a formula in $x$.
| Cross-section with side (or diameter) $s$ | Area |
|---|---|
| square | $s^{2}$ |
| semicircle with diameter $s$ | $\frac{\pi}{8}s^{2}$ |
| equilateral triangle | $\frac{\sqrt{3}}{4}s^{2}$ |
| isosceles right triangle, leg $s$ | $\frac{1}{2}s^{2}$ |
Worked example: A pyramid
A pyramid has a square base of side 4 m and height 6 m. Find its volume by slicing.
- Measure $y$ down from the top. By similar triangles the square at depth $y$ has side $4\cdot\frac{y}{6} = \frac{2y}{3}$.
- Its area is $A(y) = \left(\frac{2y}{3}\right)^{2} = \frac{4y^{2}}{9}$.
- $V = \int_0^6 \frac{4y^{2}}{9}\,dy = \frac{4}{9}\cdot\frac{6^{3}}{3} = 32$.
- Check: $\frac{1}{3}(\text{base})(\text{height}) = \frac{1}{3}\cdot 16\cdot 6 = 32$. The $\frac{1}{3}$ in that formula comes from $\int y^{2}\,dy$.
Answer: $32$ m³
Worked example: Squares on a parabolic base
The base of a solid is the region between $y = 4 - x^{2}$ and the $x$-axis. Cross-sections perpendicular to the $x$-axis are squares. Find the volume.
- At $x$, the base is $4 - x^{2}$ tall, so the square has side $s = 4 - x^{2}$ and area $(4 - x^{2})^{2}$, for $-2 \le x \le 2$.
- $$\begin{aligned} V &= \int_{-2}^{2} (4 - x^{2})^{2}\,dx \\ &= 2\int_0^2 (16 - 8x^{2} + x^{4})\,dx \end{aligned}$$ (the integrand is even).
- $$\begin{aligned} &= 2\left(32 - \frac{64}{3} + \frac{32}{5}\right) \\ &= 2\cdot\frac{256}{15} \\ &= \frac{512}{15}. \end{aligned}$$
Answer: $\frac{512}{15}$ cubic units
Worked example: Triangles on a disk
A solid has the disk $x^{2} + y^{2} \le 9$ as its base. Cross-sections perpendicular to the $x$-axis are equilateral triangles. Find the volume.
- At $x$ the disk runs from $y = -\sqrt{9 - x^{2}}$ to $y = \sqrt{9 - x^{2}}$, so the side is $s = 2\sqrt{9 - x^{2}}$ and $s^{2} = 4(9 - x^{2})$.
- $$\begin{aligned} A(x) &= \frac{\sqrt{3}}{4}\cdot 4(9 - x^{2}) \\ &= \sqrt{3}\,(9 - x^{2}). \end{aligned}$$
- $$\begin{aligned} V &= \sqrt{3}\int_{-3}^{3} (9 - x^{2})\,dx \\ &= \sqrt{3}\,(54 - 18) \\ &= 36\sqrt{3}. \end{aligned}$$
Answer: $36\sqrt{3}$ cubic units
Common mistake
Integrating the side length $s(x)$ instead of the area $A(x)$. That adds up lengths times thicknesses, which is an area, not a volume. Check the units: $A(x)$ is in square units and $dx$ in units, so $\int A(x)\,dx$ comes out in cubic units.