Volumes with Known Cross Sections

Quick answer

The volume of a solid is the integral of the area of its cross sections: V = ∫ₐᵇ A(x) dx, where A(x) is the area of the slice at x. A solid with known cross sections has a flat base region, and each slice perpendicular to the x-axis is a square, rectangle, triangle or semicircle standing on the segment across the base. The side of that shape is the width of the base at x, top boundary minus bottom, and the shape's area formula turns it into A(x).

What you'll learn

Explain why volume is the integral of cross-sectional area
Find the side of a cross section from the base region
Compute volumes with square, triangular, rectangular and semicircular sections
Slice perpendicular to the y-axis when a problem calls for it

Volume by slicing

One way to measure a loaf of bread is to slice it and add up the slices. A slice of thickness $\Delta x$ whose face has area $A$ has volume about $A\,\Delta x$ . Cut a solid lying along the $x$ -axis, from $x = a$ to $x = b$ , into thin slices, and their volumes add up to a Riemann sum:

V \approx \sum_{i=1}^{n} A(x_i)\,\Delta x \quad\longrightarrow\quad V = \int_a^b A(x)\,dx

Here $A(x)$ is the area of the cross section at $x$ : the face you would see if you cut the solid straight across there.

Why volume is the integral of area

A thin slice is nearly a slab: its two faces have almost the same area, so its volume is close to face area times thickness. Thinner slices make the error smaller, and in the limit the sum becomes the integral. Area between curves worked the same way one dimension down, adding up the lengths of thin strips. Every slicing method for volume is one formula: integrate the area of a slice.

The formula also explains an old fact, Cavalieri’s principle. Two solids whose slices have equal areas at every height have equal volumes, however different the shapes of the slices are.

Solids with known cross sections

A solid can be built on a flat base region. On each segment across the region, perpendicular to the $x$ -axis, stand a shape: a square, say. The segment at $x$ runs from the lower boundary to the upper one, so the shape’s side is the width of the base there:

s(x) = \text{top}(x) - \text{bottom}(x)

The shape’s area formula then gives $A(x)$ :

Cross section standing on a side $s$	Area
square	$s^2$
rectangle of height $h$	$sh$
semicircle, diameter $s$	$\tfrac{\pi}{8}s^2$
equilateral triangle	$\tfrac{\sqrt{3}}{4}s^2$
isosceles right triangle, leg $s$	$\tfrac{1}{2}s^2$
isosceles right triangle, hypotenuse $s$	$\tfrac{1}{4}s^2$

Square cross sections on the region between y = √x and y = −√x

Worked examples

Example 3 — Semicircles on a disk

The base of a solid is the disk $x^2 + y^2 \le 4$ . Cross sections perpendicular to the $x$ -axis are semicircles. Find the volume.

Semicircular cross sections on a disk of radius 2

The diameter is the width of the disk: $s = 2\sqrt{4 - x^2}$ . So $A(x) = \tfrac{\pi}{8}\cdot 4(4 - x^2) = \tfrac{\pi}{2}(4 - x^2)$ , and

V = \frac{\pi}{2}\int_{-2}^{2} (4 - x^2)\,dx = \frac{\pi}{2}\cdot\frac{32}{3} = \frac{16\pi}{3}

Each semicircle has radius $\sqrt{4 - x^2}$ , the same as a slice of a ball of radius $2$ , so the solid is a hemisphere. Its volume matches $\tfrac{2}{3}\pi \cdot 2^3 = \tfrac{16\pi}{3}$ .

Common mistakes

Practice problems

The base of a solid is the region between $y = x$ and $y = 0$ for $0 \le x \le 3$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Answer

$9$

Full solution

The side is $x - 0 = x$ , so $V = \int_0^3 x^2\,dx = \tfrac{27}{3} = 9$ .
The base is the region between $y = x^2$ and $y = 1$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Answer

$\tfrac{16}{15}$

Full solution

The curves meet at $x = \pm 1$ , and the side is $1 - x^2$ . $V = \int_{-1}^{1} (1 - x^2)^2\,dx = \int_{-1}^{1} (1 - 2x^2 + x^4)\,dx = 2\left(1 - \tfrac{2}{3} + \tfrac{1}{5}\right) = \tfrac{16}{15}$ .
The base is the region between $y = \sqrt{x}$ and $y = 0$ for $0 \le x \le 4$ . Cross sections perpendicular to the $x$ -axis are semicircles. Find the volume.

Answer

$\pi$

Full solution

The diameter is $\sqrt{x}$ , so $A(x) = \tfrac{\pi}{8}x$ and $V = \tfrac{\pi}{8}\int_0^4 x\,dx = \tfrac{\pi}{8} \cdot 8 = \pi$ .
The base is the region between $y = x$ and $y = 0$ for $0 \le x \le 2$ . Cross sections perpendicular to the $x$ -axis are equilateral triangles. Find the volume.

Answer

$\tfrac{2\sqrt{3}}{3} \approx 1.15$

Full solution

$A(x) = \tfrac{\sqrt{3}}{4}x^2$ , so $V = \tfrac{\sqrt{3}}{4}\int_0^2 x^2\,dx = \tfrac{\sqrt{3}}{4} \cdot \tfrac{8}{3} = \tfrac{2\sqrt{3}}{3}$ .
The base is the region between $y = 2x$ and $y = x^2$ . Cross sections perpendicular to the $x$ -axis are isosceles right triangles with a leg on the base. Find the volume.

Answer

$\tfrac{8}{15}$

Full solution

The curves meet at $x = 0$ and $x = 2$ , and the leg is $2x - x^2$ . $V = \tfrac{1}{2}\int_0^2 (2x - x^2)^2\,dx = \tfrac{1}{2}\int_0^2 (4x^2 - 4x^3 + x^4)\,dx = \tfrac{1}{2}\left(\tfrac{32}{3} - 16 + \tfrac{32}{5}\right) = \tfrac{8}{15}$ .
The base is the region between $y = \sqrt{x}$ and $y = 0$ for $0 \le x \le 4$ . Cross sections perpendicular to the $x$ -axis are rectangles of height $3$ . Find the volume.

Answer

$16$

Full solution

$A(x) = 3\sqrt{x}$ , so $V = 3\int_0^4 x^{1/2}\,dx = 3 \cdot \tfrac{2}{3} \cdot 8 = 16$ .
Use slicing to find the volume of a pyramid with a $6$ by $6$ square base and height $10$ .

Answer

$120$

Full solution

At distance $y$ below the apex the square has side $\tfrac{6}{10}y$ . $V = \int_0^{10} 0.36y^2\,dy = 0.36 \cdot \tfrac{1000}{3} = 120$ , which is $\tfrac{1}{3} \cdot 36 \cdot 10$ .
A cone has radius $3$ and height $6$ . Use circular cross sections to find its volume.

Answer

$18\pi$

Full solution

At distance $y$ below the apex the radius is $\tfrac{3}{6}y = \tfrac{y}{2}$ , so $A(y) = \tfrac{\pi y^2}{4}$ and $V = \tfrac{\pi}{4}\int_0^6 y^2\,dy = \tfrac{\pi}{4} \cdot 72 = 18\pi$ .
The base is the region between $x = y^2$ and $x = 4$ . Cross sections perpendicular to the $y$ -axis are semicircles. Find the volume.

Answer

$\tfrac{64\pi}{15} \approx 13.4$

Full solution

The diameter is $4 - y^2$ for $-2 \le y \le 2$ . $V = \tfrac{\pi}{8}\int_{-2}^{2} (4 - y^2)^2\,dy = \tfrac{\pi}{8} \cdot \tfrac{512}{15} = \tfrac{64\pi}{15}$ .
The base is the region between $y = x$ and $y = 0$ for $0 \le x \le 2$ , with semicircular cross sections perpendicular to the $x$ -axis. A student writes $V = \int_0^2 \tfrac{\pi}{2}x^2\,dx = \tfrac{4\pi}{3}$ . What went wrong?

Hint

Is $x$ the radius or the diameter of the semicircle at $x$ ?

Answer

The student used the diameter $x$ as the radius. The volume is $\tfrac{\pi}{3}$ .

Full solution

The segment across the base has length $x$ , and it is the diameter, so the radius is $\tfrac{x}{2}$ and $A(x) = \tfrac{1}{2}\pi\left(\tfrac{x}{2}\right)^2 = \tfrac{\pi}{8}x^2$ .

Then $V = \tfrac{\pi}{8}\int_0^2 x^2\,dx = \tfrac{\pi}{8} \cdot \tfrac{8}{3} = \tfrac{\pi}{3}$ , a quarter of the student’s answer.

Frequently asked questions

How do I find the volume of a solid with known cross sections?

Write the side of a slice as the width of the base at x, put it into the shape's area formula to get A(x), and integrate A(x) across the base.

What is the area of a semicircular cross section?

If its diameter s lies across the base, its radius is s/2, so its area is (1/2)π(s/2)² = πs²/8.

What is the area of an equilateral triangle cross section?

(√3/4)s², where s is the side lying across the base.

Why is the volume of a pyramid one third of base times height?

Its square cross sections grow like x², and integrating x² produces the factor 1/3: the integral from 0 to H of (sx/H)² dx is s²H/3.

When do I integrate with respect to y?

When the slices are perpendicular to the y-axis. Then the side is the right boundary minus the left, written in terms of y, with y-limits.

What to learn next

Volumes of Revolution: Disks and WashersFind the volume of a solid of revolution with disks and washers: V = π∫R² dx or π∫(R² − r²) dx, around the axes and around other lines.

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