Volumes by Cross Sections: Examples

To find volumes by cross sections, slice the solid perpendicular to an axis, find the area A(x) of a typical slice, and integrate: $V = \int_a^b A(x)\,dx$ . The slice’s size s comes from the base region. For squares A = s², for semicircles with diameter s A = πs²/8, and for equilateral triangles A = (√3/4)s².

Before this lesson: Area Between Curves, Cross Sections (Geometry) and Fundamental Theorem of Calculus, Part 2

The idea: stack thin slices

Think of a loaf of bread. Each slice is thin, with a face of area A(x) and thickness dx. Its volume is about A(x) dx. Adding up all the slices from one end to the other gives the volume of the loaf.

In these problems, the solid sits on a flat base region. Every slice stands straight up from the base, and its shape (square, semicircle, triangle) is given. The width of the base at x is the size s of that slice.

Left: top view of the base region under y = square root of x from x = 0 to 4, with one thin slice at x marked; its length is s = square root of x. Right: that slice stands up as a square with side s, so its area is A(x) = s squared = x. — Each slice of the base has length s = √x. Standing up as a square, it has area A(x) = (√x)² = x.

Volumes by cross sections formula

For slices perpendicular to the x-axis from x = a to x = b:

V = \int_a^b A(x)\,dx

For slices perpendicular to the y-axis, use $V = \int_c^d A(y)\,dy$ . Here s is the length of the slice, top minus bottom (or right minus left).

Cross section shape	Area in terms of s
Square with side s	$A = s^2$
Semicircle with diameter s	$A = \dfrac{\pi s^2}{8}$
Equilateral triangle with side s	$A = \dfrac{\sqrt{3}}{4}s^2$
Isosceles right triangle with leg s	$A = \dfrac{s^2}{2}$
Rectangle with base s and height h	$A = sh$

The semicircle formula comes from radius s/2: $\dfrac{1}{2}\pi\left(\dfrac{s}{2}\right)^2 = \dfrac{\pi s^2}{8}$ .

How to find the volume:

Sketch the base region and draw one slice perpendicular to the given axis.
Write the slice length s in terms of x (top minus bottom) or y (right minus left).
Put s into the area formula for the shape to get A(x) or A(y).
Find the limits from the base and integrate A from one end to the other.

Volumes by cross sections examples

Example 1: Square cross sections

The base of a solid is the region under $y = \sqrt{x}$ from x = 0 to x = 4. Cross sections perpendicular to the x-axis are squares. Find the volume.

s = \sqrt{x} - 0 = \sqrt{x}

Slice length: top curve minus the x-axis.

A(x) = (\sqrt{x})^2 = x

Square: A = s².

V = \int_0^4 x\,dx

Integrate A(x) across the base.

= \Big[\dfrac{x^2}{2}\Big]_0^4 = 8 - 0 = 8

Antidifferentiate and plug in the limits (see fundamental theorem of calculus, part 2).

Answer: 8 cubic units.

Example 2: Semicircle cross sections

The base is the region between $y = x$ and $y = x^2$ . Cross sections perpendicular to the x-axis are semicircles. Find the volume.

x = x^2 \text{ at } x = 0 \text{ and } x = 1

Set the curves equal: x(x − 1) = 0.

s = x - x^2

Slice length: y = x is on top on [0, 1] (see area between curves).

A(x) = \dfrac{\pi}{8}(x - x^2)^2

Semicircle: A = πs²/8.

(x - x^2)^2 = x^2 - 2x^3 + x^4

Square the binomial (see special products of binomials).

V = \dfrac{\pi}{8}\int_0^1 (x^2 - 2x^3 + x^4)\,dx

Integrate A(x) from 0 to 1.

= \dfrac{\pi}{8}\Big[\dfrac{x^3}{3} - \dfrac{x^4}{2} + \dfrac{x^5}{5}\Big]_0^1

Antidifferentiate each term.

= \dfrac{\pi}{8}\left(\dfrac{1}{3} - \dfrac{1}{2} + \dfrac{1}{5}\right)

Plug in 1; every term is 0 at 0.

= \dfrac{\pi}{8}\cdot\dfrac{1}{30} = \dfrac{\pi}{240}

10/30 − 15/30 + 6/30 = 1/30.

Answer: $\dfrac{\pi}{240} \approx 0.0131$ cubic units. The base is small and thin, so the volume is small.

Example 3: Equilateral triangle cross sections

The base is the triangle with vertices (0, 0), (4, 0) and (0, 4). Cross sections perpendicular to the x-axis are equilateral triangles. Find the volume.

y = 4 - x

The slanted side of the base runs from (0, 4) to (4, 0).

s = 4 - x

Slice length: the line minus the x-axis.

A(x) = \dfrac{\sqrt{3}}{4}(4 - x)^2

Equilateral triangle: A = (√3/4)s².

V = \dfrac{\sqrt{3}}{4}\int_0^4 (4 - x)^2\,dx

Integrate from 0 to 4.

= \dfrac{\sqrt{3}}{4}\Big[-\dfrac{(4 - x)^3}{3}\Big]_0^4

Use u = 4 − x, du = −dx (see u-substitution).

= \dfrac{\sqrt{3}}{4}\left(0 + \dfrac{64}{3}\right)

Plug in the limits and subtract.

= \dfrac{16\sqrt{3}}{3} \approx 9.24

Multiply.

Answer: $\dfrac{16\sqrt{3}}{3} \approx 9.24$ cubic units.

Example 4: Slices perpendicular to the y-axis

The base is the region between $x = y^2$ and $x = 4$ . Cross sections perpendicular to the y-axis are squares. Find the volume.

y^2 = 4 \text{ at } y = -2 \text{ and } y = 2

Find where the curves meet.

s = 4 - y^2

Slice length: right (x = 4) minus left (x = y²).

A(y) = (4 - y^2)^2 = 16 - 8y^2 + y^4

Square the slice length.

V = \int_{-2}^{2} (16 - 8y^2 + y^4)\,dy

Integrate in y from −2 to 2.

= \Big[16y - \dfrac{8y^3}{3} + \dfrac{y^5}{5}\Big]_{-2}^{2}

Antidifferentiate each term.

= 2\left(32 - \dfrac{64}{3} + \dfrac{32}{5}\right)

The value at −2 is the negative of the value at 2, so the difference is twice the value at 2.

= 2\cdot\dfrac{256}{15} = \dfrac{512}{15}

480/15 − 320/15 + 96/15 = 256/15.

Answer: $\dfrac{512}{15} \approx 34.13$ cubic units.

Example 5: A stone paperweight

A paperweight has a circular base of radius 3 cm, $x^2 + y^2 \le 9$ . Cross sections perpendicular to the x-axis are squares. Find its volume. If the stone has a density of 2.5 grams per cubic centimeter, what is its mass?

y = \pm\sqrt{9 - x^2}

Solve the circle equation for y (see equation of a circle).

s = 2\sqrt{9 - x^2}

Slice length: top half minus bottom half.

A(x) = \left(2\sqrt{9 - x^2}\right)^2 = 4(9 - x^2)

Square: A = s².

V = \int_{-3}^{3} (36 - 4x^2)\,dx

Integrate across the circle, from x = −3 to 3.

= \Big[36x - \dfrac{4x^3}{3}\Big]_{-3}^{3} = (108 - 36) - (-108 + 36)

Antidifferentiate and plug in the limits.

= 72 + 72 = 144

Simplify.

144 \times 2.5 = 360

Mass = volume × density.

Answer: the volume is 144 cm³ and the mass is 360 g.

Common mistakes

Mistake	Example of the error	Fix
Integrating s instead of A	$V = \int_0^4 \sqrt{x}\,dx$ for square slices	Integrate the area: $A(x) = s^2 = x$ .
Using s as the semicircle’s radius	$A = \dfrac{1}{2}\pi s^2$	s is the diameter, so the radius is s/2 and $A = \dfrac{\pi s^2}{8}$ .
Measuring s from the axis instead of between curves	$s = x$ for the region between y = x and y = x²	Use top minus bottom: $s = x - x^2$ .
Using x when slices are perpendicular to the y-axis	Writing s in terms of x for Example 4	Slices perpendicular to the y-axis need s in terms of y and dy.
Forgetting the shape’s constant	$A = s^2$ for equilateral triangles	Use $A = \dfrac{\sqrt{3}}{4}s^2$ .

Related methods

The disk method is a special case

If you spin a region around an axis, every slice is a circle. Then A(x) = πr², and the cross-section formula becomes the disk method. That is the next lesson, Disk Method.

Checking with geometry

The cross-section idea also explains known formulas. A pyramid with a square base of side b and height h has square slices whose side shrinks to 0 at the top, and $\int_0^h \left(\dfrac{b(h - y)}{h}\right)^2 dy = \dfrac{1}{3}b^2h$ , the formula from volume of pyramids and cones.

Practice problems

The base is the region under $y = 2x$ from x = 0 to x = 3. Cross sections perpendicular to the x-axis are squares. Find the volume.
The base is the region between $y = x^2$ and $y = 1$ . Cross sections perpendicular to the x-axis are squares. Find the volume.
The base is the region under $y = \sqrt{x}$ from x = 0 to x = 4. Cross sections perpendicular to the x-axis are semicircles. Find the volume.
The base is the triangle under $y = 3 - x$ from x = 0 to x = 3. Cross sections perpendicular to the x-axis are squares. Find the volume.
Use the same base as problem 4, with equilateral triangle cross sections. Find the volume.
The base is the region between $y = x$ and $y = x^2$ . Cross sections perpendicular to the x-axis are squares. Find the volume.
The base is the region under $y = e^x$ from x = 0 to x = 1. Cross sections perpendicular to the x-axis are squares. Find the volume.
The base is the region between $x = y^2$ and $x = 1$ . Cross sections perpendicular to the y-axis are isosceles right triangles with a leg in the base. Find the volume.
The base is the disk $x^2 + y^2 \le 4$ . Cross sections perpendicular to the x-axis are semicircles. Find the volume.
A glass paperweight has a base bounded by $y = 1 - \dfrac{x^2}{4}$ and the x-axis, in inches. Cross sections perpendicular to the x-axis are squares. Find its volume.

Show answers

s = 2x, \; A(x) = 4x^2

Slice length and square area.

V = \int_0^3 4x^2\,dx = \Big[\dfrac{4x^3}{3}\Big]_0^3 = 36

Integrate and evaluate.

Answer: 36

x^2 = 1 \text{ at } x = \pm 1

Find the limits.

A(x) = (1 - x^2)^2 = 1 - 2x^2 + x^4

s = 1 − x²; square it.

V = \int_{-1}^{1} (1 - 2x^2 + x^4)\,dx = \Big[x - \dfrac{2x^3}{3} + \dfrac{x^5}{5}\Big]_{-1}^{1}

Integrate.

= 2\left(1 - \dfrac{2}{3} + \dfrac{1}{5}\right) = 2\cdot\dfrac{8}{15} = \dfrac{16}{15}

Evaluate; the value at −1 is the negative of the value at 1.

Answer: $\dfrac{16}{15}$

A(x) = \dfrac{\pi}{8}(\sqrt{x})^2 = \dfrac{\pi}{8}x

s = √x; semicircle area.

V = \int_0^4 \dfrac{\pi}{8}x\,dx = \dfrac{\pi}{8}\Big[\dfrac{x^2}{2}\Big]_0^4 = \dfrac{\pi}{8}(8) = \pi

Integrate and evaluate.

Answer: $\pi \approx 3.14$

V = \int_0^3 (3 - x)^2\,dx = \Big[-\dfrac{(3 - x)^3}{3}\Big]_0^3

s = 3 − x; A = s².

= 0 - \left(-\dfrac{27}{3}\right) = 9

Evaluate.

Answer: 9

V = \dfrac{\sqrt{3}}{4}\int_0^3 (3 - x)^2\,dx = \dfrac{\sqrt{3}}{4}(9)

A = (√3/4)s²; the integral is 9 from problem 4.

= \dfrac{9\sqrt{3}}{4} \approx 3.90

Multiply.

Answer: $\dfrac{9\sqrt{3}}{4} \approx 3.90$

A(x) = (x - x^2)^2 = x^2 - 2x^3 + x^4

s = x − x² on [0, 1].

V = \int_0^1 (x^2 - 2x^3 + x^4)\,dx = \dfrac{1}{3} - \dfrac{1}{2} + \dfrac{1}{5} = \dfrac{1}{30}

Integrate and evaluate (same integral as Example 2).

Answer: $\dfrac{1}{30}$

A(x) = (e^x)^2 = e^{2x}

s = eˣ.

V = \int_0^1 e^{2x}\,dx = \Big[\dfrac{e^{2x}}{2}\Big]_0^1 = \dfrac{e^2 - 1}{2}

Integrate with u = 2x and evaluate.

\dfrac{e^2 - 1}{2} \approx 3.195

Use e² ≈ 7.389.

Answer: $\dfrac{e^2 - 1}{2} \approx 3.195$

s = 1 - y^2, \; A(y) = \dfrac{(1 - y^2)^2}{2}

Right minus left; isosceles right triangle area s²/2.

V = \dfrac{1}{2}\int_{-1}^{1} (1 - 2y^2 + y^4)\,dy = \dfrac{1}{2}\cdot\dfrac{16}{15}

Same integral as problem 2, in y.

= \dfrac{8}{15}

Multiply.

Answer: $\dfrac{8}{15}$

s = 2\sqrt{4 - x^2}

Top half minus bottom half of the circle.

A(x) = \dfrac{\pi}{8}\cdot 4(4 - x^2) = \dfrac{\pi}{2}(4 - x^2)

Semicircle area πs²/8.

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

Integrate and evaluate.

Answer: $\dfrac{16\pi}{3} \approx 16.76$ . The solid is half of a ball of radius 2.

10.

1 - \dfrac{x^2}{4} = 0 \text{ at } x = \pm 2

Find the limits.

A(x) = \left(1 - \dfrac{x^2}{4}\right)^2 = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{16}

Square the slice length.

V = \Big[x - \dfrac{x^3}{6} + \dfrac{x^5}{80}\Big]_{-2}^{2}

Integrate each term.

= 2\left(2 - \dfrac{4}{3} + \dfrac{2}{5}\right) = 2\cdot\dfrac{16}{15} = \dfrac{32}{15}

Evaluate: 8/6 = 4/3 and 32/80 = 2/5.

Answer: $\dfrac{32}{15} \approx 2.13$ cubic inches

Volumes by cross sections FAQ

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

Find the area A(x) of a typical slice from its shape and its length s, then integrate A(x) over the base: $V = \int_a^b A(x)\,dx$ .

What is the area of a semicircle cross section?

If the diameter is s, the radius is s/2, so the area is $\dfrac{1}{2}\pi\left(\dfrac{s}{2}\right)^2 = \dfrac{\pi s^2}{8}$ .

When do you integrate with respect to y?

When the cross sections are perpendicular to the y-axis. Then the slice length is right minus left in terms of y, and the limits are y-values.

How is this different from the disk method?

The disk method is the case where every slice is a circle, made by spinning a region around an axis. Here the slices can be any shape.

Summary

Volume by cross sections: $V = \int_a^b A(x)\,dx$ , the sum of thin slices.
s is the length of the slice in the base: top minus bottom, or right minus left.
Square: s². Semicircle: πs²/8. Equilateral triangle: (√3/4)s².
Slices perpendicular to the y-axis use A(y) and dy.
Integrate the area, not the length.

Previous lesson: Area Between Curves

Next lesson: Disk Method

Back to the Calculus 1 Index