Cross section method | Volume | Applications of integrals

What you’ll learn:

How integrals calculate volume
Set up and compute volumes of solids with known cross-sectional areas

We’ve seen that integrals compute area by adding up infinitely thin rectangles over an interval, each with width $d x$ and a height given by either the function value $f (x)$ or the distance between two curves.

A similar idea can be applied to 3D solids. Instead of thin rectangles, a solid can be sliced into infinitely thin cross sections - flat, 2D slices that are perpendicular to an axis (usually the $x$ or $y$ -axis). Generally, cross-sections in these problems are common geometrical shapes: squares, rectangles, triangles (equilateral, isosceles, or right), semi-circles, and quarter circles. Integrating the area of the specific cross section over an interval gives the volume.

To see what solids with certain cross sections look like, visit this GeoGebra application.

When cross sections are perpendicular to the $x$ -axis, use

$V = \int_{x = a}^{x = b} A (x) d x$

and when they are perpendicular to the $y$ -axis, use

$V = \int_{y = c}^{y = d} A (y) d y$

The expressions for $A (x)$ and $A (y)$ , the area of a cross section, will depend on the specific geometrical shape.

Example 1: Squares

Use the formula for the area of a square $A = s^{2}$ .

$s$ is the distance between the curves - the side length of a square at any $x$ .

The region bounded by $y = x, y = - x$ , and $x = 4$ forms the base of a solid. Find the volume of the solid where the cross sections are:

a) Squares perpendicular to the $x$ -axis.
b) Squares perpendicular to the $y$ -axis.

Solutions

a) Squares perpendicular to the $x$ -axis

(spoiler)

Here’s a visual of the curve with 2 cross-sectional slices. Practice making drawings with the curves and 1 or 2 sample slices when figuring out the side length $s$ for the integral set-up. It does not have to be a 3D drawing - a graph on the $x y$ -plane works just fine as long as you can figure out what the side length $s$ of the cross section is.

$s$ is the distance between the curves $y = x$ and $y = - x$ :

$s = x - (- x)$

$= 2 x$

The area of each cross section is

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

$= 4 x$

The interval to integrate over is from $x = 0$ to $x = 4$ , so the volume of the solid is

$V = \int_{0}^{4} (4 x) d x$

$= 2 x^{2}_{0}^{4}$

$= 32$

b) Squares perpendicular to the $y$ -axis

(spoiler)

Here’s a visual of two slices perpendicular to the $y$ -axis instead.

Since the slices are perpendicular to the $y$ -axis, the area function depends on $y$ . In this case, the two curves can be expressed as a single function of $y$ :

$x = y^{2}$

The side length $s$ is the distance between the right function/line $x = 4$ and the left function $x = y^{2}$ , or

$s = 4 - y^{2}$

The area of each cross section is

$A (y) = (4 - y^{2})^{2}$

$x = y^{2}$ meets $x = 4$ at $(4, 2)$ and $(4, - 2)$ so the interval to integrate over is from $y = - 2$ to $y = 2$ . The volume of the solid is

$V = \int_{- 2}^{2} (4 - y^{2})^{2} d y$

$= \int_{- 2}^{2} (16 - 8 y^{2} + y^{4}) d y$

$= 34.133$

Example 2: Rectangles

Use $A = s \times h$ .

$h$ will be given in the problem and $s$ is again the distance between the curves.

The region bounded by $y = x$ and $y = x^{2}$ forms the base of a solid. Find the volume when the cross sections are:

a) Rectangles perpendicular to the $x$ -axis with height $1.5$ .
b) Rectangles perpendicular to the $y$ -axis with height $2$ times the length of the base.

Solutions

a) Perpendicular to the $x$ -axis with height 1.5

(spoiler)

$s$ , the length of the base, is the distance between the curves, or the top function minus the bottom function:

$s = x - x^{2}$

The height is a constant $1.5$ , so the area of each cross section is

$A (x) = 1.5 (x - x^{2})$

The two curves intersect at $(0, 0)$ and $(1, 1)$ so the interval to integrate over is from $x = 0$ to $x = 1$ and the volume is

$V = \int_{0}^{1} 1.5 (x - x^{2}) d x$

$= 0.25$

b) Perpendicular to the $y$ -axis with height 2 times the length of the base

(spoiler)

The functions in terms of $y$ are $x = y$ and $x = y$ .

$s$ , the length of the base, is the distance between the curves, or the right function minus the left function:

$s = y - y$

The height of each cross section is twice the length of the base, or

$h = 2 (y - y)$

Then the area of each cross section is

$A (y) = (y - y) \times 2 (y - y)$

$= 2 (y - y)^{2}$

The interval to integrate over is $y = 0$ to $y = 1$ and the volume is

$V = \int_{0}^{1} 2 (y - y)^{2} d y$

$= 0.067$

Example 3: Equilateral triangles

Use area formula $A = \frac{3}{4} s^{2}$

The region bounded by $y = \frac{1}{x}$ and the lines $y = 0.5, y = 2$ , and $x = - 1$ forms the base of a solid. Find the volume of the solid if cross sections perpendicular to the $y$ -axis are equilateral triangles.

Solution

(spoiler)

Slices perpendicular to the $y$ -axis depend on $y$ . Each triangle’s side length $s$ is the distance between the right function $x = \frac{1}{y}$ and the left function/line $x = - 1$ .

$s = \frac{1}{y} + 1$

The area of each cross section is

$A (y) = \frac{3}{4} (\frac{1}{y} + 1)^{2}$

The interval to integrate over is from $y = 0.5$ to $y = 2$ and the volume is

$V = \frac{3}{4} \int_{0.5}^{2} (\frac{1}{y} + 1)^{2} d y$

$= 2.500$

Example 4: Isosceles right triangles

There are two possibilities here:

Hypotenuse as the base
Leg as the base

Case 1: Hypotenuse as base

The distance between the curves $s$ is the hypotenuse of a $45 - 45 - 90°$ triangle, so the legs both have length $\frac{s}{2}$ . Then the area of each cross section is

$A = \frac{1}{2} (\frac{s}{2})^{2}$

$= \frac{1}{4} s^{2}$

The region bounded by the line $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms the base of a solid. Find the volume if the cross sections perpendicular to the $y$ -axis are isosceles right triangles whose hypotenuses lie along the base of the region.

Solution

(spoiler)

Slices perpendicular to the $y$ -axis depend on $y$ . The length of the hypotenuse $s$ is the distance between the right function $x = y + 1$ and left function $x = 0$ (the $y$ -axis), so

$s = y + 1$

Then the area of each cross section is

$A (y) = \frac{1}{4} (y + 1)^{2}$

The interval to integrate over is from $y = - 1$ to $y = 0$ and the volume is

$V = \frac{1}{4} \int_{- 1}^{0} (y + 1)^{2} d y$

$= 0.0833$

Case 2: Leg as base

When the distance between the curves, $s$ , is the leg of an isosceles right triangle, the other leg is also $s$ and the area formula is

$A = \frac{1}{2} s^{2}$

The region bounded by the line $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms the base of a solid. Find the volume if the cross sections perpendicular to the $x$ -axis are isosceles right triangles whose legs lie along the base of the region.

Solution

(spoiler)

$s$ is the distance between the top function $y = 0$ (the $x$ -axis) and the bottom function $y = x - 1$ .

$s = 0 - (x - 1)$

$= 1 - x$

The area of each cross section is

$A (x) = \frac{1}{2} (1 - x)^{2}$

The interval to integrate over is from $x = 0$ to $x = 1$ and the volume is

$V = \frac{1}{2} \int_{0}^{1} (1 - x)^{2} d x$

$= 0.167$

Example 5: Semi-circles & quarter circles

The area of a semicircle is $A = \frac{1}{2} π r^{2}$ . The radius $r$ is half of the distance between the curves, so the formula for the area is

$A = \frac{1}{2} π (\frac{s}{2})^{2}$

$= \frac{π}{8} s^{2}$

The region bounded by the curves $y = e^{x}$ and $y = - x$ and the $y$ -axis forms the base of a solid. Find its volume if the cross sections are semi-circles perpendicular to the $x$ -axis. Use a calculator to evaluate after setting up the integral.

Solution

(spoiler)

$s$ , the distance between the curves, is the top function $y = e^{x}$ minus the bottom one, $y = - x$ .

$s = e^{x} - (- x)$

$= e^{x} + x$

The area of each cross section is

$A (x) = \frac{π}{8} (e^{x} + x)^{2}$

Graphing the two functions on Desmos shows that they intersect at the point $(- 0.56714, 0.56714)$ . The interval to integrate over is from $x = - 0.56714$ to $x = 0$ and the volume is

$V = \frac{π}{8} \int_{- 0.56714}^{0} (e^{x} + x)^{2} d x$

$= 0.0697$

The formula for the area of a quarter circle is $A = \frac{1}{4} π r^{2}$ . This time, the radius is the distance between the curves and not half. So

$A = \frac{π}{4} s^{2}$

The region bounded by the curves $y = e^{x}$ and $y = - x$ and the $y$ -axis forms the base of a solid. Find its volume if the cross sections are quarter circles perpendicular to the $y$ -axis.

Solution

(spoiler)

Since the slices are perpendicular to the $y$ -axis, the two curves must be expressed in terms of $y$ :

$x = - y$
$x = ln (y)$

The radius of each quarter circle cross section ( $s$ ) from $y = 0$ up to the intersection point of $(- 0.56714, 0.56714)$ is the distance between the $y$ -axis ( $x = 0$ ) and the line $x = - y$ (right function minus the left function), so

$s = 0 - (- y)$

$s = y$

Then the area of each cross section is

$A_{1} (y) = \frac{π}{4} y^{2}$

The volume for that portion is

$V_{1} = \frac{π}{4} \int_{0}^{0.56714} y^{2} d y$

$= 0.0478$

From $y = 0.56714$ to $y = 1$ , the radius of each quarter circle is the distance between the $y$ -axis and the curve $x = ln (y)$ instead, so

$s = 0 - (ln (y))$

$= - ln (y)$

Then the area of each cross section is

$A_{2} (y) = \frac{π}{4} (- ln (y))^{2}$

$= \frac{π}{4} ln^{2} (y)$

The volume for that portion is

$V_{2} = \frac{π}{4} \int_{0.56714}^{1} ln^{2} (y) d y$

$= 0.0314$

The total volume of the solid is

$V_{1} + V_{2} = 0.0478 + 0.0314$

$= 0.0792$

Key points

Always identify:

The direction of slicing (perpendicular to $x$ or $y$ -axis) to determine the expression for the side length $s$ .
The area formula $A (x)$ or $A (y)$ depending on the shape of the cross section
The limits of integration

Common area formulas, where $s$ is the distance between the curves and/or axes that defines the size of the shape:

Shape	Area
Square	$A = s^{2}$
Rectangle	$A = s \times h$
Equilateral triangle	$A = \frac{3}{4} s^{2}$
Isosceles right triangle, hypotenuse as base	$A = \frac{1}{4} s^{2}$
Isosceles right triangle, leg as base	$A = \frac{1}{2} s^{2}$
Semicircle	$A = \frac{π}{8} s^{2}$
Quarter circle	$A = \frac{π}{4} s^{2}$

What you’ll learn:

How integrals calculate volume
Set up and compute volumes of solids with known cross-sectional areas

To see what solids with certain cross sections look like, visit this GeoGebra application.

When cross sections are perpendicular to the $x$ -axis, use

$V = \int_{x = a}^{x = b} A (x) d x$

and when they are perpendicular to the $y$ -axis, use

$V = \int_{y = c}^{y = d} A (y) d y$

The expressions for $A (x)$ and $A (y)$ , the area of a cross section, will depend on the specific geometrical shape.

Example 1: Squares

Use the formula for the area of a square $A = s^{2}$ .

$s$ is the distance between the curves - the side length of a square at any $x$ .

The region bounded by $y = x, y = - x$ , and $x = 4$ forms the base of a solid. Find the volume of the solid where the cross sections are:

a) Squares perpendicular to the $x$ -axis.
b) Squares perpendicular to the $y$ -axis.

Solutions

a) Squares perpendicular to the $x$ -axis

(spoiler)

$s$ is the distance between the curves $y = x$ and $y = - x$ :

$s = x - (- x)$

$= 2 x$

The area of each cross section is

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

$= 4 x$

The interval to integrate over is from $x = 0$ to $x = 4$ , so the volume of the solid is

$V = \int_{0}^{4} (4 x) d x$

$= 2 x^{2}_{0}^{4}$

$= 32$

b) Squares perpendicular to the $y$ -axis

(spoiler)

Here’s a visual of two slices perpendicular to the $y$ -axis instead.

Since the slices are perpendicular to the $y$ -axis, the area function depends on $y$ . In this case, the two curves can be expressed as a single function of $y$ :

$x = y^{2}$

The side length $s$ is the distance between the right function/line $x = 4$ and the left function $x = y^{2}$ , or

$s = 4 - y^{2}$

The area of each cross section is

$A (y) = (4 - y^{2})^{2}$

$x = y^{2}$ meets $x = 4$ at $(4, 2)$ and $(4, - 2)$ so the interval to integrate over is from $y = - 2$ to $y = 2$ . The volume of the solid is

$V = \int_{- 2}^{2} (4 - y^{2})^{2} d y$

$= \int_{- 2}^{2} (16 - 8 y^{2} + y^{4}) d y$

$= 34.133$

Example 2: Rectangles

Use $A = s \times h$ .

$h$ will be given in the problem and $s$ is again the distance between the curves.

The region bounded by $y = x$ and $y = x^{2}$ forms the base of a solid. Find the volume when the cross sections are:

a) Rectangles perpendicular to the $x$ -axis with height $1.5$ .
b) Rectangles perpendicular to the $y$ -axis with height $2$ times the length of the base.

Solutions

a) Perpendicular to the $x$ -axis with height 1.5

(spoiler)

$s$ , the length of the base, is the distance between the curves, or the top function minus the bottom function:

$s = x - x^{2}$

The height is a constant $1.5$ , so the area of each cross section is

$A (x) = 1.5 (x - x^{2})$

The two curves intersect at $(0, 0)$ and $(1, 1)$ so the interval to integrate over is from $x = 0$ to $x = 1$ and the volume is

$V = \int_{0}^{1} 1.5 (x - x^{2}) d x$

$= 0.25$

b) Perpendicular to the $y$ -axis with height 2 times the length of the base

(spoiler)

The functions in terms of $y$ are $x = y$ and $x = y$ .

$s$ , the length of the base, is the distance between the curves, or the right function minus the left function:

$s = y - y$

The height of each cross section is twice the length of the base, or

$h = 2 (y - y)$

Then the area of each cross section is

$A (y) = (y - y) \times 2 (y - y)$

$= 2 (y - y)^{2}$

The interval to integrate over is $y = 0$ to $y = 1$ and the volume is

$V = \int_{0}^{1} 2 (y - y)^{2} d y$

$= 0.067$

Example 3: Equilateral triangles

Use area formula $A = \frac{3}{4} s^{2}$

The region bounded by $y = \frac{1}{x}$ and the lines $y = 0.5, y = 2$ , and $x = - 1$ forms the base of a solid. Find the volume of the solid if cross sections perpendicular to the $y$ -axis are equilateral triangles.

Solution

(spoiler)

Slices perpendicular to the $y$ -axis depend on $y$ . Each triangle’s side length $s$ is the distance between the right function $x = \frac{1}{y}$ and the left function/line $x = - 1$ .

$s = \frac{1}{y} + 1$

The area of each cross section is

$A (y) = \frac{3}{4} (\frac{1}{y} + 1)^{2}$

The interval to integrate over is from $y = 0.5$ to $y = 2$ and the volume is

$V = \frac{3}{4} \int_{0.5}^{2} (\frac{1}{y} + 1)^{2} d y$

$= 2.500$

Example 4: Isosceles right triangles

There are two possibilities here:

Hypotenuse as the base
Leg as the base

Case 1: Hypotenuse as base

The distance between the curves $s$ is the hypotenuse of a $45 - 45 - 90°$ triangle, so the legs both have length $\frac{s}{2}$ . Then the area of each cross section is

$A = \frac{1}{2} (\frac{s}{2})^{2}$

$= \frac{1}{4} s^{2}$

The region bounded by the line $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms the base of a solid. Find the volume if the cross sections perpendicular to the $y$ -axis are isosceles right triangles whose hypotenuses lie along the base of the region.

Solution

(spoiler)

Slices perpendicular to the $y$ -axis depend on $y$ . The length of the hypotenuse $s$ is the distance between the right function $x = y + 1$ and left function $x = 0$ (the $y$ -axis), so

$s = y + 1$

Then the area of each cross section is

$A (y) = \frac{1}{4} (y + 1)^{2}$

The interval to integrate over is from $y = - 1$ to $y = 0$ and the volume is

$V = \frac{1}{4} \int_{- 1}^{0} (y + 1)^{2} d y$

$= 0.0833$

Case 2: Leg as base

When the distance between the curves, $s$ , is the leg of an isosceles right triangle, the other leg is also $s$ and the area formula is

$A = \frac{1}{2} s^{2}$

The region bounded by the line $y = x - 1$ , the $x$ -axis, and the $y$ -axis forms the base of a solid. Find the volume if the cross sections perpendicular to the $x$ -axis are isosceles right triangles whose legs lie along the base of the region.

Solution

(spoiler)

$s$ is the distance between the top function $y = 0$ (the $x$ -axis) and the bottom function $y = x - 1$ .

$s = 0 - (x - 1)$

$= 1 - x$

The area of each cross section is

$A (x) = \frac{1}{2} (1 - x)^{2}$

The interval to integrate over is from $x = 0$ to $x = 1$ and the volume is

$V = \frac{1}{2} \int_{0}^{1} (1 - x)^{2} d x$

$= 0.167$

Example 5: Semi-circles & quarter circles

The area of a semicircle is $A = \frac{1}{2} π r^{2}$ . The radius $r$ is half of the distance between the curves, so the formula for the area is

$A = \frac{1}{2} π (\frac{s}{2})^{2}$

$= \frac{π}{8} s^{2}$

The region bounded by the curves $y = e^{x}$ and $y = - x$ and the $y$ -axis forms the base of a solid. Find its volume if the cross sections are semi-circles perpendicular to the $x$ -axis. Use a calculator to evaluate after setting up the integral.

Solution

(spoiler)

$s$ , the distance between the curves, is the top function $y = e^{x}$ minus the bottom one, $y = - x$ .

$s = e^{x} - (- x)$

$= e^{x} + x$

The area of each cross section is

$A (x) = \frac{π}{8} (e^{x} + x)^{2}$

Graphing the two functions on Desmos shows that they intersect at the point $(- 0.56714, 0.56714)$ . The interval to integrate over is from $x = - 0.56714$ to $x = 0$ and the volume is

$V = \frac{π}{8} \int_{- 0.56714}^{0} (e^{x} + x)^{2} d x$

$= 0.0697$

The formula for the area of a quarter circle is $A = \frac{1}{4} π r^{2}$ . This time, the radius is the distance between the curves and not half. So

$A = \frac{π}{4} s^{2}$

The region bounded by the curves $y = e^{x}$ and $y = - x$ and the $y$ -axis forms the base of a solid. Find its volume if the cross sections are quarter circles perpendicular to the $y$ -axis.

Solution

(spoiler)

Since the slices are perpendicular to the $y$ -axis, the two curves must be expressed in terms of $y$ :

$x = - y$
$x = ln (y)$

$s = 0 - (- y)$

$s = y$

Then the area of each cross section is

$A_{1} (y) = \frac{π}{4} y^{2}$

The volume for that portion is

$V_{1} = \frac{π}{4} \int_{0}^{0.56714} y^{2} d y$

$= 0.0478$

From $y = 0.56714$ to $y = 1$ , the radius of each quarter circle is the distance between the $y$ -axis and the curve $x = ln (y)$ instead, so

$s = 0 - (ln (y))$

$= - ln (y)$

Then the area of each cross section is

$A_{2} (y) = \frac{π}{4} (- ln (y))^{2}$

$= \frac{π}{4} ln^{2} (y)$

The volume for that portion is

$V_{2} = \frac{π}{4} \int_{0.56714}^{1} ln^{2} (y) d y$

$= 0.0314$

The total volume of the solid is

$V_{1} + V_{2} = 0.0478 + 0.0314$

$= 0.0792$

Key points

Always identify:

The direction of slicing (perpendicular to $x$ or $y$ -axis) to determine the expression for the side length $s$ .
The area formula $A (x)$ or $A (y)$ depending on the shape of the cross section
The limits of integration

Common area formulas, where $s$ is the distance between the curves and/or axes that defines the size of the shape:

Shape	Area
Square	$A = s^{2}$
Rectangle	$A = s \times h$
Equilateral triangle	$A = \frac{3}{4} s^{2}$
Isosceles right triangle, hypotenuse as base	$A = \frac{1}{4} s^{2}$
Isosceles right triangle, leg as base	$A = \frac{1}{2} s^{2}$
Semicircle	$A = \frac{π}{8} s^{2}$
Quarter circle	$A = \frac{π}{4} s^{2}$