8.4.1 Cross section method

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

Our AP Calculus AB course is currently in development and is a work-in-progress.

Cross section method

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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 rectangle has width $d x$ and height given by either the function value $f (x)$ or the distance between two curves.

That same “add up thin slices” idea works for 3D solids. Instead of rectangles, you slice the solid into infinitely thin cross sections - flat 2D slices perpendicular to an axis (usually the $x$ - or $y$ -axis). In these problems, the cross sections are usually familiar shapes such as squares, rectangles, triangles (equilateral, isosceles, or right), semicircles, and quarter circles. If you can write the area of a typical cross section as a function of position, then integrating that area 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) depend on the specific geometrical shape.

Example 1: squares

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

Here, $s$ is the distance between the curves. That distance becomes the side length of the square at a given $x$ (or at a given $y$ , depending on the slicing direction).

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 two sample cross-sectional slices. When you set up these problems, it helps to sketch the region in the $x y$ -plane and draw one or two representative slices. The sketch doesn’t need to be 3D - what matters is that you can identify the side length $s$ .

$s$ 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.

Since the slices are perpendicular to the $y$ -axis, the area function depends on $y$ . Rewrite the curve in terms of $y$ :

$x = y^{2}$

The side length $s$ is the horizontal distance from the left boundary $x = y^{2}$ to the right boundary $x = 4$ :

$s = 4 - y^{2}$

The area of each cross section is

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

The curve $x = y^{2}$ meets $x = 4$ at $(4, 2)$ and $(4, - 2)$ , so the interval is $y \in [- 2, 2]$ . The volume 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 $A = s \times h$ .

Here, $h$ is given in the problem, and $s$ is 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 (top minus bottom):

$s = x - x^{2}$

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

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

The curves intersect at $(0, 0)$ and $(1, 1)$ , so the interval is $x \in [0, 1]$ . 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)

Write the curves in terms of $y$ :

$x = y$
$x = y$

$s$ , the length of the base, is the horizontal distance (right minus left):

$s = y - y$

The height is twice the base length:

$h = 2 (y - y)$

So the area of each cross section is

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

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

The interval is $y \in [0, 1]$ . The volume is

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

$= 0.067$

Example 3: equilateral triangles

Use the 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 horizontal distance from the left boundary $x = - 1$ to the right boundary $x = \frac{1}{y}$ :

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

The area of each cross section is

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

The interval is from $y = 0.5$ to $y = 2$ , so 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

Here, the distance between the curves $s$ is the hypotenuse of a $45 - 45 - 90°$ triangle. Each leg has length $\frac{s}{2}$ , so 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 hypotenuse length $s$ is the horizontal distance from $x = 0$ (the $y$ -axis) to $x = y + 1$ :

$s = y + 1$

So the area of each cross section is

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

The interval is from $y = - 1$ to $y = 0$ , so the volume is

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

$= 0.0833$

Case 2: Leg as base

Here, the distance between the curves $s$ is a leg of the isosceles right triangle. The other leg is also $s$ , so 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 vertical distance from the top boundary $y = 0$ (the $x$ -axis) to the bottom boundary $y = x - 1$ :

$s = 0 - (x - 1)$

$= 1 - x$

So the area of each cross section is

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

The interval is from $x = 0$ to $x = 1$ , so 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 $A = \frac{1}{2} π r^{2}$ . Here, the radius $r$ is half the distance between the curves, so

$A = \frac{1}{2} π (\frac{s}{2})^{2}$

$= \frac{π}{8} s^{2}$

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 top minus bottom:

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

$= e^{x} + x$

So 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 $(- 0.56714, 0.56714)$ . The interval is from $x = - 0.56714$ to $x = 0$ , so the volume is

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

$= 0.0697$

The area of a quarter circle is $A = \frac{1}{4} π r^{2}$ . This time, the radius is the full distance between the curves, 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, rewrite the curves in terms of $y$ :

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

From $y = 0$ up to the intersection point $(- 0.56714, 0.56714)$ , the radius $s$ is the distance from $x = - y$ to the $y$ -axis $x = 0$ (right minus left):

$s = 0 - (- y)$

$s = y$

So 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 is the distance from $x = ln (y)$ to the $y$ -axis:

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

$= - ln (y)$

So 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

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

$= 0.0792$

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	$A = \frac{π}{4} s^{2}$

Cross section method

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) depend on the specific geometrical shape.

Example 1: squares

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

Here, $s$ is the distance between the curves. That distance becomes the side length of the square at a given $x$ (or at a given $y$ , depending on the slicing direction).

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.

Since the slices are perpendicular to the $y$ -axis, the area function depends on $y$ . Rewrite the curve in terms of $y$ :

$x = y^{2}$

The side length $s$ is the horizontal distance from the left boundary $x = y^{2}$ to the right boundary $x = 4$ :

$s = 4 - y^{2}$

The area of each cross section is

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

The curve $x = y^{2}$ meets $x = 4$ at $(4, 2)$ and $(4, - 2)$ , so the interval is $y \in [- 2, 2]$ . The volume 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 $A = s \times h$ .

Here, $h$ is given in the problem, and $s$ is 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 (top minus bottom):

$s = x - x^{2}$

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

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

The curves intersect at $(0, 0)$ and $(1, 1)$ , so the interval is $x \in [0, 1]$ . 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)

Write the curves in terms of $y$ :

$x = y$
$x = y$

$s$ , the length of the base, is the horizontal distance (right minus left):

$s = y - y$

The height is twice the base length:

$h = 2 (y - y)$

So the area of each cross section is

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

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

The interval is $y \in [0, 1]$ . The volume is

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

$= 0.067$

Example 3: equilateral triangles

Use the 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 horizontal distance from the left boundary $x = - 1$ to the right boundary $x = \frac{1}{y}$ :

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

The area of each cross section is

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

The interval is from $y = 0.5$ to $y = 2$ , so 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

Here, the distance between the curves $s$ is the hypotenuse of a $45 - 45 - 90°$ triangle. Each leg has length $\frac{s}{2}$ , so 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 hypotenuse length $s$ is the horizontal distance from $x = 0$ (the $y$ -axis) to $x = y + 1$ :

$s = y + 1$

So the area of each cross section is

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

The interval is from $y = - 1$ to $y = 0$ , so the volume is

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

$= 0.0833$

Case 2: Leg as base

Here, the distance between the curves $s$ is a leg of the isosceles right triangle. The other leg is also $s$ , so 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 vertical distance from the top boundary $y = 0$ (the $x$ -axis) to the bottom boundary $y = x - 1$ :

$s = 0 - (x - 1)$

$= 1 - x$

So the area of each cross section is

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

The interval is from $x = 0$ to $x = 1$ , so 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 $A = \frac{1}{2} π r^{2}$ . Here, the radius $r$ is half the distance between the curves, so

$A = \frac{1}{2} π (\frac{s}{2})^{2}$

$= \frac{π}{8} s^{2}$

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 top minus bottom:

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

$= e^{x} + x$

So 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 $(- 0.56714, 0.56714)$ . The interval is from $x = - 0.56714$ to $x = 0$ , so the volume is

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

$= 0.0697$

The area of a quarter circle is $A = \frac{1}{4} π r^{2}$ . This time, the radius is the full distance between the curves, 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, rewrite the curves in terms of $y$ :

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

From $y = 0$ up to the intersection point $(- 0.56714, 0.56714)$ , the radius $s$ is the distance from $x = - y$ to the $y$ -axis $x = 0$ (right minus left):

$s = 0 - (- y)$

$s = y$

So 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 is the distance from $x = ln (y)$ to the $y$ -axis:

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

$= - ln (y)$

So 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

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

$= 0.0792$

Achievable AP Calculus AB

8. Applications of integrals

8.4. Volume

Our AP Calculus AB course is currently in development and is a work-in-progress.

Cross section method

10 min read

Font

Discuss

Feedback

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) depend on the specific geometrical shape.

Example 1: squares

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

Here, $s$ is the distance between the curves. That distance becomes the side length of the square at a given $x$ (or at a given $y$ , depending on the slicing direction).

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.

Since the slices are perpendicular to the $y$ -axis, the area function depends on $y$ . Rewrite the curve in terms of $y$ :

$x = y^{2}$

The side length $s$ is the horizontal distance from the left boundary $x = y^{2}$ to the right boundary $x = 4$ :

$s = 4 - y^{2}$

The area of each cross section is

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

The curve $x = y^{2}$ meets $x = 4$ at $(4, 2)$ and $(4, - 2)$ , so the interval is $y \in [- 2, 2]$ . The volume 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 $A = s \times h$ .

Here, $h$ is given in the problem, and $s$ is 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 (top minus bottom):

$s = x - x^{2}$

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

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

The curves intersect at $(0, 0)$ and $(1, 1)$ , so the interval is $x \in [0, 1]$ . 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)

Write the curves in terms of $y$ :

$x = y$
$x = y$

$s$ , the length of the base, is the horizontal distance (right minus left):

$s = y - y$

The height is twice the base length:

$h = 2 (y - y)$

So the area of each cross section is

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

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

The interval is $y \in [0, 1]$ . The volume is

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

$= 0.067$

Example 3: equilateral triangles

Use the 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 horizontal distance from the left boundary $x = - 1$ to the right boundary $x = \frac{1}{y}$ :

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

The area of each cross section is

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

The interval is from $y = 0.5$ to $y = 2$ , so 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

Here, the distance between the curves $s$ is the hypotenuse of a $45 - 45 - 90°$ triangle. Each leg has length $\frac{s}{2}$ , so 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 hypotenuse length $s$ is the horizontal distance from $x = 0$ (the $y$ -axis) to $x = y + 1$ :

$s = y + 1$

So the area of each cross section is

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

The interval is from $y = - 1$ to $y = 0$ , so the volume is

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

$= 0.0833$

Case 2: Leg as base

Here, the distance between the curves $s$ is a leg of the isosceles right triangle. The other leg is also $s$ , so 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 vertical distance from the top boundary $y = 0$ (the $x$ -axis) to the bottom boundary $y = x - 1$ :

$s = 0 - (x - 1)$

$= 1 - x$

So the area of each cross section is

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

The interval is from $x = 0$ to $x = 1$ , so 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 $A = \frac{1}{2} π r^{2}$ . Here, the radius $r$ is half the distance between the curves, so

$A = \frac{1}{2} π (\frac{s}{2})^{2}$

$= \frac{π}{8} s^{2}$

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 top minus bottom:

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

$= e^{x} + x$

So 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 $(- 0.56714, 0.56714)$ . The interval is from $x = - 0.56714$ to $x = 0$ , so the volume is

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

$= 0.0697$

The area of a quarter circle is $A = \frac{1}{4} π r^{2}$ . This time, the radius is the full distance between the curves, 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, rewrite the curves in terms of $y$ :

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

From $y = 0$ up to the intersection point $(- 0.56714, 0.56714)$ , the radius $s$ is the distance from $x = - y$ to the $y$ -axis $x = 0$ (right minus left):

$s = 0 - (- y)$

$s = y$

So 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 is the distance from $x = ln (y)$ to the $y$ -axis:

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

$= - ln (y)$

So 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

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

$= 0.0792$

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	$A = \frac{π}{4} s^{2}$

Cross section method

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) depend on the specific geometrical shape.

Example 1: squares

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

Here, $s$ is the distance between the curves. That distance becomes the side length of the square at a given $x$ (or at a given $y$ , depending on the slicing direction).

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.

Since the slices are perpendicular to the $y$ -axis, the area function depends on $y$ . Rewrite the curve in terms of $y$ :

$x = y^{2}$

The side length $s$ is the horizontal distance from the left boundary $x = y^{2}$ to the right boundary $x = 4$ :

$s = 4 - y^{2}$

The area of each cross section is

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

The curve $x = y^{2}$ meets $x = 4$ at $(4, 2)$ and $(4, - 2)$ , so the interval is $y \in [- 2, 2]$ . The volume 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 $A = s \times h$ .

Here, $h$ is given in the problem, and $s$ is 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 (top minus bottom):

$s = x - x^{2}$

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

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

The curves intersect at $(0, 0)$ and $(1, 1)$ , so the interval is $x \in [0, 1]$ . 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)

Write the curves in terms of $y$ :

$x = y$
$x = y$

$s$ , the length of the base, is the horizontal distance (right minus left):

$s = y - y$

The height is twice the base length:

$h = 2 (y - y)$

So the area of each cross section is

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

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

The interval is $y \in [0, 1]$ . The volume is

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

$= 0.067$

Example 3: equilateral triangles

Use the 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 horizontal distance from the left boundary $x = - 1$ to the right boundary $x = \frac{1}{y}$ :

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

The area of each cross section is

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

The interval is from $y = 0.5$ to $y = 2$ , so 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

Here, the distance between the curves $s$ is the hypotenuse of a $45 - 45 - 90°$ triangle. Each leg has length $\frac{s}{2}$ , so 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 hypotenuse length $s$ is the horizontal distance from $x = 0$ (the $y$ -axis) to $x = y + 1$ :

$s = y + 1$

So the area of each cross section is

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

The interval is from $y = - 1$ to $y = 0$ , so the volume is

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

$= 0.0833$

Case 2: Leg as base

Here, the distance between the curves $s$ is a leg of the isosceles right triangle. The other leg is also $s$ , so 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 vertical distance from the top boundary $y = 0$ (the $x$ -axis) to the bottom boundary $y = x - 1$ :

$s = 0 - (x - 1)$

$= 1 - x$

So the area of each cross section is

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

The interval is from $x = 0$ to $x = 1$ , so 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 $A = \frac{1}{2} π r^{2}$ . Here, the radius $r$ is half the distance between the curves, so

$A = \frac{1}{2} π (\frac{s}{2})^{2}$

$= \frac{π}{8} s^{2}$

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 top minus bottom:

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

$= e^{x} + x$

So 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 $(- 0.56714, 0.56714)$ . The interval is from $x = - 0.56714$ to $x = 0$ , so the volume is

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

$= 0.0697$

The area of a quarter circle is $A = \frac{1}{4} π r^{2}$ . This time, the radius is the full distance between the curves, 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, rewrite the curves in terms of $y$ :

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

From $y = 0$ up to the intersection point $(- 0.56714, 0.56714)$ , the radius $s$ is the distance from $x = - y$ to the $y$ -axis $x = 0$ (right minus left):

$s = 0 - (- y)$

$s = y$

So 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 is the distance from $x = ln (y)$ to the $y$ -axis:

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

$= - ln (y)$

So 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	$A = \frac{π}{4} s^{2}$