8.9 - Volume with Disc Method: Revolving Around the x- or y-Axis

What Is a Solid of Revolution?

A solid of revolution is a 3D object created by rotating a 2D region around an axis, called the axis of revolution.

Take a bounded region in the plane and spin it 360° around a line (usually the

x

-axis or

y

-axis). As it rotates, it sweeps out a solid.

For example, the region bounded by

y = x^2

y = 0

, and

x = 2

, rotated around the

x

-axis, creates a funnel-like solid.

8.9 - Volume with Disc Method: Revolving Around the x- or y-Axis

Sketching Solids of Revolution on Paper

Step 1: Draw the original region.

Step 2: Draw the mirror image across the axis of revolution.

Step 3: Connect the edges with curved lines to suggest the 3D shape. Draw elliptical curves at the ends, use dashed lines for hidden back edges, and add shading or perspective lines.

Important: even though the middle looks empty in a 2D sketch, the solid is completely filled. There are no hollow spaces.

Cross-Sections: The Key to the Volume

At any position along the axis of revolution, slicing perpendicular to the axis gives you a circle (a disk). Why circles? Because we rotated around an axis, so every point traces a circular path.

The area of one circular cross-section is:

A = \pi r^2

Finding the radius. The radius runs from the axis of revolution out to the outer edge of the solid. For a region bounded by

y = f(x)

rotated around the

x

-axis:

r = f(x)

The radius at position

x

is just the function value, the distance from the axis to the curve.

The Disk Method Formula

\boxed{V = \int_{a}^{b} \pi R^2\,dx}

where: -

R

is the radius function (distance from the axis of revolution to the outer edge) -

[a, b]

is the interval along the axis of revolution - We use capital

R

for the outer radius

We’re summing up the areas of infinitely many circular disks from

x = a

x = b

Example 1: Revolution Around the x-axis

Find the volume of the solid formed by rotating the region bounded by

y = x^2

y = 0

, and

x = 2

around the

x

-axis.

Step 1: radius function. The distance from the

x

-axis to the curve:

R = x^2

Step 2: bounds. From

x = 0

x = 2

Step 3: set up

V = \int_{0}^{2} \pi (x^2)^2\,dx = \int_{0}^{2} \pi x^4\,dx

Step 4: factor out the constant and integrate

= \pi \int_{0}^{2} x^4\,dx = \pi \left[\frac{x^5}{5}\right]_{0}^{2} = \pi \left(\frac{32}{5} - 0\right) = \frac{32\pi}{5}

Volume:

\frac{32\pi}{5}

cubic units

Example 2: Exponential Functions

Find the volume formed by rotating the region bounded by

y = e^x

y = 0

x = 0

, and

x = 3

around the

x

-axis.

Step 1: set up. Radius

R = e^x

V = \int_{0}^{3} \pi (e^x)^2\,dx

Step 2: simplify.

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

V = \pi \int_{0}^{3} e^{2x}\,dx

Step 3: substitution. Let

u = 2x

, so

du = 2\,dx

and

dx = \frac{du}{2}

. When

x = 0

u = 0

. When

x = 3

u = 6

V = \pi \int_{0}^{6} e^u \cdot \frac{du}{2} = \frac{\pi}{2} \int_{0}^{6} e^u\,du

Step 4: integrate

= \frac{\pi}{2} [e^u]_{0}^{6} = \frac{\pi}{2} (e^6 - e^0) = \frac{\pi}{2} (e^6 - 1)

Volume:

\frac{\pi(e^6 - 1)}{2}

cubic units

Revolution Around the y-axis

When you rotate around the

y

-axis instead:

Integrate with respect to $y$ (use $dy$ )

Convert equations to $x = f(y)$ form

Use $y$ -boundaries as your limits

The radius is now horizontal, measured in the $x$ -direction

\boxed{V = \int_{c}^{d} \pi R^2\,dy}

where

R

is expressed as a function of

y

Example 3: Revolution Around the y-axis

Set up (but don’t evaluate) an integral for the volume formed by rotating the region bounded by

y = 4 - 2x

y = 0

, and the

y

-axis around the

y

-axis.

Step 1: convert to $x = f(y)$

y = 4 - 2x \quad \Rightarrow \quad 2x = 4 - y \quad \Rightarrow \quad x = 2 - \frac{y}{2}

Step 2: find the $y$ -boundaries. When

x = 0

y = 4

(the top). The bottom is at

y = 0

. So

y

runs from 0 to 4.

Step 3: radius. The distance from the

y

-axis (where

x = 0

) out to the line:

R = 2 - \frac{y}{2}

Step 4: set up

V = \int_{0}^{4} \pi \left(2 - \frac{y}{2}\right)^2\,dy

To evaluate, you’d expand the squared term and integrate, but the setup is the skill being tested.

Another y-axis Example

Find the volume formed by rotating the region bounded by

x = \sqrt{y}

x = 0

, and

y = 4

around the

y

-axis.

Step 1: radius and bounds. Already in

x = f(y)

form, so

R = \sqrt{y}

, with

y

from 0 to 4.

Step 2: set up and evaluate

V = \int_{0}^{4} \pi (\sqrt{y})^2\,dy = \pi \int_{0}^{4} y\,dy = \pi \left[\frac{y^2}{2}\right]_{0}^{4} = \pi \left(\frac{16}{2}\right) = 8\pi

Volume:

8\pi

cubic units

Problem-Solving Strategy

Step 1: sketch the region and identify the axis of revolution.

Step 2: determine the radius function. Measure from the axis of revolution to the curve.

Step 3: identify the limits. Find where the region starts and ends along the axis of revolution.

Step 4: set up the integral. Use

\int \pi R^2\,dx

for the

x

-axis,

\int \pi R^2\,dy

for the

y

-axis.

Step 5: evaluate. Factor out

\pi

, expand squared terms, integrate, and evaluate at the bounds.

Common Patterns and Tips

When to use $dx$ vs. $dy$

Axis of Revolution	Integration Variable	Function Form Needed
$x$ -axis	$dx$	$y = f(x)$
$y$ -axis	$dy$	$x = f(y)$

Simplifying $(f(x))^2$

Always simplify the squared function: -

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

(x^2)^2 = x^4

(\sqrt{x})^2 = x

(ax + b)^2 = a^2x^2 + 2abx + b^2

Working with $\pi$

On non-calculator sections, leave

\pi

in the answer (

\frac{32\pi}{5}

, not

20.106\ldots

), and factor it out of the integral early.

On calculator sections, you can approximate:

\frac{32\pi}{5} \approx 20.106

Visualization Tips

Imagine the sweeping motion. The region traces out a path as it spins

Think in slices. Every thin vertical (or horizontal) strip becomes a disk

The radius varies. As you move along the axis, each disk’s radius changes with the function

It’s solid throughout. Not hollow. Every point in the region traces a circle

Common Mistakes to Avoid

Forgetting to square the radius. It’s $\pi R^2$ , not $\pi R$

Using the wrong variable. $x$ -axis rotation uses $dx$ ; $y$ -axis rotation uses $dy$

Not converting for $y$ -axis problems. Always solve for $x = f(y)$

Squaring incorrectly. $(x^2)^2 = x^4$ , not $x^2$

Forgetting $\pi$ , which happens easily once you factor it out

Wrong bounds. For the $x$ -axis use $x$ -values; for the $y$ -axis use $y$ -values

Summary Table

Scenario	Volume Formula	Notes
Around the $x$ -axis	$V = \int_{a}^{b} \pi [f(x)]^2\,dx$	Radius = $f(x)$
Around the $y$ -axis	$V = \int_{c}^{d} \pi [f(y)]^2\,dy$	Need $x = f(y)$ form
General form	$V = \int \pi R^2\,d(\text{var})$	$R$ = radius function

Key Takeaways

Solid of revolution: rotate a 2D region around an axis to make a 3D solid

Disk method: every cross-section perpendicular to the axis is a circle

Volume formula: $V = \int_{a}^{b} \pi R^2\,dx$ (or $dy$ for the $y$ -axis)

Radius $R$ : the distance from the axis of revolution to the outer edge

$x$ -axis rotation: use $dx$ , radius is $R = f(x)$ , bounds are $x$ -values

$y$ -axis rotation: use $dy$ , convert to $x = f(y)$ , bounds are $y$ -values

Always square the radius. A disk’s area is $\pi R^2$ , not $\pi R$

Sketching helps. Drawing the region and its mirror image clarifies the setup

$\pi$ is a constant. Factor it out for easier computation

Leave answers in terms of $\pi$ unless you’re asked to approximate

This is the foundation for disks around other axes and the washer method, coming next

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