Topic 8.10 Notes – Volume with Disc Method: Revolving Around Other Axes

What the Disc Method Is Really Doing

When you rotate a region around a line, you create a 3D solid. If your slices are perpendicular to the axis of rotation, each slice forms a solid circle (a disc).

Each tiny piece of volume looks like:

$$\text{Volume} = \pi r^2 (\text{thickness})$$

A definite integral adds up all those tiny circular volumes over an interval.

So every disc-method problem boils down to:

$$V = \int \pi (\text{radius})^2$$

The only question is: what is the radius?

Horizontal vs. Vertical Axes

Everything depends on the direction of the axis of rotation.

Rotating Around a Horizontal Line $y = k$

• Use vertical slices
• Integrate with respect to x
• Radius = vertical distance from the curve to the line

$$V = \int_a^b \pi \big(f(x) - k\big)^2 \, dx$$

You’re measuring up-and-down distance.

Rotating Around a Vertical Line $x = h$

• Use horizontal slices
• Integrate with respect to y
• Radius = horizontal distance from the curve to the line

$$V = \int_c^d \pi \big(f(y) - h\big)^2 \, dy$$

You’re measuring left-and-right distance.

Visualizing What’s Happening

Here’s the geometric idea when you rotate a region around a horizontal line.

Vertical slice forming a disk when rotated about $y = 1$

The vertical slice has height $f(x) - 1$. When that slice rotates around $y = 1$, it forms a circular disk.

Notice the radius is the distance between the curve and the line.

That distance becomes squared inside πr².

The Most Important Idea: Radius Is a Distance

Distance to a horizontal line:

$$r = f(x) - k$$

Distance to a vertical line:

$$r = f(y) - h$$

Be careful with negatives.

If rotating around $y = -2$, then:

$$r = f(x) - (-2) = f(x) + 2$$

Always ask yourself:

How far is the function from the axis?

That mental question prevents almost every mistake.

Quick Example (Horizontal Axis)

Suppose the region under $f(x) = x^2$ from 0 to 2 is rotated around $y = 3$.

Axis is horizontal → integrate in x.

Radius is distance from curve to line:

$$r = 3 - x^2$$

Volume:

$$V = \int_0^2 \pi (3 - x^2)^2 \, dx$$

Notice the entire radius is squared. On quizzes, students often square only part of it. The parentheses matter.

Choosing the Limits

Your bounds must match your variable.

• If integrating in x, limits are x-values.
• If integrating in y, limits are y-values.

You may need to:

• Solve intersections
• Rewrite equations in terms of y

On FRQs, forgetting to change bounds when switching variables costs easy setup points.

Common Errors I See Every Year

• Forgetting to subtract the axis (using just $f(x)^2$)
• Squaring only the function and not the whole expression
• Integrating in the wrong variable
• Dropping parentheses with negatives
• Skipping the sketch and misidentifying the radius

Most disc-method errors happen before the integration even starts.

When Disc Method Is the Right Tool

Use it when:

• The region touches the axis of rotation
• Cross-sections are solid circles (no hole)

If there’s a gap between the region and the axis, that becomes a washer problem (next topic).