Topic 8.11 Notes – Volume with Washer Method: Revolving Around the x- or y-Axis

What the Washer Method Is

When you rotate a region between two curves around an axis, the cross-sections may have a hole in the middle. Each slice looks like a washer (a disk with a smaller disk removed).

The area of one washer is:

$$\text{Area} = \pi(R^2 - r^2)$$

• $R$ = outer radius (farther from the axis)
• $r$ = inner radius (closer to the axis)

To get volume, we integrate those areas:

$$V = \int \pi(R^2 - r^2)\,dx \quad \text{or} \quad \int \pi(R^2 - r^2)\,dy$$

Two things matter immediately:

• Radius means distance to the axis, not just “top function.”
• You square first, then subtract.
  $\pi(R^2 - r^2) \neq \pi(R - r)^2$

Visualizing a Washer

Here’s what one cross-section looks like when the region between two curves is revolved around the x-axis.

A single washer formed by revolving the region between two curves around the x-axis

That shaded ring is what your integral is stacking infinitely many times to build the full solid.

Revolving Around the x-Axis

If you rotate around the x-axis (which is $y = 0$), you usually use vertical slices and integrate with respect to $x$.

• Radii are vertical distances.
• Outer radius = top function.
• Inner radius = bottom function.

So the setup looks like:

$$V = \int_a^b \pi\big[(\text{top})^2 - (\text{bottom})^2\big]\,dx$$

Quick Example Setup (no full evaluation)

Suppose the region between
$y = 4 - x^2$ and $y = x$
is rotated around the x-axis.

1. Find intersections:
   $4 - x^2 = x \Rightarrow x^2 + x - 4 = 0$
   (calculator is fine here on AP if needed)

2. Outer radius = $4 - x^2$
   Inner radius = $x$

3. Volume integral:

   $$V = \int_{x_1}^{x_2} \pi\big[(4 - x^2)^2 - x^2\big]\,dx$$

That’s a complete correct setup. On FRQs, that alone earns major credit.

Revolving Around the y-Axis

Now the axis is vertical. That changes everything.

You use horizontal slices and integrate with respect to $y$.

• Radii are horizontal distances.
• You must rewrite equations as $x =$ functions of $y$.

Focus on the right-hand diagram below. The slice has thickness $\Delta y$, and the radius $R$ is measured horizontally to the vertical axis of rotation.

Rotation about a vertical axis using horizontal slices

The structure becomes:

$$V = \int_c^d \pi\big[(\text{right function})^2 - (\text{left function})^2\big]\,dy$$

Students lose points here because they forget to:

• Solve for $x$ in terms of $y$
• Change the bounds to $y$-values

If your bounds are still x-values but you wrote $dy$, something’s wrong.

Shifted Axes

If the axis is not $x=0$ or $y=0$, your radius is a distance formula.

• Around $y = k$: radius = (function − k)
• Around $x = h$: radius = (function − h)

Example structure:

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

You must subtract the axis before squaring.

Classic mistake: squaring first, then subtracting the constant.

Full Setup Checklist

Before integrating, you should clearly know:

1. Axis of rotation
2. Bounds (usually intersection points)
3. Outer radius
4. Inner radius
5. Correct variable ($dx$ or $dy$)

If your final answer is negative, you flipped outer and inner.

When the Washer Method Is Used

Use washers when:

• There are two bounding curves
• Rotation creates a hole
• Cross-sections are perpendicular to the axis

If there’s no hole, that’s the disk method from earlier.

On the AP exam, washer problems often appear in multi-part FRQs. Sometimes the integral isn’t meant to be evaluated by hand. The scoring focuses heavily on whether you:

• Chose correct radii
• Used correct bounds
• Set up the integral properly

Structure is everything.