💡 Key Idea: Use washers (disks with holes) instead of solid disks! Subtract the inner disk from the outer disk.

The Washer Method Formula

Rotating around the x-axis

When rotating the region between $f(x)$ (outer) and $g(x)$ (inner) from $x = a$ to $x = b$:

$V = \pi \int_a^b \left[(R(x))^2 - (r(x))^2\right]\,dx$

where:

• $R(x)$ = outer radius (distance from axis to outer curve)
• $r(x)$ = inner radius (distance from axis to inner curve)

Why This Works

Volume of a Washer

A washer is a disk with a circular hole:

Outer disk area: $\pi R^2$ Inner disk area: $\pi r^2$ Washer area: $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

Multiply by thickness $\Delta x$: $\text{Volume} = \pi(R^2 - r^2)\Delta x$

Sum and take the limit → integral!

Example 1: Region Between Two Curves

Find the volume when the region between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$ is rotated around the x-axis.

Step 1: Identify outer and inner radii

On $[0, 1]$, the line $y = x$ is above $y = x^2$.

• Outer radius: $R(x) = x$ (distance to line)
• Inner radius: $r(x) = x^2$ (distance to parabola)

Step 2: Set up the integral

$V = \pi \int_0^1 [(x)^2 - (x^2)^2]\,dx$

$= \pi \int_0^1 (x^2 - x^4)\,dx$

Step 3: Integrate

$= \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1$

$= \pi \left(\frac{1}{3} - \frac{1}{5}\right) - 0$

$= \pi \left(\frac{5}{15} - \frac{3}{15}\right) = \frac{2\pi}{15}$

Answer: $\frac{2\pi}{15}$ cubic units

Example 2: Region Between Curve and Axis

Find the volume when the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ is rotated around the line $y = -1$.

Step 1: Identify the radii

The axis of rotation is $y = -1$ (horizontal line below x-axis).

• Outer radius: $R(x) = \sqrt{x} - (-1) = \sqrt{x} + 1$
• Inner radius: $r(x) = 0 - (-1) = 1$

(Distance from axis to each curve!)

Step 2: Set up the integral

$V = \pi \int_0^4 [(\sqrt{x} + 1)^2 - 1^2]\,dx$

Step 3: Expand $(\sqrt{x} + 1)^2$

$(\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1$

$V = \pi \int_0^4 [(x + 2\sqrt{x} + 1) - 1]\,dx$

$= \pi \int_0^4 (x + 2\sqrt{x})\,dx$

$= \pi \int_0^4 (x + 2x^{1/2})\,dx$

Step 4: Integrate

$= \pi \left[\frac{x^2}{2} + 2 \cdot \frac{x^{3/2}}{3/2}\right]_0^4$

$= \pi \left[\frac{x^2}{2} + \frac{4x^{3/2}}{3}\right]_0^4$

$= \pi \left(\frac{16}{2} + \frac{4(8)}{3}\right)$

$= \pi \left(8 + \frac{32}{3}\right) = \pi \left(\frac{24 + 32}{3}\right) = \frac{56\pi}{3}$

Answer: $\frac{56\pi}{3}$ cubic units

Rotating Around Vertical Lines

Around the y-axis (or $x = 0$)

$V = \pi \int_c^d \left[(R(y))^2 - (r(y))^2\right]\,dy$

• Outer radius: $R(y)$ (rightmost curve)
• Inner radius: $r(y)$ (leftmost curve)

Around the line $x = k$

Adjust radii by distance from the line $x = k$:

• If curve is at $x = f(y)$ and $f(y) > k$: radius = $f(y) - k$
• If curve is at $x = f(y)$ and $f(y) < k$: radius = $k - f(y)$

Always: radius = |distance from axis to curve|

Example 3: Rotating Around y-axis

Find the volume when the region between $x = y^2$ and $x = 4$ from $y = -2$ to $y = 2$ is rotated around the y-axis.

Step 1: Identify radii

The line $x = 4$ is farther from the y-axis than $x = y^2$.

• Outer radius: $R(y) = 4$
• Inner radius: $r(y) = y^2$

Step 2: Set up the integral

$V = \pi \int_{-2}^2 [4^2 - (y^2)^2]\,dy$

$= \pi \int_{-2}^2 (16 - y^4)\,dy$

Step 3: Integrate

$= \pi \left[16y - \frac{y^5}{5}\right]_{-2}^2$

At $y = 2$: $16(2) - \frac{32}{5} = 32 - \frac{32}{5} = \frac{128}{5}$

At $y = -2$: $16(-2) - \frac{-32}{5} = -32 + \frac{32}{5} = -\frac{128}{5}$

Step 4: Subtract

$V = \pi \left(\frac{128}{5} - \left(-\frac{128}{5}\right)\right)$

$= \pi \cdot \frac{256}{5} = \frac{256\pi}{5}$

Answer: $\frac{256\pi}{5}$ cubic units

Disk vs Washer: When to Use Which

Use Disk Method when:

• Region is bounded by one curve and an axis
• Rotation creates a solid (no hole)
• Formula: $V = \pi \int [R(x)]^2\,dx$

Use Washer Method when:

• Region is between two curves
• Rotation creates a hollow solid
• Formula: $V = \pi \int [(R(x))^2 - (r(x))^2]\,dx$

Finding Radii

Key Questions:

1. What is the axis of rotation?
2. Which curve is farther from the axis? (outer radius)
3. Which curve is closer to the axis? (inner radius)
4. What is the distance from axis to each curve? (that's the radius)

Example 4: Non-Standard Axis

Find the volume when the region between $y = x^2$ and $y = 2x$ from $x = 0$ to $x = 2$ is rotated around the line $y = 8$.

Step 1: Find which is closer to $y = 8$

At $x = 1$:

• Line: $y = 2(1) = 2$, distance from $y = 8$ is $8 - 2 = 6$
• Parabola: $y = 1^2 = 1$, distance from $y = 8$ is $8 - 1 = 7$

Line is closer (inner), parabola is farther (outer)? No, wait...

Actually, the line is ABOVE the parabola (farther from axis below).

Let me reconsider: $y = 8$ is ABOVE both curves.

Better approach:

• Both curves below $y = 8$
• Line $y = 2x$ is closer to $y = 8$ (smaller distance)
• Parabola $y = x^2$ is farther from $y = 8$ (larger distance)

Radii:

• Outer radius: $R(x) = 8 - x^2$ (to parabola)
• Inner radius: $r(x) = 8 - 2x$ (to line)

Step 2: Set up and integrate

$V = \pi \int_0^2 [(8-x^2)^2 - (8-2x)^2]\,dx$

Expand: $(8-x^2)^2 = 64 - 16x^2 + x^4$

$(8-2x)^2 = 64 - 32x + 4x^2$

$V = \pi \int_0^2 [(64 - 16x^2 + x^4) - (64 - 32x + 4x^2)]\,dx$

$= \pi \int_0^2 (32x - 20x^2 + x^4)\,dx$

Step 3: Integrate

$= \pi \left[16x^2 - \frac{20x^3}{3} + \frac{x^5}{5}\right]_0^2$

$= \pi \left(64 - \frac{160}{3} + \frac{32}{5}\right)$

$= \pi \left(\frac{960 - 800 + 96}{15}\right) = \frac{256\pi}{15}$

Answer: $\frac{256\pi}{15}$ cubic units

⚠️ Common Mistakes

Mistake 1: Confusing Inner and Outer

Check: Which curve is farther from the axis of rotation?

That's your outer radius $R(x)$!

Mistake 2: Not Squaring Correctly

WRONG: $[(R(x))^2 - (r(x))^2] = [R(x) - r(x)]^2$

RIGHT: Square each radius separately, then subtract!

$R^2 - r^2 \neq (R - r)^2$

Mistake 3: Forgetting the Axis of Rotation

If rotating around $y = k$ (not the x-axis):

• Radius ≠ just $f(x)$
• Radius = distance from curve to line $y = k$

Mistake 4: Sign Errors with Distance

Distance is always positive!

If axis is at $y = k$ and curve is at $y = f(x)$:

• Radius = $|f(x) - k|$

Usually: if curve below axis, $r = k - f(x)$ If curve above axis, $r = f(x) - k$

Disk Method as Special Case

Notice: If inner radius $r(x) = 0$ (one curve is the axis):

$V = \pi \int_a^b [R^2 - 0^2]\,dx = \pi \int_a^b R^2\,dx$

This is just the disk method!

Washer method is the general case.

Summary: Washer Method

Around x-axis or $y = k$

$V = \pi \int_a^b \left[(R(x))^2 - (r(x))^2\right]\,dx$

Around y-axis or $x = h$

$V = \pi \int_c^d \left[(R(y))^2 - (r(y))^2\right]\,dy$

Remember:

• $R$ = outer (farther from axis)
• $r$ = inner (closer to axis)
• Both are distances (positive)
• Square each separately!

📝 Practice Strategy

1. Sketch the region and axis of rotation
2. Draw a sample washer (perpendicular to axis)
3. Identify outer and inner radii
4. Calculate distances from axis to each curve
5. Set up: $V = \pi \int [(R)^2 - (r)^2]\,dx$ or $dy$
6. Expand if needed (watch for $(a+b)^2$!)
7. Integrate and evaluate
8. Check: Answer should be positive!

Step 3: Apply washer method.

$V = \pi \int_{-2}^2 [R(x)^2 - r(x)^2] \, dx$

$= \pi \int_{-2}^2 [16 - x^4] \, dx$

By symmetry:

$= 2\pi \int_0^2 (16 - x^4) \, dx$

$= 2\pi \left[16x - \frac{x^5}{5}\right]_0^2$

$= 2\pi \left(32 - \frac{32}{5}\right)$

$= 2\pi \left(\frac{160 - 32}{5}\right)$

$= 2\pi \cdot \frac{128}{5}$

$= \frac{256\pi}{5}$ cubic units

Distance from $y = -2$ to $y = 0$: $0 - (-2) = 2$

Step 2: Identify radii

• Outer radius: $R(x) = x^2 + 3$ (to upper curve)
• Inner radius: $r(x) = 2$ (to x-axis)

Step 3: Set up the integral

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

Step 4: Expand $(x^2+3)^2$

$(x^2+3)^2 = x^4 + 6x^2 + 9$

$V = \pi \int_0^2 (x^4 + 6x^2 + 9 - 4)\,dx$

$= \pi \int_0^2 (x^4 + 6x^2 + 5)\,dx$

Step 5: Integrate

$= \pi \left[\frac{x^5}{5} + 2x^3 + 5x\right]_0^2$

$= \pi \left(\frac{32}{5} + 16 + 10\right)$

$= \pi \left(\frac{32}{5} + 26\right)$

$= \pi \left(\frac{32 + 130}{5}\right) = \frac{162\pi}{5}$

Answer: $\frac{162\pi}{5}$ cubic units

Line is farther from y-axis (outer).

Step 2: Identify radii

• Outer radius: $R(y) = 2y$
• Inner radius: $r(y) = y^2$

Step 3: Set up the integral

$V = \pi \int_0^2 [(2y)^2 - (y^2)^2]\,dy$

$= \pi \int_0^2 (4y^2 - y^4)\,dy$

Step 4: Integrate

$= \pi \left[\frac{4y^3}{3} - \frac{y^5}{5}\right]_0^2$

$= \pi \left(\frac{32}{3} - \frac{32}{5}\right)$

$= \pi \left(\frac{160 - 96}{15}\right) = \frac{64\pi}{15}$

Answer: $\frac{64\pi}{15}$ cubic units