Washer Method
Revolving Around Other Axes
Master the technique for finding volumes of solids of revolution when the axis of rotation is not the x-axis or y-axis.
AP Calculus AB
What Is the Washer Method?
When you rotate a region between two curves around an axis, each cross-section perpendicular to that axis is a washer: a disk with a hole in the middle. The volume of each thin washer is the area of the outer disk minus the inner disk, times the thickness.
Why Other Axes Matter
Standard Axis
Revolving around \( y=0 \) or \( x=0 \). Radius is simply the function value \( f(x) \) or \( g(y) \).
Other Axis
Revolving around \( y=k \) or \( x=h \). Radius is the distance from the curve to that line, e.g., \( |f(x)-k| \) or \( |g(y)-h| \).
Visualizing the Washer Cross-Section
The Core Formula
\( R(x) \)
Outer radius: distance from the farther curve to the axis of rotation.
\( r(x) \)
Inner radius: distance from the closer curve to the axis of rotation.
\([a, b]\)
Integration bounds: the interval over which the two curves form the region.
Radius Formulas for Each Axis Type
| Axis of Rotation | Outer Radius \( R \) | Inner Radius \( r \) |
|---|---|---|
| \( y = k \) (below region) | \( y_{upper} - k \) | \( y_{lower} - k \) |
| \( y = k \) (above region) | \( k - y_{lower} \) | \( k - y_{upper} \) |
| \( x = h \) (left of region) | \( x_{right} - h \) | \( x_{left} - h \) |
| \( x = h \) (right of region) | \( h - x_{left} \) | \( h - x_{right} \) |
Disk vs. Washer
Disk Method
One curve bounds the region and the axis of rotation is the other boundary. The inner radius is zero: $$ V = \pi \int_a^b R(x)^2\, dx $$
Washer Method
Two curves bound the region, neither of which is the axis. Both radii are nonzero: $$ V = \pi \int_a^b \big[R(x)^2 - r(x)^2\big]\, dx $$
How to Apply the Washer Method
1
Sketch the region and identify the axis of rotation. Mark which curve is farther from and which is closer to the axis.
2
Write the radii. Outer radius \( R \) = distance from farther curve to axis. Inner radius \( r \) = distance from closer curve to axis.
3
Find the bounds by solving for the intersection points of the two curves (or the given interval).
4
Set up the integral: \( V = \pi \displaystyle\int_a^b [R^2 - r^2]\, dx \) (or \( dy \) for vertical axes).
5
Evaluate the integral. Expand, simplify, then integrate term by term.
Example 1: Revolve Around \( y = -1 \)
Find the volume when the region between \( f(x) = \sqrt{x} \) and \( g(x) = x \) on \([0,1]\) is revolved about the line \( y = -1 \).
Example 1: Solution
1
Outer radius: \( R = \sqrt{x} - (-1) = \sqrt{x} + 1 \)
2
Inner radius: \( r = x - (-1) = x + 1 \)
3
Set up: \( V = \pi \displaystyle\int_0^1 \big[(\sqrt{x}+1)^2 - (x+1)^2\big]\,dx \)
4
Expand: \( (\sqrt{x}+1)^2 = x + 2\sqrt{x} + 1 \) and \( (x+1)^2 = x^2 + 2x + 1 \)
5
Simplify integrand: \( (x + 2\sqrt{x} + 1) - (x^2 + 2x + 1) = -x^2 - x + 2x^{1/2} \)
Example 2: Revolve Around \( y = 4 \)
Find the volume when the region between \( y = x^2 \) and \( y = 2x \) is revolved about the line \( y = 4 \).
Example 2: Solution
1
Bounds: \( x^2 = 2x \implies x=0,\; x=2 \)
2
Radii: \( R = 4 - x^2 \), \( r = 4 - 2x \)
3
Integral: \( V = \pi \displaystyle\int_0^2 \big[(4-x^2)^2 - (4-2x)^2\big]\,dx \)
4
Expand: \( (4-x^2)^2 = 16 - 8x^2 + x^4 \) and \( (4-2x)^2 = 16 - 16x + 4x^2 \)
5
Integrand: \( (16 - 8x^2 + x^4) - (16 - 16x + 4x^2) = x^4 - 12x^2 + 16x \)
Example 3: Revolve Around \( x = 5 \)
Find the volume when the region bounded by \( x = y^2 \) and \( x = 2y \) is revolved about the line \( x = 5 \).
Example 3: Solution
1
Integrate with respect to \( y \) since the axis is vertical. Bounds: \( y^2=2y \implies y=0, y=2 \).
2
Axis is right of the region. \( R = 5 - y^2 \) (farther, since \( y^2 < 2y \) on \((0,2)\)). \( r = 5 - 2y \).
3
Integral: \( V = \pi \displaystyle\int_0^2 \big[(5 - y^2)^2 - (5 - 2y)^2\big]\,dy \)
4
Integrand: \( (25 - 10y^2 + y^4) - (25 - 20y + 4y^2) = y^4 - 14y^2 + 20y \)
Common Mistakes to Avoid
Forgetting to Shift
Using \( f(x) \) as the radius instead of \( f(x) - k \) when the axis is \( y = k \). Always subtract the axis value.
Swapping R and r
The outer radius \( R \) must be the larger distance. If the axis is above the region, the lower curve gives \( R \).
Squaring Before Subtracting
Writing \( \pi\!\int (R - r)^2\, dx \) instead of \( \pi\!\int (R^2 - r^2)\, dx \). Square each radius separately, then subtract.
Summary Comparison
| Feature | Standard Axis | Other Axis |
|---|---|---|
| Axis | \( y=0 \) or \( x=0 \) | \( y=k \) or \( x=h \) |
| Outer Radius | \( y_{upper} \) | e.g., \( y_{upper}-k \) or \( k-y_{lower} \) |
| Inner Radius | \( y_{lower} \) | e.g., \( y_{lower}-k \) or \( k-y_{upper} \) |
| Key Difference | Direct function value | Distance to shifted axis |
| Common Error | Rare | Forgetting to shift by \( k \) or \( h \) |
Question 1 of 5
The region between \( y = x \) and \( y = x^2 \) on \([0,1]\) is revolved about \( y = -2 \). What is the outer radius?
The axis \( y=-2 \) is below the region. The farther curve from the axis is \( y=x \) (the upper curve). So \( R = x - (-2) = x + 2 \).
Question 2 of 5
When revolving a region about a horizontal line above the region, the outer radius is determined by:
When the axis is above the region, the lower curve is farther away. So \( R = k - y_{lower} \) where \( y_{lower} \) is the lower curve and \( k \) is the axis. The upper curve, being closer, gives the inner radius \( r = k - y_{upper} \).
Question 3 of 5
Which expression is incorrect for the washer method?
You must square each radius individually, then subtract: \( R^2 - r^2 \). The expression \( (R - r)^2 \) squares the difference, which is mathematically different and produces wrong volumes.
Question 4 of 5
The region between \( y = \sqrt{x} \) and \( y = x \) on \([0,1]\) is revolved about \( y = -1 \). What is the volume?
With \( R = \sqrt{x}+1 \) and \( r = x+1 \), the integral \( \pi\!\int_0^1[(x+2\sqrt{x}+1)-(x^2+2x+1)]\,dx = \pi\!\int_0^1(-x^2 - x + 2x^{1/2})\,dx = \pi(-\frac{1}{3} - \frac{1}{2} + \frac{4}{3}) = \frac{\pi}{2} \).
Question 5 of 5
When revolving the region bounded by \( x = y^2 \) and \( x = 2y \) about the line \( x = 5 \), you should integrate with respect to:
When the axis of rotation is vertical (\( x = h \)), cross-sectional washers are horizontal. Their thickness is \( dy \), so you integrate with respect to \( y \). The radii are horizontal distances: \( R = 5 - y^2 \) and \( r = 5 - 2y \).