8.12 Washer Method Revolve Around Other Axes

The washer method is a cornerstone of integral calculus, providing a powerful technique for calculating the volume of solids of revolution that possess a hollow center. But 12 in standard calculus curricula—require revolving regions around horizontal lines (y = c) or vertical lines (x = c) that are not the coordinate axes. While introductory problems typically revolve regions around the x-axis or y-axis, advanced applications—often cataloged as section 8.Mastering this shift is essential for solving real-world engineering problems and excelling in advanced mathematics courses.

This article provides a complete walkthrough to the washer method when the axis of revolution is translated away from the origin. We will explore the geometric intuition, the algebraic adjustments required for radii, the step-by-step procedural framework, and work through detailed examples to solidify your understanding.

This is where a lot of people lose the thread Worth keeping that in mind..

Understanding the Geometry: Translating the Axis

When a region is revolved around the x-axis (y = 0), the radius of a typical washer is simply the y-value of the function. When the axis shifts to a horizontal line y = k, the geometry changes fundamentally. The radius is no longer the function value itself; it becomes the vertical distance between the function curve and the line y = k Still holds up..

Imagine a region bounded by an upper curve $y = f(x)$ and a lower curve $y = g(x)$. If we revolve this around a horizontal axis $y = k$:

• Outer Radius ($R$): The distance from the axis $y = k$ to the curve farther from the axis.
• Inner Radius ($r$): The distance from the axis $y = k$ to the curve closer to the axis.

Because distance is always positive, we calculate these radii using absolute values or, more practically, by determining which function is "above" the other relative to the axis. The standard formula for volume using the washer method integrated with respect to $x$ (vertical slices) is:

$V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] dx$

Where:

• $R(x) = |f(x) - k|$ (Distance from axis to outer boundary)
• $r(x) = |g(x) - k|$ (Distance from axis to inner boundary)

Crucial Insight: Do not simply plug $f(x)$ and $g(x)$ into the formula. You must subtract the axis constant $k$ from the function values. If the axis is above the region, the radii are $k - f(x)$ and $k - g(x)$. If the axis is below the region, the radii are $f(x) - k$ and $g(x) - k$. Drawing a diagram is non-negotiable for avoiding sign errors.

Vertical Axes of Revolution: Integrating with Respect to y

The logic mirrors the horizontal case when revolving around a vertical line $x = h$. Here, we typically use horizontal slices and integrate with respect to $y$. The region is defined by a right curve $x = f(y)$ and a left curve $x = g(y)$ Surprisingly effective..

This is where a lot of people lose the thread Simple, but easy to overlook..

The volume formula becomes:

$V = \pi \int_{c}^{d} \left[ R(y)^2 - r(y)^2 \right] dy$

Where the radii represent horizontal distances:

• $R(y) = |f(y) - h|$ (Distance from axis $x = h$ to rightmost boundary)
• $r(y) = |g(y) - h|$ (Distance from axis $x = h$ to leftmost boundary)

Again, the relative position of the axis $x = h$ to the region dictates the subtraction order. If the axis is to the right of the region, radii are $h - f(y)$ and $h - g(y)$. If the axis is to the left, radii are $f(y) - h$ and $g(y) - h$ That's the whole idea..

Step-by-Step Problem-Solving Strategy

To consistently solve these problems correctly, follow this structured workflow:

1. Sketch the Region: Plot the bounding curves and shade the enclosed region. Identify intersection points to determine limits of integration ($a, b$ or $c, d$).
2. Draw the Axis of Revolution: Draw the line $y = k$ or $x = h$ clearly on your sketch. Label it.
3. Identify the Slice Direction:
   • Horizontal axis ($y = k$) $\rightarrow$ Vertical slices ($dx$).
   • Vertical axis ($x = h$) $\rightarrow$ Horizontal slices ($dy$).
4. Determine Outer and Inner Boundaries: For a representative slice, identify which curve forms the outer edge of the washer and which forms the inner hole relative to the axis.
5. Set Up Radii Expressions: Write $R$ and $r$ as (Boundary Function) - (Axis Constant) or (Axis Constant) - (Boundary Function), ensuring the result is positive. Pro Tip: Since the radius is squared in the formula, $(f(x) - k)^2 = (k - f(x))^2$. You can subtract in either order if you are consistent, but conceptualizing it as "Top minus Axis" or "Axis minus Bottom" prevents logic errors when setting up the integrand.
6. Construct the Integral: Substitute $R^2$ and $r^2$ into $V = \pi \int (R^2 - r^2) , d\text{variable}$.
7. Evaluate: Compute the definite integral using the Fundamental Theorem of Calculus.

Detailed Example 1: Horizontal Axis ($y = k$)

Problem: Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ about the line $y = 2$ Took long enough..

Solution:

1. Sketch: The region is under the square root curve from $x=0$ to $x=4$, above the x-axis. The axis $y=2$ is a horizontal line above the region.
2. Slice: Vertical slices ($dx$). Limits: $x = 0$ to $x = 4$.
3. Boundaries:
   • Top curve (farther from axis $y=2$): $y = 0$ (the x-axis).
   • Bottom curve (closer to axis $y=2$): $y = \sqrt{x}$.
   • Wait, check distances: Distance from $y=2$ to $y=0$ is $2$. Distance from $y=2$ to $y=\sqrt{x}$ is $2 - \sqrt{x}$. Since $\sqrt{x} \ge 0$, $2 > 2-\sqrt{x}$. The "outer" radius corresponds to the curve farther from the axis. Here, $y=0$ is farther away.
   • Outer Radius $R(x)$: Distance from axis $y=2$ to $y=0 \rightarrow R = 2 - 0 = 2$.
   • Inner Radius $r(x)$: Distance from axis $y=2$ to $y=\sqrt{x} \rightarrow r = 2 - \sqrt{x}$.
4. Integral Setup: $V = \pi \int_{0}^{4} \left[ (2)^2 - (2 - \sqrt{x})^2 \right] dx$
5. Simplify and Integrate: $(2 - \sqrt{x})^2 = 4 - 4\sqrt{x} + x$ $R^2 - r^2 = 4 - (4 - 4\sqrt{x

$+ x) = 4\sqrt{x} - x$ $V = \pi \int_{0}^{4} (4x^{1/2} - x) , dx$ $V = \pi \left[ \frac{4x^{3/2}}{3/2} - \frac{x^2}{2} \right]_0^4 = \pi \left[ \frac{8}{3}x^{3/2} - \frac{1}{2}x^2 \right]_0^4$ $V = \pi \left[ \left( \frac{8}{3}(8) - \frac{16}{2} \right) - 0 \right] = \pi \left( \frac{64}{3} - 8 \right) = \pi \left( \frac{64 - 24}{3} \right) = \frac{40\pi}{3} \text{ cubic units.}$

Detailed Example 2: Vertical Axis ($x = h$)

Problem: Find the volume of the solid generated by revolving the region bounded by $y = x^2$ and $y = 2x$ about the line $x = -1$.

Solution:

1. Sketch: The region is bounded by a parabola and a line. Setting $x^2 = 2x$ gives $x(x-2) = 0$, so the intersection points are $(0,0)$ and $(2,4)$. The axis $x = -1$ is a vertical line to the left of the region.
2. Slice: Since the axis is vertical, we use horizontal slices ($dy$). We must rewrite the functions in terms of $y$:
   • $y = x^2 \rightarrow x = \sqrt{y}$ (Right boundary)
   • $y = 2x \rightarrow x = y/2$ (Left boundary)
   • Limits: $y = 0$ to $y = 4$.
3. Boundaries:
   • The outer radius $R(y)$ is the distance from the axis $x = -1$ to the curve farther away.
   • The inner radius $r(y)$ is the distance from the axis $x = -1$ to the curve closer.
   • Outer Radius $R(y)$: Distance from $x = -1$ to $x = \sqrt{y} \rightarrow R = \sqrt{y} - (-1) = \sqrt{y} + 1$.
   • Inner Radius $r(y)$: Distance from $x = -1$ to $x = y/2 \rightarrow r = y/2 - (-1) = y/2 + 1$.
4. Integral Setup: $V = \pi \int_{0}^{4} \left[ (\sqrt{y} + 1)^2 - (y/2 + 1)^2 \right] dy$
5. Simplify and Integrate: $(\sqrt{y} + 1)^2 = y + 2\sqrt{y} + 1$ $(y/2 + 1)^2 = \frac{y^2}{4} + y + 1$ $R^2 - r^2 = (y + 2\sqrt{y} + 1) - (\frac{y^2}{4} + y + 1) = 2\sqrt{y} - \frac{y^2}{4}$ $V = \pi \int_{0}^{4} (2y^{1/2} - \frac{1}{4}y^2) , dy$ $V = \pi \left[ \frac{2y^{3/2}}{3/2} - \frac{y^3}{12} \right]_0^4 = \pi \left[ \frac{4}{3}y^{3/2} - \frac{y^3}{12} \right]_0^4$ $V = \pi \left[ \left( \frac{4}{3}(8) - \frac{64}{12} \right) - 0 \right] = \pi \left( \frac{32}{3} - \frac{16}{3} \right) = \frac{16\pi}{3} \text{ cubic units.}$

Summary and Conclusion

Mastering the washer method requires a disciplined approach to geometry before attempting the calculus. On top of that, the most common errors occur not during integration, but during the setup of the radii. By consistently sketching the region and identifying the distance from the axis of revolution—rather than simply subtracting the functions from one another—you make sure the "hole" of the washer is correctly accounted for.

Whether revolving around the $x$-axis, $y$-axis, or an arbitrary line $x=h$ or $y=k$, the logic remains the same: Volume = (Outer Volume) - (Inner Volume). By visualizing the representative slice and defining $R$ and $r$ as the distance from the axis to the boundaries, you can solve any problem of this type with confidence.

People argue about this. Here's where I land on it.