8.2 Volumes of Solids of Revolution

8.2 Volumes of Solids of Revolution

Scope note — read this first: The College Board's published Calculus outline lists exactly three Applications of the Definite Integral: average value of a function on an interval; area, including area between curves; and "Other (e.g., accumulated change from a rate of change)." Volumes of solids of revolution are not named in that outline. Treat this section as enrichment, not as a guaranteed exam topic — finish Sections 8.1, 8.3, and 7.3 first. It is included because the "Other" category is open-ended, because rotating a region is a direct extension of the area work in Section 8.1, and because the Calculus I courses that CLEP credit substitutes for almost always cover it.

The method itself: A solid of revolution is a three-dimensional geometric body generated by rotating a two-dimensional planar region about a fixed straight line called the axis of revolution. Success requires choosing between the Disk Method and the Washer Method, setting up correct radius expressions relative to both standard coordinate axes and shifted lines ($y = k$ or $x = h$), and avoiding algebraic squaring traps.

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1. The General Slicing Principle

Before specializing to solids of revolution, consider the overarching Slicing Method for computing the volume of any solid. If a solid extends along an axis from $x = a$ to $x = b$, and its cross-sectional area perpendicular to the $x$-axis at any point $x$ is given by a continuous function $A(x)$, we slice the solid into $n$ thin slabs of thickness $\Delta x$.

          Cross-Sectional Area A(x)
                 .-''''-.
               .'        '.
              /    A(x)    \
             :      O       :
              \            /
               '.        .'
                 `-....-'
                 |<-dx->|
        ===+=========+=========+===> x
           a         x         b

The volume of each approximating cylindrical slab is $\Delta V_i = A(x_i^*) , \Delta x$. Summing over all $n$ slabs and taking the limit as $n \to \infty$ yields the fundamental volume integral:

$V = \lim_{n \to \infty} \sum_{i=1}^n A(x_i^*) \, \Delta x = \int_a^b A(x) \, dx$

For solids of revolution, every cross-section sliced perpendicular to the axis of revolution is either a solid circular disk or a hollow annular washer.

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2. The Disk Method (No Central Void / Flush Boundary)

When the region being rotated is flush against the axis of revolution throughout the entire interval of integration, the solid possesses no internal cavity. Slicing perpendicular to the axis produces solid circular disks.

Cross-Sectional Geometry of a Disk

• Shape: Solid circle of radius $R$
• Area: $A = \pi R^2$
• Differential Volume Element: $dV = \pi [R]^2 , d(\text{variable})$

           Solid Circular Disk
                .--------.
              .'          '.
             /      |       \   Radius R
            |       o--------| ---------> R
             \              /
              '.          .'
                `--------'
                |<- dx ->|

Disk Method Formulas Across All Axes

1. Rotation around the $x$-axis ($y = 0$): $V = \pi \int_a^b [f(x)]^2 \, dx$

2. Rotation around a horizontal line $y = k$ (flush boundary): $V = \pi \int_a^b [f(x) - k]^2 \, dx$

3. Rotation around the $y$-axis ($x = 0$): $V = \pi \int_c^d [g(y)]^2 \, dy$

4. Rotation around a vertical line $x = h$ (flush boundary): $V = \pi \int_c^d [g(y) - h]^2 \, dy$

Step-by-Step Worked Example 1: Disk Method About $x$-Axis

Problem: Find the volume of the solid generated by rotating the region bounded by $y = \sqrt{x}$, the $x$-axis ($y = 0$), and the line $x = 4$ about the $x$-axis.

Step 1: Identify region and axis. The region is bounded by $x \in [0, 4]$. The axis of revolution is the $x$-axis ($y = 0$). Slices are perpendicular to the $x$-axis $\implies dx$ integration.

Step 2: Determine the radius $R(x)$. Since the bottom of the region rests directly on the $x$-axis, the radius is the vertical distance from the axis to the curve: $R(x) = \sqrt{x} - 0 = \sqrt{x}$

Step 3: Set up and evaluate the integral. $V = \pi \int_0^4 [R(x)]^2 \, dx = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$ $V = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \left( \frac{16}{2} - 0 \right) = 8\pi$

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3. The Washer Method (Hollow Solids / Inner Void)

When the rotated region does not touch the axis of revolution along one of its boundaries, revolving the region sweeps out an interior hole or channel. Slicing perpendicular to the axis produces a washer (the region between two concentric circles).

                 Annular Washer Cross-Section
                      .------------.
                    .'      R       '.
                   /    .-------.     \
                  /   .'    r    '.    \
                 |   /      o------|----| ----> Outer Radius R
                 |   |      |<---->|    | ----> Inner Radius r
                  \   '.         .'    /
                   \    '-------'     /
                    '.              .'
                      `------------'
                      |<-  dx   ->|

Cross-Sectional Geometry of a Washer

• Outer Radius ($R$): Distance from axis of revolution to the far boundary curve.
• Inner Radius ($r$): Distance from axis of revolution to the near boundary curve.
• Washer Area: $A = \text{Area}{\text{outer}} - \text{Area}{\text{inner}} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2)$
• Differential Volume Element: $dV = \pi [R^2 - r^2] , d(\text{variable})$

⚠️ Critical CLEP Warning: The Difference of Squares Trap

$\mathbf{V = \pi \int \left( R^2 - r^2 \right) dx \quad \neq \quad \pi \int (R - r)^2 \, dx}$

Why this fails algebraically: $(R - r)^2 = R^2 - 2Rr + r^2$. Subtracting the areas of two circles requires $A = \pi R^2 - \pi r^2$. Squaring the difference $(R - r)$ computes the area of a tiny circle with radius equal to the thickness $(R - r)$, severely underestimating the true volume!

Washer Method Formulas Across Standard Axes

1. Rotation around the $x$-axis ($y = 0$): $V = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) dx \quad \text{where } |f(x)| \ge |g(x)|$

2. Rotation around the $y$-axis ($x = 0$): $V = \pi \int_c^d \left( [f(y)]^2 - [g(y)]^2 \right) dy \quad \text{where } |f(y)| \ge |g(y)|$

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4. Revolutions Around Non-Standard Axes ($y = k$ and $x = h$)

On challenging CLEP questions, the axis of revolution is shifted to an arbitrary line. The universal rule for setting up radius expressions is based on absolute geometric distance:

$\text{Radius} = |\text{Boundary Curve} - \text{Axis of Revolution}|$

Horizontal Axis $y = k$ (Slices Perpendicular $\implies dx$)

Vertical Axis $x = h$ (Slices Perpendicular $\implies dy$)

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5. Step-by-Step Worked Example 2: Washer Around Shifted Axis $y = 5$

Problem: Find the volume of the solid generated by rotating the region enclosed by $y = x^2$ and $y = 2x$ about the horizontal line $y = 5$.

Step 1: Find intersection points. $x^2 = 2x \implies x^2 - 2x = 0 \implies x(x - 2) = 0 \implies x = 0 \quad \text{and} \quad x = 2$

Step 2: Determine spatial layout relative to axis $y = 5$. On the interval $[0, 2]$, $2x \ge x^2$. The axis of revolution is $y = 5$, which lies strictly above the entire region (since the maximum $y$-value is $2(2) = 4 < 5$).

Step 3: Identify Outer and Inner radii.

• The curve furthest from $y = 5$ is the lower curve $y = x^2$: $R(x) = 5 - x^2$
• The curve nearest to $y = 5$ is the upper curve $y = 2x$: $r(x) = 5 - 2x$

Step 4: Set up the definite integral. $V = \pi \int_0^2 \left[ (R(x))^2 - (r(x))^2 \right] dx = \pi \int_0^2 \left[ (5 - x^2)^2 - (5 - 2x)^2 \right] dx$

Step 5: Expand the algebraic expressions. $(5 - x^2)^2 = 25 - 10x^2 + x^4$ $(5 - 2x)^2 = 25 - 20x + 4x^2$ $\left[ (5 - x^2)^2 - (5 - 2x)^2 \right] = (25 - 10x^2 + x^4) - (25 - 20x + 4x^2) = x^4 - 14x^2 + 20x$

Step 6: Antidifferentiate and evaluate. $V = \pi \int_0^2 (x^4 - 14x^2 + 20x) \, dx = \pi \left[ \frac{x^5}{5} - \frac{14x^3}{3} + 10x^2 \right]_0^2$ $V = \pi \left( \frac{32}{5} - \frac{14(8)}{3} + 10(4) \right) = \pi \left( \frac{32}{5} - \frac{112}{3} + 40 \right)$ $V = \pi \left( \frac{96 - 560 + 600}{15} \right) = \frac{136\pi}{15}$

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6. Step-by-Step Worked Example 3: Rotation Around the $y$-Axis ($dy$)

Problem: Find the volume of the solid generated by revolving the region bounded by $y = x^3$, $y = 8$, and the $y$-axis ($x = 0$) about the $y$-axis.

Step 1: Determine orientation and variable of integration. Rotation is around the vertical $y$-axis. Slicing perpendicular to the $y$-axis requires horizontal slices, meaning integration with respect to $y$ ($dy$).

Step 2: Express curves as functions of $y$.

• Bounding curve: $y = x^3 \implies x = y^{1/3} = \sqrt[3]{y}$.
• Inner boundary: $y$-axis ($x = 0$).
• Integration bounds in $y$: from $y = 0$ (origin) to $y = 8$.

Step 3: Setup Disk integral ($r = 0$). $V = \pi \int_0^8 [R(y)]^2 \, dy = \pi \int_0^8 \left( y^{1/3} \right)^2 dy = \pi \int_0^8 y^{2/3} \, dy$

Step 4: Evaluate. $V = \pi \left[ \frac{y^{5/3}}{5/3} \right]_0^8 = \pi \cdot \frac{3}{5} \left( 8^{5/3} - 0 \right) = \frac{3\pi}{5} \left( (\sqrt[3]{8})^5 \right) = \frac{3\pi}{5}(32) = \frac{96\pi}{5}$

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7. The 5-Step Master Protocol for Volume Problems

+-------------------------------------------------------------------------+
|                   VOLUME PROBLEM MASTER PROTOCOL                        |
+-------------------------------------------------------------------------+
| 1. SKETCH & IDENTIFY AXIS: Draw region and highlight axis of rotation.  |
| 2. CHOOSE VARIABLE: Axis horizontal -> dx; Axis vertical -> dy.        |
| 3. DETERMINE DISK VS WASHER: Is region flush against axis?              |
|    - Flush everywhere -> Disk: A = pi * R^2                             |
|    - Gap/void exists  -> Washer: A = pi * (R_out^2 - r_in^2)            |
| 4. CONSTRUCT RADII: R = |Curve_far - Axis|; r = |Curve_near - Axis|.    |
| 5. INTEGRATE: Factor out pi, expand polynomials, evaluate boundaries.   |
+-------------------------------------------------------------------------+

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8. Master Method & Formula Comparison Table

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9. Common CLEP Pitfalls & Traps

1. The Squaring-Difference Trap: Writing $\pi \int (R - r)^2 dx$ instead of $\pi \int (R^2 - r^2) dx$. This is the single most common distractor on the CLEP.
2. Forgetting the Constant $\pi$: Omitting $\pi$ from the front of the integral leads directly to incorrect numerical options.
3. Variable Mismatch on Rotations: Integrating with respect to $x$ when revolving about a vertical axis. Remember: Slices are perpendicular to the axis of revolution. A vertical axis requires horizontal slices ($dy$).
4. Incorrect Radius on Shifted Axes: When revolving about $y = k$ where $k$ is above the region, students often write $R(x) = f(x) - k$ (which is negative) instead of $k - f(x)$. While squaring hides the sign error in a single term, mixing signs in washers ($R^2 - r^2$) destroys the calculation.
5. Interchanging Outer and Inner Radii: The outer radius $R$ is always measured to the boundary furthest from the axis of rotation; $r$ is always measured to the boundary closest to the axis.