Volume of Solids of Revolution Using Integrals
Besides calculating area, integrals are also used to calculate volumes of solids of revolution. A solid of revolution is formed when a region bounded by curves is rotated around a specific line (usually the X-axis or Y-axis) through a full rotation of 360°.
Visualization: Rotating a triangular region around the X-axis forms a cone. Rotating a semicircular region around the X-axis forms a sphere.
Two Main Methods
Characteristic: Rotation direction matches integration boundaries.
Example: Rotate around X-axis, integrate with X-boundaries.
Forms solid disks/washers perpendicular to axis of rotation.
Characteristic: Rotation direction differs from integration boundaries.
Example: Rotate around Y-axis, integrate with X-boundaries.
Forms cylindrical shells parallel to axis of rotation.
Disk Method
Region bounded by \(y = f(x)\), X-axis, \(x = a\), \(x = b\), rotated 360° around X-axis:
Find volume when region bounded by \(y = x\), X-axis, and \(x = 3\) is rotated 360° around X-axis.
Solution:
\(V = \pi \int_0^3 [x]^2 dx = \pi \int_0^3 x^2 dx\)
\(= \pi \left[ \frac{1}{3}x^3 \right]_0^3\)
\(= \pi \left( \frac{1}{3}(27) - 0 \right)\)
\(= \pi (9) = 9\pi\)
Answer: Volume = \(9\pi\) cubic units.
Region bounded by \(x = f(y)\), Y-axis, \(y = a\), \(y = b\), rotated 360° around Y-axis:
Find volume when region bounded by \(y = x^2\), Y-axis, \(y = 2\), and \(y = 5\) is rotated 360° around Y-axis.
Solution:
First, rewrite: \(y = x^2 \Rightarrow x = \sqrt{y}\)
\(V = \pi \int_2^5 [\sqrt{y}]^2 dy = \pi \int_2^5 y dy\)
\(= \pi \left[ \frac{1}{2}y^2 \right]_2^5\)
\(= \pi \left( \frac{25}{2} - \frac{4}{2} \right) = \pi \left( \frac{21}{2} \right)\)
Answer: Volume = \(\frac{21}{2}\pi\) cubic units.
Around X-axis: Region between \(y_1 = f(x)\) and \(y_2 = g(x)\) where \(|f(x)| \geq |g(x)|\) on \([a, b]\):
Around Y-axis: Region between \(x_1 = f(y)\) and \(x_2 = g(y)\) where \(|f(y)| \geq |g(y)|\) on \([a, b]\):
Always subtract: curve farther from rotation axis minus curve closer to rotation axis.
Find volume when region bounded by \(y = 6x - x^2\) and \(y = x\) is rotated 360° around X-axis.
Solution:
1. Find intersections: \(6x - x^2 = x \Rightarrow x^2 - 5x = 0 \Rightarrow x(x-5) = 0\)
Intersections at \(x = 0\) and \(x = 5\)
2. On [0,5], \(6x - x^2 \geq x\) (check at \(x=2\): \(8 > 2\))
So \(f(x) = 6x - x^2\), \(g(x) = x\)
3. \(V = \pi \int_0^5 \left( (6x - x^2)^2 - x^2 \right) dx\)
\(= \pi \int_0^5 (36x^2 - 12x^3 + x^4 - x^2) dx\)
\(= \pi \int_0^5 (x^4 - 12x^3 + 35x^2) dx\)
\(= \pi \left[ \frac{1}{5}x^5 - 3x^4 + \frac{35}{3}x^3 \right]_0^5\)
\(= \pi \left( \frac{3125}{5} - 1875 + \frac{4375}{3} \right)\)
\(= \pi \left( 625 - 1875 + \frac{4375}{3} \right) = \pi \left( \frac{625}{3} \right)\)
Answer: Volume = \(208\frac{1}{3}\pi\) cubic units.
Shell Method
Rotate around Y-axis (X-boundaries):
Rotate around X-axis (Y-boundaries):
When to use: When it's difficult to rewrite \(y = f(x)\) as \(x = f(y)\) or vice versa.
Find volume when region bounded by \(y = -x^3 + 4x\), \(x = 0\), \(x = 1\), and X-axis is rotated 360° around Y-axis.
Solution:
Why use shell method? \(y = -x^3 + 4x\) is difficult to rewrite as \(x = f(y)\), so shell method is easier.
\(V = 2\pi \int_0^1 x \cdot (-x^3 + 4x) dx\)
\(= 2\pi \int_0^1 (-x^4 + 4x^2) dx\)
\(= 2\pi \left[ -\frac{1}{5}x^5 + \frac{4}{3}x^3 \right]_0^1\)
\(= 2\pi \left( -\frac{1}{5} + \frac{4}{3} \right) = 2\pi \left( \frac{-3 + 20}{15} \right)\)
\(= 2\pi \left( \frac{17}{15} \right) = \frac{34}{15}\pi = 2\frac{4}{15}\pi\)
Answer: Volume = \(2\frac{4}{15}\pi\) cubic units.
Rotate around Y-axis (X-boundaries):
Where \(|f(x)| \geq |g(x)|\) on \([a, b]\)
Rotate around X-axis (Y-boundaries):
Find volume when region bounded by \(y = \frac{1}{3}x^2\), \(y = x\), \(x = 0\), \(x = 2\), and X-axis is rotated 360° around Y-axis.
Solution:
On [0,2], \(x \geq \frac{1}{3}x^2\) (check at \(x=1\): \(1 > \frac{1}{3}\))
So \(f(x) = x\), \(g(x) = \frac{1}{3}x^2\)
\(V = 2\pi \int_0^2 x \cdot \left( x - \frac{1}{3}x^2 \right) dx\)
\(= 2\pi \int_0^2 \left( x^2 - \frac{1}{3}x^3 \right) dx\)
\(= 2\pi \left[ \frac{1}{3}x^3 - \frac{1}{12}x^4 \right]_0^2\)
\(= 2\pi \left( \frac{8}{3} - \frac{16}{12} \right) = 2\pi \left( \frac{8}{3} - \frac{4}{3} \right)\)
\(= 2\pi \left( \frac{4}{3} \right) = \frac{8}{3}\pi = 2\frac{2}{3}\pi\)
Answer: Volume = \(2\frac{2}{3}\pi\) cubic units.
Method Selection Guide
• Rotation axis matches integration variable (rotate around X, integrate dx)
• Easy to express functions in terms of rotation axis variable
• Forms solid perpendicular slices to rotation axis
• Rotation axis differs from integration variable (rotate around Y, integrate dx)
• Difficult to rewrite functions in terms of rotation axis variable
• Forms cylindrical shells parallel to rotation axis
To master volume calculations using integrals, you need:
- Graphing skills: Sketch curves and identify regions
- Integration skills: Calculate definite integrals of algebraic functions
- Analytical thinking: Determine which method applies
- Visualization: Imagine the 3D solid formed by rotation
Even for difficult problems, the process remains the same as in these examples.
| Method | Rotation | Formula |
|---|---|---|
| Disk | Around X-axis | \(V = \pi \int_a^b [f(x)]^2 dx\) |
| Disk | Around Y-axis | \(V = \pi \int_a^b [f(y)]^2 dy\) |
| Washer | Around X-axis | \(V = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) dx\) |
| Washer | Around Y-axis | \(V = \pi \int_a^b ([f(y)]^2 - [g(y)]^2) dy\) |
| Shell | Around Y-axis | \(V = 2\pi \int_a^b x f(x) dx\) |
| Shell | Around X-axis | \(V = 2\pi \int_a^b y f(y) dy\) |