Introduction to Solids of Revolution
Learn how to find the volume of a solid of revolution generated by revolving a region bounded by curves around the x-axis or y-axis. This tutorial uses definite integrals and covers both the disk method (for solid shapes) and the washer method (for shapes with hollow centers), complete with formulas, visual figures, and step-by-step practice problems.
In what follows, \( C \) represents the constant of integration.
Volume Formulas
Formula 1: Disk Method (Around the x-axis)
If \( f \) is a function such that \( f(x) \geq 0 \) for all \( x \) in the interval \( [x_1 , x_2] \), the volume of the solid generated by revolving the region bounded by the graph of \( f \), the x-axis (\( y = 0 \)), and the vertical lines \( x = x_1 \) and \( x = x_2 \) around the x-axis is given by:
\[ \text{Volume} = \int_{x_1}^{x_2} \pi [f(x)]^2 \, dx \]
Formula 2: Washer Method (Around the x-axis)
If \( f \) and \( h \) are functions such that \( f(x) \geq h(x) \) for all \( x \) in the interval \( [x_1 , x_2] \), the volume of the solid generated by revolving the region bounded by the graphs of \( f \) and \( h \) around the x-axis is given by:
\[ \text{Volume} = \int_{x_1}^{x_2} \pi [f(x)^2 - h(x)^2] \, dx \]
Formula 3: Disk Method (Around the y-axis)
If \( z \) is a function of \( y \) such that \( x = z(y) \) and \( z(y) \geq 0 \) for all \( y \) in the interval \( [y_1 , y_2] \), the volume generated by revolving the region bounded by \( z \), the y-axis (\( x = 0 \)), and the horizontal lines \( y = y_1 \) and \( y = y_2 \) around the y-axis is:
\[ \text{Volume} = \int_{y_1}^{y_2} \pi [z(y)]^2 \, dy \]
Formula 4: Washer Method (Around the y-axis)
If \( z \) and \( w \) are functions of \( y \) such that \( z(y) \geq w(y) \) for all \( y \) in the interval \( [y_1 , y_2] \), the volume generated by revolving the region bounded by \( z \) and \( w \) around the y-axis is given by:
\[ \text{Volume} = \int_{y_1}^{y_2} \pi [z(y)^2 - w(y)^2] \, dy \]
Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Find the volume of the solid generated by revolving the region bounded by the graph of \( y = x \), \( y = 0 \), \( x = 0 \), and \( x = 2 \) around the x-axis.
Show Solution
Method 1: Geometric Formula
Because \( y = x \) forms a triangle, revolving it creates a right circular cone with a radius of 2 and a height of 2. Using the geometric formula:
\[ \text{Volume} = \frac{1}{3} \pi (\text{radius})^2 (\text{height}) = \frac{1}{3} \pi (2)^2 (2) = \frac{8\pi}{3} \]Method 2: Definite Integral (Disk Method)
Let \( f(x) = x \). Using Formula 1, the volume is given by the definite integral from 0 to 2:
\[ \text{Volume} = \int_{0}^{2} \pi x^2 \, dx = \left[ \pi \frac{x^3}{3} \right]_0^2 = \frac{8\pi}{3} \]Example 2
Find the volume of the solid generated by revolving the semicircle \( y = \sqrt{r^2 - x^2} \) around the x-axis, where \( r > 0 \).
Show Solution
The graph of \( y = \sqrt{r^2 - x^2} \) is bounded from \( x = -r \) to \( x = r \). The volume is given by the disk method (Formula 1):
\[ \text{Volume} = \int_{-r}^{r} \pi \left(\sqrt{r^2 - x^2}\right)^2 \, dx = \int_{-r}^{r} \pi (r^2 - x^2) \, dx \]Find the antiderivative and evaluate at the limits:
\[ = \pi \left[ r^2 x - \frac{x^3}{3} \right]_{-r}^r \] \[ = \pi \left[ \left(r^3 - \frac{r^3}{3}\right) - \left(-r^3 + \frac{r^3}{3}\right) \right] = \frac{4}{3} \pi r^3 \]This derives the well-known formula for the volume of a sphere!
Example 3
Find the volume of the solid generated by revolving the shaded region bounded by \( y = x \), \( y = -x + 2 \), and the x-axis about the y-axis.
Show Solution
Since the solid is generated by revolving around the y-axis, we must rewrite our equations as functions of \( y \) (\( x = z(y) \)). The limits of integration along the y-axis are from \( y = 0 \) to \( y = 1 \) (where the lines intersect).
The right curve is \( y = -x + 2 \implies x = 2 - y \).
The left curve is \( y = x \implies x = y \).
Using the washer method (Formula 4):
\[ \text{Volume} = \int_{0}^{1} \pi [ (2-y)^2 - y^2 ] \, dy \]Expand the binomial and simplify the integrand:
\[ = \int_{0}^{1} \pi [ 4 - 4y + y^2 - y^2 ] \, dy = \int_{0}^{1} \pi [ 4 - 4y ] \, dy \]Integrate and evaluate:
\[ = \pi \left[ 4y - 2y^2 \right]_0^1 = \pi (4 - 2) = 2\pi \]Example 4
Find the volume of the solid generated by the rotation of the region bounded by the curves \( y = 1 + x^2 \) and \( y = \sqrt{x} \), around the x-axis from \( x = 0 \) to \( x = 2 \).
Show Solution
We use the washer method along the x-axis (Formula 2). Let \( f(x) = 1 + x^2 \) (the outer radius) and \( h(x) = \sqrt{x} \) (the inner radius).
\[ \text{Volume} = \int_{0}^{2} \pi ( f(x)^2 - h(x)^2 ) \, dx \] \[ = \int_{0}^{2} \pi \left( (1+x^2)^2 - (\sqrt{x})^2 \right) \, dx \]Expand the squares:
\[ = \pi \int_{0}^{2} ( 1 + 2x^2 + x^4 - x ) \, dx \]Integrate using the power rule:
\[ = \pi \left[ x + \frac{2x^3}{3} + \frac{x^5}{5} - \frac{x^2}{2} \right]_0^2 \]Evaluate at \( x = 2 \):
\[ = \pi \left( 2 + \frac{16}{3} + \frac{32}{5} - 2 \right) = \frac{176\pi}{15} \]Example 5
Find the volume of the torus generated when the circle with a center at \( (0, R) \) and radius \( r \) is rotated around the x-axis.
Show Solution
The equation of the circle is given by \( x^2 + (y - R)^2 = r^2 \).
Solve the equation for \( y \) to obtain the upper and lower bounds of the circle:
- Upper semicircle: \( f(x) = R + \sqrt{r^2 - x^2} \)
- Lower semicircle: \( h(x) = R - \sqrt{r^2 - x^2} \)
Because of the symmetry with respect to the y-axis, we can integrate from \( x = 0 \) to \( x = r \), then double the answer to find the total volume. Using the washer method (Formula 2):
\[ \frac{1}{2} \text{Volume} = \int_{0}^{r} \pi [ f(x)^2 - h(x)^2 ] \, dx \] \[ = \pi \int_{0}^{r} \left[ (R + \sqrt{r^2 - x^2})^2 - (R - \sqrt{r^2 - x^2})^2 \right] \, dx \]When you expand and subtract the squares, the \( R^2 \) and \( (r^2 - x^2) \) terms cancel out, leaving:
\[ = 4 R \pi \int_{0}^{r} \sqrt{r^2 - x^2} \, dx \]The integral of \( \sqrt{r^2 - x^2} \) yields the area of a quarter circle, which we can solve using standard trigonometric substitution rules:
\[ = 4 R \pi \left( \frac{1}{2} \right) \left[ x \sqrt{r^2-x^2} + r^2\arcsin\left(\frac{x}{r}\right) \right]_0^r \] \[ = 2 R \pi \left( r^2 \arcsin(1) - 0 \right) = 2 R \pi \left( r^2 \frac{\pi}{2} \right) = \pi^2 R r^2 \]Since this was only half the torus, the total volume is twice the above result:
\[ \text{Volume} = 2 \pi^2 R r^2 \](Note: A minor typographical error in the original calculus derivation for the arcsin term has been corrected here for mathematical accuracy).
Practice Exercises
Click on each exercise to reveal the step-by-step solution.
Exercise 1
Find the volume of the solid generated when the region between the graphs of \( f(x) = x^2 + 2 \) and \( h(x) = x \) is revolved about the x-axis over the interval \( [0, 1] \).
Show Solution
Use the washer method around the x-axis:
\[ \text{Volume} = \pi \int_{0}^{1} \left[ (x^2 + 2)^2 - (x)^2 \right] \, dx \] \[ = \pi \int_{0}^{1} (x^4 + 4x^2 + 4 - x^2) \, dx = \pi \int_{0}^{1} (x^4 + 3x^2 + 4) \, dx \]Integrate and evaluate at the bounds:
\[ = \pi \left[ \frac{x^5}{5} + x^3 + 4x \right]_0^1 = \pi \left( \frac{1}{5} + 1 + 4 \right) = \frac{26\pi}{5} \]Exercise 2
Find the volume generated when the finite region bounded by the curves \( y = x^3 \) and \( y = x^2 \) is revolved about the y-axis.
Show Solution
First, find the points of intersection by setting \( x^3 = x^2 \). The curves intersect at \( x = 0 \) and \( x = 1 \), which correspond to \( y = 0 \) and \( y = 1 \).
Since we are rotating around the y-axis, we need to express \( x \) as functions of \( y \). In the interval \( y \in [0, 1] \), \( y = x^3 \implies x = y^{1/3} \) is the outer radius, and \( y = x^2 \implies x = y^{1/2} \) is the inner radius.
Use the washer method around the y-axis:
\[ \text{Volume} = \pi \int_{0}^{1} \left[ (y^{1/3})^2 - (y^{1/2})^2 \right] \, dy = \pi \int_{0}^{1} (y^{2/3} - y) \, dy \]Integrate and evaluate:
\[ = \pi \left[ \frac{3}{5}y^{5/3} - \frac{y^2}{2} \right]_0^1 = \pi \left( \frac{3}{5} - \frac{1}{2} \right) = \frac{\pi}{10} \]