Find the volume of a solid of revolution generated by revolving a region bounded by the graph of a function around one of the axes using definite integrals and the method of cylindrical shells where the integration is perpendicular to the axis of rotation. Exercises with their answers are presented at the bottom of the page.
In what follows, \( C \) represents the constant of integration.
Formula - Method of Cylindrical Shells
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 around the \( y \)-axis the region bounded by the graph of \( f \), the \( x \)-axis (\( y = 0 \)), and the vertical lines \( x = x_1 \) and \( x = x_2 \) is given by the integral:
\[ \Large \text{Volume} = \int_{x_1}^{x_2} 2\pi x f(x) \, dx \]
If \( g \) is a function such that \( g(y) \geq 0 \) for all \( y \) in the interval \( [y_1, y_2] \), the volume of the solid generated by revolving around the \( x \)-axis the region bounded by the graph of \( g \), the \( y \)-axis (\( x = 0 \)), and the horizontal lines \( y = y_1 \) and \( y = y_2 \) is given by the integral:
\[ \Large \text{Volume} = \int_{y_1}^{y_2} 2\pi y g(y) \, dy \]Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1: Volume of solid generated by revolving a triangle about the \( y \)-axis
Solution to Example 1:
Note that this problem has been solved in Volume of a Solid of Revolution using the washers method. Let's now solve it using the cylindrical shells method.
The region to be rotated is bounded by \( x = 0 \), \( y = 0 \), and two curves \( y = x \) and \( y = -x + 2 \) with an intersection point at \( (1, 1) \). We therefore split the integral into two parts:
\[ \text{Text} \implies \text{Volume} = \int_{0}^{1} 2\pi x(x) \, dx + \int_{1}^{2} 2\pi x(-x + 2) \, dx \]Integrate:
\[ = 2\pi \left[\dfrac{x^3}{3}\right]_0^1 + 2\pi \left[-\dfrac{x^3}{3} + x^2\right]_1^2 \]Evaluate:
\[ = \dfrac{2\pi}{3} + 2\pi \left[\left(-\dfrac{8}{3} + 4\right) - \left(-\dfrac{1}{3} + 1\right)\right] \] \[ = \dfrac{2\pi}{3} + 2\pi \left[\dfrac{4}{3} - \dfrac{2}{3}\right] = \dfrac{2\pi}{3} + \dfrac{4\pi}{3} = 2\pi \]Example 2: Volume of solid generated by a cubic curve region in the first quadrant
Solution to Example 2:
The graphs of \( y = -x^3 + 2x^2 - x + 2 \) and \( y = -x + 1 \) are shown below.
The \( x \)-intercept of \( y = -x^3 + 2x^2 - x + 2 \) is found by solving:
\[ -x^3 + 2x^2 - x + 2 = 0 \implies x^2(-x + 2) + 1(-x + 2) = 0 \implies (-x + 2)(x^2 + 1) = 0 \]This gives \( x = 2 \). The \( x \)-intercept of \( y = -x + 1 \) is \( x = 1 \).
From \( x = 0 \) to \( x = 1 \), the region is bounded above by \( f(x) = -x^3 + 2x^2 - x + 2 \) and below by \( h(x) = -x + 1 \). The volume \( V_1 \) is:
\[ V_1 = \int_{0}^{1} 2\pi x (f(x) - h(x)) \, dx = 2\pi \int_{0}^{1} x (-x^3 + 2x^2 - x + 2 - (-x + 1)) \, dx \] \[ = 2\pi \int_{0}^{1} (-x^4 + 2x^3 + x) \, dx = 2\pi \left[ -\dfrac{x^5}{5} + \dfrac{x^4}{2} + \dfrac{x^2}{2} \right]_0^1 = \dfrac{8\pi}{5} \]From \( x = 1 \) to \( x = 2 \), the region is bounded above by \( f(x) \) and below by \( y = 0 \). The volume \( V_2 \) is:
\[ V_2 = \int_{1}^{2} 2\pi x f(x) \, dx = 2\pi \int_{1}^{2} x (-x^3 + 2x^2 - x + 2) \, dx \] \[ = 2\pi \left[ -\dfrac{x^5}{5} + \dfrac{x^4}{2} - \dfrac{x^3}{3} + x^2 \right]_1^2 = \dfrac{59\pi}{15} \]Total volume \( V = V_1 + V_2 = \dfrac{8\pi}{5} + \dfrac{59\pi}{15} = \dfrac{83\pi}{15} \).
Example 3: Volume generated by a sine curve region around the \( y \)-axis
Solution to Example 3:
Find a formula for the volume enclosed by \( y = 0 \), \( x = 0 \), and \( y = a + \sin(x) \) for \( 0 \leq x \leq 2\pi \) and \( a \geq 1 \) about the \( y \)-axis:
\[ V = \int_{0}^{2\pi} 2\pi x (a + \sin(x)) \, dx = 2\pi \int_{0}^{2\pi} ax \, dx + 2\pi \int_{0}^{2\pi} x \sin(x) \, dx \]Evaluating the first integral \( I_1 \):
\[ I_1 = 2\pi a \left[ \dfrac{x^2}{2} \right]_0^{2\pi} = 4\pi^3 a \]Evaluating the second integral \( I_2 \) using integration by parts (\( u = x \), \( dv = \sin(x) \, dx \)):
\[ I_2 = 2\pi \left[ -x \cos(x) \right]_0^{2\pi} + 2\pi \int_{0}^{2\pi} \cos(x) \, dx = -4\pi^2 + 0 = -4\pi^2 \]Total volume: \( V = I_1 + I_2 = 4\pi^3 a - 4\pi^2 \).
Example 4: Volume generated by an ellipse in the first quadrant around the \( x \)-axis
Solution to Example 4:
Use cylindrical shells to find a formula for the volume of the solid generated by revolving the area enclosed by \( y = 0 \), \( x = 0 \), and \( \left(\dfrac{x}{a}\right)^2 + \left(\dfrac{y}{b}\right)^2 = 1 \) in the first quadrant about the \( x \)-axis (\( a, b > 0 \)).
Solving for \( x \): \( x = a \sqrt{1 - \left(\dfrac{y}{b}\right)^2} \).
Since rotation is around the \( x \)-axis, shells are parallel to the \( x \)-axis:
\[ V = \int_{0}^{b} 2\pi y \left( a \sqrt{1 - \left(\dfrac{y}{b}\right)^2} \right) dy \]Using substitution \( u = 1 - \left(\dfrac{y}{b}\right)^2 \) and evaluating yields:
\[ V = \dfrac{2\pi a b^2}{3} \]Exercises
Calculate the following integrals. Click each exercise to reveal the step-by-step solution.
Exercise 1: Find the volume of the solid generated when part of the graph of \( f(x) = -x^4 + 3x^3 - x + 3 \) in quadrant I is revolved about the \( y \)-axis.
Solution:
Find the \( x \)-intercept by solving \( -x^4 + 3x^3 - x + 3 = 0 \implies x^3(3 - x) + 1(3 - x) = (x^3 + 1)(3 - x) = 0 \implies x = 3 \) (in quadrant I).
Using the cylindrical shells formula from \( x = 0 \) to \( x = 3 \):
\[ V = \int_{0}^{3} 2\pi x (-x^4 + 3x^3 - x + 3) \, dx = 2\pi \int_{0}^{3} (-x^5 + 3x^4 - x^2 + 3x) \, dx \]Integrating term by term:
\[ = 2\pi \left[ -\dfrac{x^6}{6} + \dfrac{3x^5}{5} - \dfrac{x^3}{3} + \dfrac{3x^2}{2} \right]_0^3 = \dfrac{288\pi}{5} \]