7.3 | Volume: The Shell Method


The volume of the solid of revolution generated by revolving the region bounded by
$$
y = f(x), y = g(x), x = a, x = b
$$
about the \(y\)-axis is given by
$$
\int_a^b 2 \pi x (f(x)-g(x)) \; dx
$$
provided that \(f(x) \geq g(x)\) on \([a,b]\). \(x\) represents the distance from the axis of revolution to the base of the rectangle and \(f(x)-g(x)\) represents the height of the rectangle at \(x\). This method is referred to as the shell method

Find the volume of the solid of revolution obtained by revolving the region bounded by the graphs of \(y=2+ \sin x, y = 0,x = 0,\) and \(x = 2 \pi \) around the (a) \(y\)-axis, (b) \(x\)-axis.
Hint:
$$ \int x \sin x \; dx = \sin x – x \cos x + C
$$


  1. Using the shell method, the volume of the solid generated by revolving the bounded region about the \(y\)-axis is:
    $$
    \begin{aligned}
    V & = \int_0^{2 \pi} 2 \pi x (2+\sin x) \; dx \\
    & = \int_0^{2 \pi} 4 \pi x+2 \pi x \sin x \; dx \\
    & = \int_0^{2 \pi} 4 \pi x\; dx +\int_0^{2 \pi} 2 \pi x \sin x \; dx \\
    & =2 \pi x ^2 \Big|_0^{2 \pi} \; dx +\int_0^{2 \pi} 2 \pi x \sin x \; dx \\
    & =2 \pi x ^2 \Big|_0^{2 \pi} \; dx + 2 \pi (\sin x – x \cos x ) \Big|_0^{2 \pi} \; dx \\
    & = 2 \pi (2 \pi)^2 + 2 \pi (0 – 2 \pi) \\
    & = 8 \pi^3 – 4 \pi^2
    \end{aligned}
    $$

    where

    • in the first equality, we apply the formula for shell method
    • in the second equality, we simplify the integrand
    • in the third equality, we distribute the definite integral over addition
    • in the fourth equality, we apply the Fundamental Theorem of Calculus (Part 2) to the first definite integral
    • in the fourth equality, we apply the hint and the Fundamental Theorem of Calculus (Part 2) to the remaining definite integral
    • in the remaining equalities, we evaluate and simplify
  2. Using the disk method, the volume of the solid generated by revolving the bounded region about the \(x\)-axis is:
    $$
    \begin{aligned}
    V & = \int_0^{2 \pi} \pi (2+\sin x)^2 \; dx \\
    & = \int_0^{2 \pi} 4 \pi + 4 \pi \sin x + \pi \sin^2 x\; dx \\
    & = 4 \pi \int_0^{2 \pi} 1 + \sin x \; dx + \pi \int_0^{2 \pi}\sin^2 x\; dx \\
    & = 4 \pi(x -\cos x ) \Big|_0^{2 \pi} + \int_0^{2 \pi}\sin^2 x\; dx \\
    & = 8\pi^2 + \int_0^{2 \pi}\sin^2 x\; dx \\
    & = 8\pi^2 + \int_0^{2 \pi}\frac{1-\cos(2x)}{2}\; dx \\
    & = 8\pi^2 + \frac{1}{2}x-\frac{\sin(2x)}{4}\Big|_0^{2 \pi} \\
    & = 8\pi^2 + \pi\\
    \end{aligned}
    $$

    where

    • in the first equality, we apply the formula for disk method
    • in the second equality, we simplify the integrand
    • in the third equality, we distribute the definite integral over addition and pull out constant multipliers
    • in the fourth equality, we apply the Fundamental Theorem of Calculus (Part 2) to the first definite integral
    • in the fifth equality, we simplify
    • in the sixth equality, we apply a trigonometric identity for \(\sin^2 x\)
    • in the seventh equality, we apply the Fundamental Theorem of Calculus (Part 2) to the remaining definite integral
    • in the last few steps, we simplify.
Find the volume of the solid formed by revolving the region bounded by the graph of \(y = x^{2/3}\), \(x= 0\), and \(y = 1\), about the \(y\)-axis.


Placing a vertical rectangle within the bounded region, and revolving it around the \(y\)-axis, we get a shell. The little bit of volume, \(\Delta V\), that this shell contributes to the volume of the solid of revolution is
$$
\begin{aligned}
\Delta V & = 2 \pi r h \Delta x \\
& = 2 \pi x (1- x^{2/3}) \Delta x \\
\end{aligned}
$$
Adding up all such volumes, infinitely many of them, starting at \(x = 0\) and going to \(x = 1\) is equal to the definite integral
$$
V = \int_0^1 2 \pi x (1- x^{2/3}) \; dx = \pi/4,
$$
which represents the volume of the solid of revolution.

E 7.3 Exercises