# 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.