We use the terms “revolve” and “rotate” interchangeably.

This method for computing the volume is referred to as the disk method because the vertical slices of the solid are circular disks.

The cross sections of a solid of revolution are circles of radius \(R = f(x)\) and area \(\pi R^2 = \pi [f(x)]^2\). The volume, given by Eq. (1), is the integral of cross-sectional area.

Volume of Revolution: Disk Method

If \(f(x)\) is continuous and \(f(x)\ge 0\) on \([a,b]\), then the solid obtained by rotating the region under the graph about the \(x\)-axis has volume [with \(R=f(x)\)] \[ \begin{equation} \boxed{\bbox[#FAF8ED,5pt]{V = \pi \int_a^b R^2\, dx= \pi \int_a^b [f(x)]^2\, dx}} \tag {1} \end{equation} \]

EXAMPLE 1

Calculate the volume \(V\) of the solid obtained by rotating the region under \(y = x^2\) about the \(x\)-axis for \(0\le x \le 2\).

Figure 6.26: Region under \(y=x^2\) rotated about the \(x\)-axis.

Solution The solid is shown in Figure 6.26. By Eq. (1) with \(f(x) = x^2\), its volume is \[ V = \pi\int_0^2 R^2\, dx = \pi\int_0^2 (x^2)^2\, dx = \pi\int_0^2 x^4\, dx = \pi\,\frac{x^5}{5}\bigg|_0^2 = \pi\frac{2^5}{5} = \frac{32}5\,\pi \]

EXAMPLE 2 Region Between Two Curves

Find the volume \(V\) obtained by revolving the region between \(y=x^2+4\) and \(y = 2\) about the \(x\)-axis for \(1 \leq x \leq 3\).

Solution The graph of \(y=x^2+4\) lies above the graph of \(y = 2\) (Figure 6.29). Therefore, \(R_{\textrm{outer}}=x^2+4\) and \(R_{\textrm{inner}}=2\). By Eq. (2), \[ \begin{align*} V &=\pi\int_1^3 \big(R_{\textrm{outer}}^2-R_{\textrm{inner}}^2\big) \, dx = \pi\int_1^3 \big((x^2+4)^2-2^2\big) \,dx\\ &= \pi\int_1^3 \big(x^4+8x^2+12\big) \, dx =\pi \left(\frac15 x ^5+ \frac 83 x^3+12x \right)\bigg|_1^3 = \frac{2126}{15}\,\pi \end{align*} \]

EXAMPLE 3 Revolving About a Horizontal Axis

Find the volume \(V\) of the “wedding band” [Figure 6.30(C)] obtained by rotating the region between the graphs of \(f(x)= x^2+ 2\) and \(g(x) = 4-x^2\) about the horizontal line \(y=-3\).

Solution First, let's find the points of intersection of the two graphs by solving \[ f(x) = g(x)\quad\Rightarrow\quad x^2+2=4-x^2 \quad\Rightarrow\quad x^2=1\quad\Rightarrow\quad x=\pm 1 \] Figure 6.30(A) shows that \(g(x)\ge f(x)\) for \(-1\le x\le 1\).

If we wanted to revolve about the \(x\)-axis, we would use Eq. (2). Since we want to revolve around \(y=-3\), we must determine how the radii are affected. Figure 6.30(B) shows that when we rotate about \(y = -3\), \(\overline{AB}\) generates a washer whose outer and inner radii are both \(3\) units longer:

• \(R_{\textrm{outer}}=g(x)- (-3)=(4 -x^2) + 3 = 7-x^2\)
• \(R_{\textrm{inner}} =f(x)- (-3)= (x^2 +2) + 3 =x^2+5\)

The volume of revolution is equal to the integral of the area of this washer: \[ \begin{align*} \textrm{\(V\) (about \(y=-3\))}&= \pi\int_{-1}^1\big(R_{\textrm{outer}}^2 - R_{\textrm{inner}}^2\big)\, dx= \pi \int_{-1}^1 \Bigl((g(x)+3)^2- (f(x)+3)^2\Bigr)\,dx\\ & = \pi \int_{-1}^1\big((7-x^2)^2-(x^2+5)^2 \big)\,dx\\ &= \pi \int_{-1}^1\big((49-14x^2+x^4) - (x^4+10x^2+25)\big)\,dx\\ & = \pi\int_{-1}^1(24-24x^2)\,dx =\pi(24x-8x^3 )\Big|_{-1}^1 = 32\pi \end{align*} \]

EXAMPLE 4

Find the volume obtained by rotating the graphs of \(f(x) = 9 - x^2\) and \(y = 12\) for \(0 \leq x \leq 3\) about

1. the line \(y=12\)
2. the line \(y = 15\).

Solution To set up the integrals, let's visualize the cross section. Is it a disk or a washer?

1. Figure 6.31(B) shows that \(\overline{AB}\) rotated about \(y=12\) generates a disk of radius \[ R = \textrm{length of \(\overline{AB}\)} = 12 - f(x) = 12 - (9-x^2)=3+x^2 \]

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   Figure 6.31: Segment \(\overline{AB}\) generates a disk when rotated about \(y=12\), but it generates a washer when rotated about \(y=15\).

   The volume when we rotate about \(y=12\) is \[ \begin{align*} V = \pi\int_0^3 R^2\, dx &=\pi\int_0^3(3+x^2)^2\, dx=\pi\int_0^3(9+6x^2+x^4)\, dx \\ &=\pi \left(9x + 2 x^3 + \frac15x^5\right)\bigg|_0^3 = \frac{648}5\,\pi \end{align*} \]

2. Figure 6.31(C) shows that \(\overline{AB}\) rotated about \(y=15\) generates a washer. The outer radius of this washer is the distance from \(B\) to the line \(y=15\): \[ R_{\textrm{outer}} = 15-f(x)=15-(9-x^2)=6+x^2 \] The inner radius is \(R_{\textrm{inner}}=3\), so the volume of revolution about \(y=15\) is \[ \begin{align*} V = \pi\int_0^3 \big(R_{\textrm{outer}}^2-R_{\textrm{inner}}^2\big)\,dx &= \pi\int_0^3 \big((6+x^2)^2 - 3^2\big)\, dx\\ & = \pi\int_0^3 (36 +12 x^2 + x^4 - 9)\, dx\\ &=\pi\left(27x +4x^3 + \frac15x^5\right)\bigg|_0^3 = \frac{1188}5\,\pi \end{align*} \]

EXAMPLE 5 Revolving About a Vertical Axis

Find the volume of the solid obtained by rotating the region under the graph of \(f(x)=9-x^2\) for \(0 \leq x \leq 3\) about the vertical axis \(x=-2\).

Solution Figure 6.32 shows that \(\overline{AB}\) sweeps out a horizontal washer when rotated about the vertical line \(x=-2\). We are going to integrate with respect to \(y\), so we need the inner and outer radii of this washer as functions of \(y\). Solving for \(x\) in \(y=9-x^2\), we obtain \(x^2=9-y\), or \(x=\sqrt{9-y}\) (since \(x \geq 0\)). Therefore, \[ \begin{align*} R_{\textrm{outer}} &= \sqrt{9-y}+2,\qquad R_{\textrm{inner}}= 2\\ R_{\textrm{outer}}^2 - R_{\textrm{inner}}^2 &= \big(\sqrt{9-y}+2\big)^2 - 2^2 = (9-y)+4\sqrt{9-y}+4-4\\& = 9-y +4\sqrt{9-y} \end{align*} \] The region extends from \(y=0\) to \(y=9\) along the \(y\)-axis, so \[ \begin{align*} V = \pi\int_0^9 \big(R_{\textrm{outer}}^2-R_{\textrm{inner}}^2\big) \, dy &= \pi\int_0^9 \Big({9-y}+4\sqrt{9-y}\Big)\,dy\\ &= \pi \left(9y-\frac12y^2 -\frac83(9-y)^{3/2}\right)\bigg|_0^9=\frac{225}2\,\pi \end{align*} \]