Volumes Disk And Washer Methods

A solid of revolution is what you get when a plane region is spun around a fixed line, the axis of revolution. Slice the solid perpendicular to that axis and every slice is a circle: a full disk if the region touches the axis, a washer (a disk with a hole) if it does not. Integrating the area of that circular slice through the solid gives the volume.

Item	Statement
Element of volume, disk	$d V = π R^{2} d x$ , where $R$ is the radius of the slice
Element of volume, washer	$d V = π (R^{2} - r^{2}) d x$ , outer radius $R$ , inner radius $r$
Disk, about the $x$ -axis	$V = π \int_{a}^{b} [f (x)]^{2} d x$
Disk, about the $y$ -axis	$V = π \int_{c}^{d} [h (y)]^{2} d y$
Washer, about the $x$ -axis	$V = π \int_{a}^{b} ([f (x)]^{2} - [g (x)]^{2}) d x$
Washer, about the $y$ -axis	$V = π \int_{c}^{d} ([h_{out} (y)]^{2} - [h_{in} (y)]^{2}) d y$
About the line $y = k$	Radius is the vertical distance $\| y - k \|$ ; integrate in $x$
About the line $x = k$	Radius is the horizontal distance $\| x - k \|$ ; integrate in $y$
Rectangle orientation	The typical rectangle is drawn perpendicular to the axis of revolution
Variable of integration	$d x$ for a horizontal axis, $d y$ for a vertical axis
Cylinder volume	$π R^{2} h$ (base area $\times$ thickness)
Hollow cylinder volume	$π (R^{2} - r^{2}) h$
Sanity checks	Cone: $\frac{1}{3} π r^{2} h$ . Sphere: $\frac{4}{3} π r^{3}$

Solids of Revolution

In this section we learn two methods for computing the volumes of solids of revolution. A solid of revolution is a solid figure obtained by revolving a region in the plane around some straight line, called the axis of revolution.

Some examples of solids of revolution in engineering and manufacturing are pistons, axles, and funnels. Solids of revolution are also widely used in architecture and in the decorative arts: a vase turned on a potter's wheel, a column, a bell, and a wine glass are all surfaces swept out by a single profile curve. Renaissance draftsmen such as Paolo Uccello drew vases precisely this way, one profile revolved about a vertical axis.

Study of a vase as a solid of revolution by Paolo Uccello. 15th century.

Both methods of this section come from the same two-step idea used for areas in Section 8.1: write down the contribution $d V$ of one infinitesimally thin slice, then add up the slices with an integral.

The Disk Method

Consider the region bounded by the curve $y = f (x)$ , the $x$ -axis, and the vertical lines $x = a$ and $x = b$ , shown in the figure below.

The region bounded above by the curve y = f(x), below by the x-axis, and on the sides by the vertical lines x = a and x = b. — The region under $y = f (x)$ between $x = a$ and $x = b$ .

If this region is revolved around the $x$ -axis, it generates a three-dimensional body called a solid of revolution. The profile curve $y = f (x)$ sweeps out the curved surface, and the two vertical segments at $x = a$ and $x = b$ sweep out the flat circular ends.

The three-dimensional solid obtained by revolving the region under y = f(x) about the x-axis. — Revolving the region about the $x$ -axis generates a solid of revolution.

The figure below shows the region itself together with a typical infinitesimally thin rectangle of height $f (x)$ and width $d x$ .

The region under the curve with a typical thin vertical rectangle of height f(x) and width dx drawn inside it. — A typical infinitesimally thin rectangle of height $f (x)$ and width $d x$ .

When this rectangle is revolved, we obtain a circular disk in the form of a thin right circular cylinder with base radius $f (x)$ and thickness $d x$ . The volume of this disk, which is our element of volume $d V$ , is then

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{dV=\pi[f(x)]^{2}\,dx.}

Recall that the volume of a right circular cylinder of radius $R$ and height (or thickness) $h$ is area of the base $\times$ thickness $= π R^{2} h$ .

The thin circular disk swept out when the typical rectangle is revolved about the x-axis, with radius f(x) and thickness dx. — $d V = π [f (x)]^{2} d x$

The solid is composed of infinitely many infinitesimally thin disks like this. So its total volume is obtained by adding up all the elements of volume as the typical disk sweeps through the solid from left to right. That is,

V = \int_{a}^{b} d V = \int_{a}^{b} π [f (x)]^{2} d x .

This method of finding volume is called the disk method.

Volume of a Solid of Revolution: The Disk Method. If the region bounded by the curve $y = f (x)$ , the $x$ -axis, $x = a$ , and $x = b$ is revolved about the $x$ -axis, the volume of the solid generated this way is

V = π \int_{a}^{b} [f (x)]^{2} d x .

Revolving About the y-Axis

If the region bounded by the curve $x = h (y)$ , the $y$ -axis, and the horizontal lines $y = c$ and $y = d$ (with $c < d$ ) is revolved about the $y$ -axis, we compute the volume of the resulting solid in exactly the same way. The only change is bookkeeping: the typical rectangle is now horizontal, because it must stay perpendicular to the axis of revolution, its length is $h (y)$ , its thickness is $d y$ , and everything must be expressed in terms of $y$ .

A region bounded by the curve x = h(y) and the y-axis between y = c and y = d, revolved about the y-axis into a solid built from horizontal disks. — The element of volume is $d V = π [h (y)]^{2} d y$ and the total volume is $V = π \int_{c}^{d} [h (y)]^{2} d y = π \int_{c}^{d} x^{2} d y$ .

If the region bounded by the curve $x = h (y)$ , the $y$ -axis, $y = c$ , and $y = d$ is revolved about the $y$ -axis, then the element of volume is $d V = π [h (y)]^{2} d y$ and the total volume is

V = \int_{c}^{d} d V = π \int_{c}^{d} [h (y)]^{2} d y = π \int_{c}^{d} x^{2} d y .

The region under $y = 2 x$ on $[0, 1]$ is revolved about the $x$ -axis. Find the volume of the resulting solid.

The triangular region under the line y = 2x from x = 0 to x = 1, above the x-axis. — The region under $y = 2 x$ on $[0, 1]$ .

Solution

The element of volume is

\begin{aligned} dV &= \pi[f(x)]^{2}\,dx\\ &= \pi(2x)^{2}\,dx\\ &= 4\pi x^{2}\,dx. \end{aligned}

So the volume of the solid is

\begin{aligned} V &= \int dV\\ &= \int_{0}^{1}4\pi x^{2}\,dx\\ &= \left.4\pi\frac{x^{3}}{3}\right|_{0}^{1}=\frac{4\pi}{3}\approx 4.1888. \end{aligned}

This is the result we expected, because the solid is a cone whose radius is $r = 2$ and whose height is $h = 1$ :

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (2)^{2} (1) = \frac{4}{3} π .

The cone of radius 2 and height 1 swept out when the triangular region under y = 2x is revolved about the x-axis. — The solid is a cone of radius $r = 2$ and height $h = 1$ .

The region bounded by $y = 2 x$ , $y = 2$ , and $x = 0$ is rotated about the $y$ -axis. Compute the volume of the resulting solid.

The triangular region bounded by the line y = 2x, the horizontal line y = 2, and the y-axis. — The region bounded by $y = 2 x$ , $y = 2$ , and $x = 0$ .

Solution

The axis of revolution is vertical, so the typical rectangle is horizontal and the element of volume is

d V = π x^{2} d y .

To integrate, we have to express $x$ in terms of $y$ . From $y = 2 x$ we get $x = y / 2$ , so

d V = π {(\frac{y}{2})}^{2} d y = \frac{π}{4} y^{2} d y .

The rectangle sweeps from $y = 0$ up to $y = 2$ , so

\begin{aligned} V &= \int_{0}^{2}\pi\left(\frac{y}{2}\right)^{2}dy\\ &= \frac{\pi}{4}\int_{0}^{2}y^{2}\,dy\\ &= \left.\frac{\pi}{4}\cdot\frac{y^{3}}{3}\right|_{0}^{2}\\ &= \frac{\pi}{4}\cdot\frac{8}{3}=\frac{2\pi}{3}\approx 2.0944. \end{aligned}

Again the answer agrees with elementary geometry. The solid is a circular cone whose radius is $r = 1$ (the value of $x$ when $y = 2$ ) and whose height is $h = 2$ :

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (1)^{2} (2) = \frac{2}{3} π .

The cone of radius 1 and height 2 swept out when that triangular region is revolved about the y-axis. — The solid is a cone of radius $r = 1$ and height $h = 2$ .

Compute the volume of the solid obtained by revolving the region under the curve $y = x^{3}$ on $[1, 2]$ about the $x$ -axis.

The region under the curve y = x cubed between x = 1 and x = 2, above the x-axis. — The region under $y = x^{3}$ on $[1, 2]$ .

Solution

The element of volume is

\begin{aligned} dV &= \pi y^{2}\,dx\\ &= \pi[f(x)]^{2}\,dx\\ &= \pi\left(x^{3}\right)^{2}dx\\ &= \pi x^{6}\,dx, \end{aligned}

and the total volume is

\begin{aligned} V &= \int dV\\ &= \int_{1}^{2}\pi x^{6}\,dx\\ &= \left.\frac{\pi}{7}x^{7}\right|_{1}^{2}\\ &= \frac{\pi}{7}\left(2^{7}-1\right)=\frac{127\pi}{7}\approx 56.9975. \end{aligned}

The solid obtained by revolving the region under y = x cubed on the interval from 1 to 2 about the x-axis. — The solid generated by revolving the region about the $x$ -axis.

Compute the volume of the solid obtained by revolving the region bounded by $y = x^{3}$ , $y = 8$ , $y = 1$ , and the $y$ -axis about the $y$ -axis.

The region bounded by the curve y = x cubed, the y-axis, and the horizontal lines y = 1 and y = 8. — The region bounded by $y = x^{3}$ , $y = 1$ , and $y = 8$ , to be revolved about the $y$ -axis.

Solution

The axis is the $y$ -axis, so the slices are horizontal and the radius of a slice is the $x$ -coordinate of the curve. Solving $y = x^{3}$ for $x$ gives $x = y^{1 / 3}$ , and the element of volume is

\begin{aligned} dV &= \pi x^{2}\,dy\\ &= \pi\left(y^{1/3}\right)^{2}dy\\ &= \pi y^{2/3}\,dy. \end{aligned}

The total volume is

\begin{aligned} V &= \int dV\\ &= \int_{1}^{8}\pi y^{2/3}\,dy\\ &= \left.\pi\frac{3}{5}y^{5/3}\right|_{1}^{8}\\ &= \frac{3\pi}{5}\left(8^{5/3}-1\right)\\ &= \frac{3\pi}{5}\left(2^{5}-1\right)=\frac{93\pi}{5}\approx 58.4336. \end{aligned}

Here we used $8^{5 / 3} = {(8^{1 / 3})}^{5} = 2^{5} = 32$ .

The Washer Method

The washer method is a generalization of the disk method. Actually, a washer is a disk with a hole. Suppose $R$ is a region enclosed by the curves $y = f (x)$ and $y = g (x)$ (with $f (x) \geq g (x)$ ) on $[a, b]$ , and $R$ is revolved about the $x$ -axis.

The region enclosed between an upper curve y = f(x) and a lower curve y = g(x) from x = a to x = b, with a typical vertical rectangle. — The region between $y = f (x)$ and $y = g (x)$ on $[a, b]$ .

To compute the volume of this solid, consider an infinitely thin vertical rectangle. When this rectangle is revolved about the $x$ -axis, it does not sweep out a full disk, because it does not reach down to the axis: it generates an infinitely thin circular hollow cylinder whose outer radius is $f (x)$ and whose inner radius is $g (x)$ . The volume of this hollow cylinder, which is our element of volume, is

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{dV=\pi\left([f(x)]^{2}-[g(x)]^{2}\right)dx}

Recall that the volume of a right circular hollow cylinder of outer radius $R$ , inner radius $r$ , and height (or thickness) $h$ is the area of the base $\times$ thickness $= π (R^{2} - r^{2}) h$ .
A common error is to write $d V = π [f (x) - g (x)]^{2} d x$ . The square of the difference is not the difference of the squares: it is the two areas $π R^{2}$ and $π r^{2}$ that subtract, not the radii.

The washer swept out by the typical rectangle: a thin hollow cylinder of outer radius f(x), inner radius g(x), and thickness dx. — $d V = π ([f (x)]^{2} - [g (x)]^{2}) d x$

Adding up the washers gives the total volume

V = \int_{a}^{b} d V = π \int_{a}^{b} ([f (x)]^{2} - [g (x)]^{2}) d x .

In fact, the volume of this solid is the volume of the solid generated by revolving the region bounded by the curve $y = f (x)$ and the $x$ -axis minus the volume of the solid generated by revolving the region bounded by the curve $y = g (x)$ and the $x$ -axis. That is,

V = π \int_{a}^{b} [f (x)]^{2} d x - π \int_{a}^{b} [g (x)]^{2} d x .

The revolved region $g$ sweeps out is precisely the tunnel drilled through the middle of the bigger solid, so subtracting it leaves the hollow solid we want.

The hollow solid shown as the solid generated by the upper curve with the solid generated by the lower curve removed from its middle. — The washer solid is the big solid minus the small one.

Volume of a Solid of Revolution: The Washer Method. If the region between $y = f (x)$ and $y = g (x)$ on $[a, b]$ , where $f (x) \geq g (x) \geq 0$ , is revolved about the $x$ -axis, then

V = π \int_{a}^{b} ([f (x)]^{2} - [g (x)]^{2}) d x .

For a region between $x = h_{out} (y)$ and $x = h_{in} (y)$ on $[c, d]$ revolved about the $y$ -axis,

V = π \int_{c}^{d} ([h_{out} (y)]^{2} - [h_{in} (y)]^{2}) d y .

Let $R$ be the region in the first quadrant enclosed by the curve $y = 4 - x^{2}$ and the line $y = 3 x$ (bounded on the left by the $y$ -axis). If $R$ is revolved about the $x$ -axis, compute the volume of the resulting solid.

The first-quadrant region enclosed between the parabola y = 4 minus x squared and the line y = 3x. — The region in the first quadrant enclosed by $y = 4 - x^{2}$ and $y = 3 x$ .

Solution

The same first-quadrant region with a typical vertical rectangle running from the line up to the parabola, and the two curves crossing at x = 1. — The curves cross at $(1, 3)$ , so the slices run from $x = 0$ to $x = 1$ .

If a vertical rectangle is revolved about the $x$ -axis, it will generate a thin hollow disk with outer radius $4 - x^{2}$ and inner radius $3 x$ . The volume of the thin hollow disk (a washer) is

d V = π [(4 - x^{2})^{2} - (3 x)^{2}] d x .

To find the limits of the integral, we have to find where the curves $y = 4 - x^{2}$ and $y = 3 x$ intersect:

4 - x^{2} = 3 x \Rightarrow x^{2} + 3 x - 4 = 0

\Rightarrow (x + 4) (x - 1) = 0

\Rightarrow x = 1, x = - 4.

Because the point $(- 4, - 12)$ does not lie in the first quadrant, the limits of integration are $x = 0$ and $x = 1$ . Expanding the integrand first,

\begin{aligned} (4-x^{2})^{2}-(3x)^{2} &= 16-8x^{2}+x^{4}-9x^{2}\\ &= x^{4}-17x^{2}+16, \end{aligned}

so the total volume is

\begin{aligned} V &= \int_{0}^{1}dV\\ &= \pi\int_{0}^{1}\left[(4-x^{2})^{2}-(3x)^{2}\right]dx\\ &= \pi\int_{0}^{1}\left(x^{4}-17x^{2}+16\right)dx\\ &= \pi\left[\frac{x^{5}}{5}-\frac{17}{3}x^{3}+16x\right]_{x=0}^{x=1}\\ &= \pi\left[\frac{1}{5}-\frac{17}{3}+16\right]\\ &= \pi\cdot\frac{3-85+240}{15}\\ &= \frac{158\pi}{15}\approx 33.0914. \end{aligned}

Notice the two extreme slices. At $x = 0$ the inner radius is $0$ , so the washer there is a full disk of radius $4$ ; at $x = 1$ the two radii are equal ( $4 - 1 = 3$ and $3 \cdot 1 = 3$ ), so the washer has shrunk to a circle of zero area. Both are consistent with $x^{4} - 17 x^{2} + 16$ , which equals $16$ at $x = 0$ and $0$ at $x = 1$ .

Find the volume of the solid generated by revolving about the $x$ -axis the region in the first quadrant bounded between the circle $x^{2} + y^{2} = 9$ and the ellipse

\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1.

The first-quadrant region between the quarter circle x squared plus y squared equals 9 and the quarter ellipse x squared over 9 plus y squared over 4 equals 1. — The region between the circle $x^{2} + y^{2} = 9$ and the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ in the first quadrant.

Solution

If we revolve an infinitesimally thin vertical rectangle as shown in the figure below, it will generate a washer with outer radius $M P$ and inner radius $N P$ .

A vertical rectangle in the region, with its top point M on the circle, its bottom point N on the ellipse, and its foot P on the x-axis. — The washer has outer radius $M P$ and inner radius $N P$ .

To find the length of $M P$ , we solve $x^{2} + y^{2} = 9$ for $y$ (with $y > 0$ ):

x^{2} + y^{2} = 9 \Rightarrow y = \sqrt{9 - x^{2}}

∴ M P = \sqrt{9 - x^{2}} .

For the length of $N P$ , we solve $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ for $y$ (with $y > 0$ ):

\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \Rightarrow y^{2} = 4 (1 - \frac{x^{2}}{9})

\Rightarrow y = 2 \sqrt{1 - \frac{x^{2}}{9}}

∴ N P = 2 \sqrt{1 - \frac{x^{2}}{9}} .

Notice that only the squares of these radii are needed, so no square roots survive. The volume of the infinitesimally thin washer is

\begin{aligned} dV &= \pi\left(MP^{2}-NP^{2}\right)dx\\ &= \pi\left(9-x^{2}-4\left(1-\frac{x^{2}}{9}\right)\right)dx\\ &= \pi\left(9-x^{2}-4+\frac{4}{9}x^{2}\right)dx\\ &= \pi\left(5-\frac{5}{9}x^{2}\right)dx. \end{aligned}

Both curves meet the $x$ -axis at $x = 3$ , so the slices run from $x = 0$ to $x = 3$ and the total volume is

\begin{aligned} V &= \int_{0}^{3}dV\\ &= \pi\int_{0}^{3}\left(5-\frac{5}{9}x^{2}\right)dx\\ &= \pi\left[5x-\frac{5}{27}x^{3}\right]_{x=0}^{x=3}\\ &= \pi\left[15-\frac{5}{27}\cdot 27\right]\\ &= 10\pi\approx 31.4159. \end{aligned}

As a check, the full solid of revolution generated by the quarter disk is a hemisphere of radius $3$ , of volume $\frac{2}{3} π (3)^{3} = 18 π$ , and the quarter ellipse generates half an ellipsoid of volume $\frac{2}{3} π (3) (2)^{2} = 8 π$ . The difference is $18 π - 8 π = 10 π$ .

Revolving About a Line That Is Not a Coordinate Axis

Nothing in the derivation used the fact that the axis was a coordinate axis. All that matters is the distance from the axis to the near and far edges of the typical rectangle, because that distance is the radius of the circle traced by a point of the rectangle.

Rotation about $y = k$ (a horizontal line). Slice vertically. A point at height $y$ traces a circle of radius $| y - k |$ , so

V = π \int_{a}^{b} ({[f (x) - k]}^{2} - {[g (x) - k]}^{2}) d x,

where $y = f (x)$ is the edge of the region farther from the line and $y = g (x)$ the edge nearer to it.

Rotation about $x = k$ (a vertical line). Slice horizontally and integrate in $y$ , with radii measured as horizontal distances from $x = k$ .

In practice the only thing that changes is the radius: for a region lying above the line $y = k$ the radius of the far edge is $f (x) - k$ , and for a region lying below that line the radius is $k - f (x)$ . Everything else, the choice of variable, the limits, and the integral itself, works exactly as before. Exercise 8 below carries this out twice.

Exercises

In each problem, sketch the region first, decide whether the typical rectangle should be vertical or horizontal (it must be perpendicular to the axis of revolution), and then decide whether the slice is a disk or a washer.

The region bounded by $y = \sqrt{x}$ , the $x$ -axis, and the line $x = 4$ is revolved about the $x$ -axis. Find the volume of the solid generated.

Solution

The region touches the axis of revolution, so each slice is a full disk of radius $R = \sqrt{x}$ and thickness $d x$ :

d V = π {(\sqrt{x})}^{2} d x = π x d x .

Therefore

\begin{aligned} V &= \pi\int_{0}^{4}x\,dx\\ &= \left.\pi\frac{x^{2}}{2}\right|_{0}^{4}\\ &= \pi\cdot\frac{16}{2}=8\pi\approx 25.1327. \end{aligned}

Find the volume of the solid obtained by revolving the region under $y = \frac{1}{x}$ on $[1, 3]$ about the $x$ -axis.

Solution

The slice is a disk of radius $1 / x$ :

d V = π {(\frac{1}{x})}^{2} d x = π x^{- 2} d x .

Hence

\begin{aligned} V &= \pi\int_{1}^{3}x^{-2}\,dx\\ &= \pi\left[-\frac{1}{x}\right]_{1}^{3}\\ &= \pi\left(-\frac{1}{3}+1\right)=\frac{2\pi}{3}\approx 2.0944. \end{aligned}

Check the antiderivative by differentiating: $\frac{d}{d x} (- x^{- 1}) = x^{- 2}$ , as required.

The region bounded by the curve $x = y^{2}$ , the $y$ -axis, and the horizontal lines $y = 0$ and $y = 2$ is revolved about the $y$ -axis. Find the volume.

Solution

The axis is vertical, so the rectangle is horizontal, its length is the radius $x = y^{2}$ , and its thickness is $d y$ :

d V = π {(y^{2})}^{2} d y = π y^{4} d y .

Then

\begin{aligned} V &= \pi\int_{0}^{2}y^{4}\,dy\\ &= \left.\pi\frac{y^{5}}{5}\right|_{0}^{2}\\ &= \frac{32\pi}{5}\approx 20.1062. \end{aligned}

Note that the answer would be different if this same region were revolved about the $x$ -axis, because then the rectangle would be perpendicular to a horizontal line and the roles of the variables would swap.

Find the volume of the solid generated by revolving the region under one arch of $y = \sin x$ , that is on $[0, π]$ , about the $x$ -axis.

Solution

The element of volume is $d V = π \sin^{2} x d x$ . To integrate $\sin^{2} x$ , use the half-angle identity $\sin^{2} x = \frac{1 - \cos 2 x}{2}$ :

\begin{aligned} V &= \pi\int_{0}^{\pi}\sin^{2}x\,dx\\ &= \frac{\pi}{2}\int_{0}^{\pi}\left(1-\cos 2x\right)dx\\ &= \frac{\pi}{2}\left[x-\frac{\sin 2x}{2}\right]_{0}^{\pi}\\ &= \frac{\pi}{2}\left[\left(\pi-0\right)-\left(0-0\right)\right]\\ &= \frac{\pi^{2}}{2}\approx 4.9348. \end{aligned}

The squaring is essential: the area under the same arch is $\int_{0}^{π} \sin x d x = 2$ , which has nothing to do with this volume.

The region bounded by $y = 2 x$ and $y = x^{2}$ is revolved about the $x$ -axis. Find the volume of the solid generated.

Solution

First find the intersections: $2 x = x^{2}$ gives $x (x - 2) = 0$ , so $x = 0$ and $x = 2$ . On $[0, 2]$ the line lies above the parabola (at $x = 1$ , $2 > 1$ ), so the outer radius is $R = 2 x$ and the inner radius is $r = x^{2}$ :

d V = π [(2 x)^{2} - {(x^{2})}^{2}] d x = π (4 x^{2} - x^{4}) d x .

Therefore

\begin{aligned} V &= \pi\int_{0}^{2}\left(4x^{2}-x^{4}\right)dx\\ &= \pi\left[\frac{4x^{3}}{3}-\frac{x^{5}}{5}\right]_{0}^{2}\\ &= \pi\left[\frac{32}{3}-\frac{32}{5}\right]\\ &= 32\pi\cdot\frac{5-3}{15}=\frac{64\pi}{15}\approx 13.4041. \end{aligned}

Revolve the region of the previous exercise, bounded by $y = 2 x$ and $y = x^{2}$ , about the $y$ -axis instead, and find the volume.

Solution

The axis is now vertical, so slice horizontally and express both boundaries as functions of $y$ . The two curves meet at $(0, 0)$ and $(2, 4)$ , so $y$ runs from $0$ to $4$ . Solving,

y = x^{2} \Rightarrow x = \sqrt{y}, y = 2 x \Rightarrow x = \frac{y}{2} .

For $0 < y < 4$ we have $\sqrt{y} > \frac{y}{2}$ (at $y = 1$ , $1 > 0.5$ ), so the parabola is the outer boundary and the line is the inner one:

d V = π [{(\sqrt{y})}^{2} - {(\frac{y}{2})}^{2}] d y = π (y - \frac{y^{2}}{4}) d y .

Hence

\begin{aligned} V &= \pi\int_{0}^{4}\left(y-\frac{y^{2}}{4}\right)dy\\ &= \pi\left[\frac{y^{2}}{2}-\frac{y^{3}}{12}\right]_{0}^{4}\\ &= \pi\left[8-\frac{64}{12}\right]\\ &= \pi\left[8-\frac{16}{3}\right]=\frac{8\pi}{3}\approx 8.3776. \end{aligned}

The same region gives a different volume about a different axis, and the roles of "outer" and "inner" are exchanged.

The region bounded by $y = 4 - x^{2}$ and the $x$ -axis is revolved about the $x$ -axis. Find the volume of the resulting solid.

Solution

The parabola crosses the $x$ -axis where $4 - x^{2} = 0$ , that is at $x = \pm 2$ . Each slice is a disk of radius $4 - x^{2}$ :

d V = π {(4 - x^{2})}^{2} d x = π (16 - 8 x^{2} + x^{4}) d x .

Therefore

\begin{aligned} V &= \pi\int_{-2}^{2}\left(16-8x^{2}+x^{4}\right)dx\\ &= 2\pi\int_{0}^{2}\left(16-8x^{2}+x^{4}\right)dx\\ &= 2\pi\left[16x-\frac{8x^{3}}{3}+\frac{x^{5}}{5}\right]_{0}^{2}\\ &= 2\pi\left[32-\frac{64}{3}+\frac{32}{5}\right]\\ &= 2\pi\cdot\frac{480-320+96}{15}\\ &= \frac{512\pi}{15}\approx 107.2330. \end{aligned}

The integrand is even, which is why we could double the integral over $[0, 2]$ .

Revolve each of the following regions about the line $y = - 1$ and find the volume of the solid generated.

(a) the region bounded by $y = \sqrt{x}$ , the $x$ -axis, and $x = 4$ ;
(b) the region under $y = x^{2}$ on $[0, 2]$ .

Solution

In both parts the region lies above the line $y = - 1$ but does not touch it, so revolving the typical rectangle produces a washer, not a disk. Measure both radii from that line: the far edge of the rectangle is the curve and the near edge is the $x$ -axis, at distance $0 - (- 1) = 1$ .

(a) The outer radius is $\sqrt{x} - (- 1) = \sqrt{x} + 1$ and the inner radius is $1$ , so

\begin{aligned} dV &= \pi\left[\left(\sqrt{x}+1\right)^{2}-1^{2}\right]dx\\ &= \pi\left[x+2\sqrt{x}+1-1\right]dx\\ &= \pi\left(x+2x^{1/2}\right)dx. \end{aligned}

Therefore

\begin{aligned} V &= \pi\int_{0}^{4}\left(x+2x^{1/2}\right)dx\\ &= \pi\left[\frac{x^{2}}{2}+\frac{4}{3}x^{3/2}\right]_{0}^{4}\\ &= \pi\left[8+\frac{4}{3}\cdot 8\right]\\ &= \pi\left[8+\frac{32}{3}\right]=\frac{56\pi}{3}\approx 58.6431. \end{aligned}

Compare with Exercise 1, where the same region about the $x$ -axis gave only $8 π$ .

(b) Now the outer radius is $x^{2} - (- 1) = x^{2} + 1$ and the inner radius is again $1$ :

d V = π [{(x^{2} + 1)}^{2} - 1^{2}] d x = π (x^{4} + 2 x^{2}) d x .

Therefore

\begin{aligned} V &= \pi\int_{0}^{2}\left(x^{4}+2x^{2}\right)dx\\ &= \pi\left[\frac{x^{5}}{5}+\frac{2x^{3}}{3}\right]_{0}^{2}\\ &= \pi\left[\frac{32}{5}+\frac{16}{3}\right]\\ &= \pi\cdot\frac{96+80}{15}=\frac{176\pi}{15}\approx 36.8614. \end{aligned}

Had we carelessly revolved this region about the $x$ -axis instead, we would have found $π \int_{0}^{2} x^{4} d x = \frac{32 π}{5}$ , which is much smaller: moving the axis away from the region enlarges every radius.

Let $R$ be the region in the first quadrant bounded by $y = x^{2}$ , the line $y = 4$ , and the $y$ -axis. Find the volume of the solid obtained by revolving $R$ about the vertical line $x = 3$ .

Solution

The axis is vertical, so slice horizontally and integrate in $y$ , with $y$ running from $0$ to $4$ . The typical rectangle stretches from the $y$ -axis ( $x = 0$ ) to the curve $x = \sqrt{y}$ . Distances are measured from $x = 3$ :

outer radius: $R = 3 - 0 = 3$ (the far edge is the $y$ -axis),
inner radius: $r = 3 - \sqrt{y}$ (the near edge is the curve).

\begin{aligned} dV &= \pi\left[3^{2}-\left(3-\sqrt{y}\right)^{2}\right]dy\\ &= \pi\left[9-\left(9-6\sqrt{y}+y\right)\right]dy\\ &= \pi\left(6\sqrt{y}-y\right)dy, \end{aligned}

and

\begin{aligned} V &= \pi\int_{0}^{4}\left(6y^{1/2}-y\right)dy\\ &= \pi\left[6\cdot\frac{2}{3}y^{3/2}-\frac{y^{2}}{2}\right]_{0}^{4}\\ &= \pi\left[4\cdot 8-8\right]\\ &= 24\pi\approx 75.3982. \end{aligned}

Since $4^{3 / 2} = 8$ , the first bracketed term is $32$ . The solid is a cylinder of radius $3$ and height $4$ , of volume $π (3)^{2} (4) = 36 π$ , with the region's "shadow" solid of volume $12 π$ removed.

Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$ and $y = x$ about the $x$ -axis.

Solution

The curves meet where $\sqrt{x} = x$ , that is $x = x^{2}$ , so $x = 0$ and $x = 1$ . On $(0, 1)$ we have $\sqrt{x} > x$ (at $x = \frac{1}{4}$ , $\frac{1}{2} > \frac{1}{4}$ ), so $R = \sqrt{x}$ and $r = x$ :

d V = π [{(\sqrt{x})}^{2} - x^{2}] d x = π (x - x^{2}) d x .

Hence

\begin{aligned} V &= \pi\int_{0}^{1}\left(x-x^{2}\right)dx\\ &= \pi\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}\\ &= \pi\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi}{6}\approx 0.5236. \end{aligned}

The washer is degenerate at both ends: at $x = 0$ and $x = 1$ the two radii coincide, so those slices contribute nothing.

Let $r > 0$ be a constant. By revolving the region under the semicircle $y = \sqrt{r^{2} - x^{2}}$ , $- r \leq x \leq r$ , about the $x$ -axis, prove that the volume of a sphere of radius $r$ is $\frac{4}{3} π r^{3}$ . Then find the volume of the solid generated by revolving only the part of that region over $[0, \frac{r}{2}]$ .

Solution

Revolving the semicircular region about the $x$ -axis produces a solid ball of radius $r$ . Each slice is a disk of radius $y = \sqrt{r^{2} - x^{2}}$ , so

d V = π {(\sqrt{r^{2} - x^{2}})}^{2} d x = π (r^{2} - x^{2}) d x .

Since $r$ is a constant, integrate in $x$ from $- r$ to $r$ :

\begin{aligned} V &= \pi\int_{-r}^{r}\left(r^{2}-x^{2}\right)dx\\ &= 2\pi\int_{0}^{r}\left(r^{2}-x^{2}\right)dx\\ &= 2\pi\left[r^{2}x-\frac{x^{3}}{3}\right]_{0}^{r}\\ &= 2\pi\left(r^{3}-\frac{r^{3}}{3}\right)\\ &= 2\pi\cdot\frac{2r^{3}}{3}=\frac{4}{3}\pi r^{3}. \end{aligned}

For the second part, the same element of volume is integrated from $0$ to $r / 2$ :

\begin{aligned} V_{1} &= \pi\int_{0}^{r/2}\left(r^{2}-x^{2}\right)dx\\ &= \pi\left[r^{2}x-\frac{x^{3}}{3}\right]_{0}^{r/2}\\ &= \pi\left(\frac{r^{3}}{2}-\frac{r^{3}}{24}\right)\\ &= \pi r^{3}\cdot\frac{12-1}{24}=\frac{11\pi r^{3}}{24}. \end{aligned}

As a check, $\frac{11}{24} π r^{3}$ is more than half of the hemisphere's $\frac{2}{3} π r^{3} = \frac{16}{24} π r^{3}$ , which is right: the slab near $x = r$ is thin and contributes little.

A bowl is made by revolving the curve $y = x^{2}$ , $0 \leq x \leq 2$ , about the $y$ -axis; the inside of the bowl is the solid of revolution swept out by the region between the curve and the $y$ -axis. Find the volume of liquid the bowl holds when the liquid is $h$ units deep ( $0 \leq h \leq 4$ ), find the total capacity of the bowl, and find the depth at which the bowl is exactly half full.

Solution

The liquid is a solid of revolution about the $y$ -axis, so slice it horizontally. At height $y$ the radius of the circular cross-section is the $x$ -value on the curve, $x = \sqrt{y}$ , so

d V = π {(\sqrt{y})}^{2} d y = π y d y .

The liquid occupies heights $0 \leq y \leq h$ , so

V (h) = π \int_{0}^{h} y d y = {π \frac{y^{2}}{2} |}_{0}^{h} = \frac{π h^{2}}{2} .

The rim of the bowl is at $x = 2$ , that is $y = 4$ , so the total capacity is

V (4) = \frac{π \cdot 16}{2} = 8 π \approx 25.1327 .

The bowl is half full when $V (h) = 4 π$ :

\frac{π h^{2}}{2} = 4 π \Rightarrow h^{2} = 8 \Rightarrow h = 2 \sqrt{2} \approx 2.8284 .

The half-full level is well above the midpoint $h = 2$ , and that is exactly what the shape predicts: the bowl is narrow near the bottom and wide near the top, so the upper half of the height holds much more than the lower half.

Solids of Revolution

The Disk Method

Revolving About the y-Axis

The Washer Method

Revolving About a Line That Is Not a Coordinate Axis

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