Slice the region parallel to the axis, spin each strip into a thin tube, and add up $2\pi\,(\text{radius})(\text{height})(\text{thickness})$.
In Section 6.2 we built solids of revolution out of disks and washers: we sliced the solid perpendicular to the axis of revolution and added up the areas of circular slices. That works beautifully when each slice has a radius we can write down. But try revolving the region under $y=2x^2-x^3$ $(0\le x\le2)$ about the $y$-axis. A horizontal slice at height $y$ is a washer whose inner and outer radii are the two solutions of the cubic equation $2x^2-x^3=y$ โ and solving a cubic for $x$ in terms of $y$ is not something anyone wants to do by hand.
There is a second way to cut the same solid. Slice the region into thin strips parallel to the axis. When a vertical strip is spun around the $y$-axis it sweeps out a thin-walled tube โ a cylindrical shell, like one layer of an onion or one ring of a tree trunk. Each shell's volume is easy to write down, and adding the shells gives an integral in $x$ that uses $f(x)$ exactly as it is given. This is the shell method. It does not replace washers; it complements them, and for many solids it turns an awkward problem into a two-line calculation.
The Volume of a Cylindrical Shell
๐ Definition 6.4 โ Cylindrical shell
A cylindrical shell is the solid between two coaxial circular cylinders of the same height $h$, with inner radius $r_1$ and outer radius $r_2$. Its thickness is $\Delta r=r_2-r_1$ and its average radius is $r=\tfrac12(r_1+r_2)$. Its volume is
$$V_{\text{shell}}=\pi r_2^2h-\pi r_1^2h=\pi(r_2+r_1)(r_2-r_1)h=2\pi\,r\,h\,\Delta r .$$
The formula $V_{\text{shell}}=2\pi r h\,\Delta r$ is easy to remember if you imagine cutting the shell straight down one side and unrolling it flat. You get (very nearly) a rectangular slab whose length is the circumference $2\pi r$, whose height is $h$ and whose thickness is $\Delta r$:
$$\text{volume}\approx(\text{circumference})\times(\text{height})\times(\text{thickness})=2\pi r\cdot h\cdot\Delta r .$$
Remarkably, when $r$ is the average radius the formula is not an approximation at all โ the algebra above shows it is exact.
From shells to an integral
Let $f$ be continuous with $f(x)\ge0$ on $[a,b]$, where $0\le a\lt b$, and let $\mathcal R$ be the region under $y=f(x)$ above the $x$-axis. Partition $[a,b]$ into $n$ subintervals $[x_{i-1},x_i]$ of width $\Delta x$ and let $x_i^*=\tfrac12(x_{i-1}+x_i)$ be the midpoint. Revolving the rectangle of height $f(x_i^*)$ on $[x_{i-1},x_i]$ about the $y$-axis produces a shell of average radius $x_i^*$, height $f(x_i^*)$ and thickness $\Delta x$, so
$$V\approx\sum_{i=1}^{n}2\pi\,x_i^*\,f(x_i^*)\,\Delta x .$$
This is a Riemann sum for the continuous function $2\pi x f(x)$; as $n\to\infty$ it converges to an integral.
๐ Theorem 6.5 โ The shell method (revolving about the $y$-axis)
Let $f$ be continuous and nonnegative on $[a,b]$, where $0\le a\lt b$. The solid obtained by revolving the region under the graph of $f$ (above the $x$-axis, between $x=a$ and $x=b$) about the $y$-axis has volume
$$V=\int_a^b 2\pi\,x\,f(x)\,dx .$$
More generally, if $g$ is continuous and $g(x)\le f(x)$ on $[a,b]$, the region between the curves gives $V=\displaystyle\int_a^b2\pi\,x\,\big(f(x)-g(x)\big)\,dx$. Read it as $\displaystyle\int 2\pi\,(\text{radius})(\text{height})\,d(\text{thickness})$.
โ๏ธ Example 1 โ The cubic that washers cannot handle
Find the volume of the solid obtained by revolving the region bounded by $y=2x^2-x^3$ and $y=0$ about the $y$-axis.
๐ก Show solutionโถ
Region $2x^2-x^3=x^2(2-x)=0$ at $x=0$ and $x=2$, and $2x^2-x^3\ge0$ on $[0,2]$. So the region lies over $0\le x\le2$.
Shell A vertical strip at $x$ has radius $x$ (its distance from the $y$-axis), height $2x^2-x^3$ and thickness $dx$.
Check Interactive 1 below adds $n$ shells for this very region; with $n=24$ the sum is $10.068$, already within $0.02$ of $10.053$. With washers we would have needed both roots of $x^3-2x^2+y=0$ for every $y$ โ Interactive 2 does that numerically, and gets the same number.
Interactive 1 โ Nested Shells, Unrolled
The solid is built from $n$ nested shells (blue). One shell at a time (red) is cut down its front edge and unrolled into a flat slab of length $2\pi r$, height $h$ and thickness $\Delta x$ โ so its volume is $2\pi r h\,\Delta x$. Increase $n$ and watch the sum of the shells settle on the exact volume.
6
red shell: radius $r$โ
height $h=f(r)$โ
$2\pi r h\,\Delta x$โ
sum of $n$ shellsโ
exact volumeโ
โ๏ธ Example 2 โ The same solid computed both ways
The region bounded by $y=\sqrt x$, $y=0$ and $x=4$ is revolved about the $y$-axis. Find the volume (a) with shells and (b) with washers, and confirm that the answers agree.
(b) Washers Slice horizontally at height $y$, $0\le y\le2$. The strip runs from the curve $x=y^2$ to the line $x=4$, so the washer has outer radius $4$ and inner radius $y^2$: $$V=\int_0^2\pi\left(4^2-\left(y^2\right)^2\right)dy=\pi\left[16y-\frac{y^5}{5}\right]_0^2=\pi\left(32-\frac{32}{5}\right)=\frac{128\pi}{5}.$$
Check Both methods give $\dfrac{128\pi}{5}\approx80.425$. They are two different Riemann sums for the same volume, so they must agree โ computing a volume both ways is an excellent way to catch a wrong radius or a wrong height.
Revolving about Other Lines
Nothing in the shell argument needs the axis to be the $y$-axis. What matters is the distance from the strip to the axis (the radius) and the length of the strip (the height). If the axis is the vertical line $x=c$, a vertical strip at $x$ has radius $|x-c|$. If the axis is horizontal, we use horizontal strips, describe the region with $x$ as a function of $y$, and integrate in $y$.
๐ Theorem 6.6 โ Shells about any vertical or horizontal line
Vertical axis $x=c$. Let $f$ and $g$ be continuous on $[a,b]$ with $g(x)\le f(x)$, and suppose the axis does not cut through the region: $c\le a$ or $c\ge b$. Revolving the region between the curves about $x=c$ gives
$$V=\int_a^b2\pi\,|x-c|\,\big(f(x)-g(x)\big)\,dx .$$
Horizontal axis $y=k$. Let $u$ and $v$ be continuous on $[p,q]$ with $v(y)\le u(y)$ (right curve $x=u(y)$, left curve $x=v(y)$, as in Section 6.1), so the region is $\{(x,y):p\le y\le q,\ v(y)\le x\le u(y)\}$, and suppose $k\le p$ or $k\ge q$. Revolving the region about $y=k$ gives
$$V=\int_p^q2\pi\,|y-k|\,\big(u(y)-v(y)\big)\,dy .$$
In practice the absolute value is replaced by $x-c$ or $c-x$ (or $y-k$, $k-y$), whichever is nonnegative on the interval.
โ๏ธ Example 3 โ Revolving about the line $x=2$
The region between $y=x$ and $y=x^2$ is revolved about the vertical line $x=2$. Find the volume.
๐ก Show solutionโถ
Region The curves meet where $x^2=x$, i.e. $x=0$ and $x=1$; on $[0,1]$ the line is on top ($x\ge x^2$). The axis $x=2$ lies to the right of the region, so it does not cut through it.
Shell Radius $=2-x$ (distance from $x$ to $2$), height $=x-x^2$.
Check With washers (horizontal slices, $0\le y\le1$): at height $y$ the region runs from $x=y$ to $x=\sqrt y$. The point $x=y$ is farther from the axis, so the outer radius is $2-y$ and the inner radius is $2-\sqrt y$: $$\pi\int_0^1\left[(2-y)^2-\left(2-\sqrt y\right)^2\right]dy=\pi\int_0^1\left(4\sqrt y-5y+y^2\right)dy=\pi\left(\frac83-\frac52+\frac13\right)=\frac\pi2.\ โ$$
โ๏ธ Example 4 โ Horizontal shells about the line $y=-1$
The region bounded by $y=\sqrt x$, $y=0$ and $x=4$ is revolved about the horizontal line $y=-1$. Use shells to find the volume, then check with washers.
๐ก Show solutionโถ
Describe in $y$ The axis is horizontal, so shells come from horizontal strips. For $0\le y\le2$ a horizontal strip runs from $x=y^2$ (the curve) to $x=4$ (the line): length $4-y^2$.
Shell Radius $=y-(-1)=y+1$, height (length of strip) $=4-y^2$, thickness $dy$.
1. Radius measured from the wrong place. The radius is the distance to the axis of revolution, not to the $y$-axis. About $x=2$ a strip at $x$ has radius $2-x$, not $x$. About $x=-1$ it has radius $x+1$.
2. Wrong variable. Shells always use strips parallel to the axis. A horizontal axis means horizontal strips, so the integral is in $y$ and every length must be written in terms of $y$.
3. Height not "top minus bottom". For the region between two curves the height of a vertical strip is $f(x)-g(x)$, never just $f(x)$.
4. Squaring. Washers square the radii; shells do not. $2\pi r h$ is a lateral area (circumference times height), and it is multiplied by a thickness.
Shells or Washers?
Both methods compute the same number, so the choice is purely about convenience. The deciding question is: which description of the region is easier to integrate?
๐งญ Problem-solving strategy โ choosing a method
Sketch the region and the axis of revolution, and find where the boundary curves meet.
Draw a representative strip in each direction. A strip perpendicular to the axis gives a disk or washer; a strip parallel to the axis gives a shell.
Ask which strip has a single formula for its length over the whole interval. If a strip's ends switch from one curve to another part-way, that direction needs two (or more) integrals.
Ask whether you can solve for the variable you need. Washers about the $y$-axis need $x$ as a function of $y$; shells about the $y$-axis use $y=f(x)$ as given.
Set up: shells $\displaystyle\int2\pi(\text{radius})(\text{height})$, washers $\displaystyle\int\pi\left(R^2-r^2\right)$ ($R$ outer, $r$ inner radius), integrating in the variable that runs along the strip thickness.
Disks / washers
Shells
strip vs axis
perpendicular
parallel
axis = $y$-axis
integrate in $y$; need $x(y)$
integrate in $x$; use $y=f(x)$
axis = $x$-axis
integrate in $x$; use $y=f(x)$
integrate in $y$; need $x(y)$
element
$\pi\left(R^2-r^2\right)\,(\text{thickness})$
$2\pi r h\,(\text{thickness})$
Interactive 2 โ Washers versus Shells for the Same Solid
Choose a region and an axis (dashed). The red strip sweeps across the region; its mirror image across the axis shows the circle it traces. A strip perpendicular to the axis makes a washer, one parallel to the axis makes a shell. Switch methods: the running totals take different routes but always finish at the same volume.
strip positionโ
element $dV$ per unit thicknessโ
running totalโ
shells totalโ
washers totalโ
โ๏ธ Example 5 โ When shells are the only sensible choice
Find the volume of the solid formed by revolving the region under $y=\sin\left(x^2\right)$, $0\le x\le\sqrt\pi$, about the $y$-axis.
๐ก Show solutionโถ
Why shells On $[0,\sqrt\pi]$ we have $x^2\in[0,\pi]$, so $\sin(x^2)\ge0$. Washers would need $x$ in terms of $y$ on two branches ($x=\sqrt{\sin^{-1}y}$ and $x=\sqrt{\pi-\sin^{-1}y}$). With shells, the factor $x$ in $2\pi x$ is exactly what a substitution needs.
Set up $$V=\int_0^{\sqrt\pi}2\pi x\sin\left(x^2\right)dx .$$
Note The integral $\int\sin(x^2)\,dx$ has no elementary antiderivative, yet the volume is elementary โ the radius factor $x$ from the shell rescues us.
โ๏ธ Example 6 โ The napkin ring (an applied problem)
A cylindrical hole of radius $3$ cm is drilled straight through the centre of a wooden ball of radius $5$ cm. Find the volume of wood that remains.
๐ก Show solutionโถ
Model Put the centre of the ball at the origin and drill along the $y$-axis. The ball is the circle $x^2+y^2=25$ revolved about the $y$-axis. The remaining ring is generated by the part of the disk with $3\le x\le5$, whose vertical strip at $x$ runs from $y=-\sqrt{25-x^2}$ to $y=\sqrt{25-x^2}$.
Check The ring is $h=2\sqrt{25-9}=8$ cm tall. Repeating the calculation with ball radius $R$ and hole radius $r$ gives $\tfrac{4\pi}{3}\left(R^2-r^2\right)^{3/2}=\tfrac{\pi h^3}{6}$, which here is $\tfrac{512\pi}{6}=\tfrac{256\pi}{3}$ โ. Surprisingly, the volume depends only on the height of the ring, not on the size of the ball!
Exercises
I
Use the shell method to find the volume of the solid obtained by revolving the region about the $y$-axis. Each region lies above the $x$-axis and under the given curve unless stated otherwise.
1 $y=x^2$, $0\le x\le2$
2 $y=\dfrac1x$, $1\le x\le3$
3 $y=\sqrt{9-x^2}$, $0\le x\le3$ (what solid is this?)
4 $y=e^{-x^2}$, $0\le x\le1$
5 $y=\sin x$, $0\le x\le\pi$
6 $y=4x-x^2$, $0\le x\le4$
7 the region between $y=\sqrt x$ and $y=x$
8 $y=\dfrac{1}{1+x^2}$, $0\le x\le1$
9 $y=\ln x$, $1\le x\le e$
Check: the region, its shell integrand $2\pi x\,h(x)$, and a numerical value of the volume
the region (shaded)
shell integrand $2\pi x\,h(x)$; area under it $=V$
2 $V=\displaystyle\int_1^32\pi x\cdot\frac1x\,dx=2\pi\int_1^3dx=4\pi\approx12.566$. (Every shell has the same "lateral area" $2\pi$.)
3 $V=\displaystyle\int_0^32\pi x\sqrt{9-x^2}\,dx$. With $w=9-x^2$: $V=\pi\displaystyle\int_0^9w^{1/2}dw=\frac{2\pi}{3}\cdot27=18\pi\approx56.549$. This is a hemisphere of radius $3$: $\tfrac23\pi(3)^3=18\pi$ โ.
4 $V=\displaystyle\int_0^12\pi xe^{-x^2}dx$. With $w=x^2$: $V=\pi\displaystyle\int_0^1e^{-w}dw=\pi\left(1-e^{-1}\right)\approx1.986$.
5 $V=\displaystyle\int_0^{\pi}2\pi x\sin x\,dx$. By parts ($u=x$, $dv=\sin x\,dx$): $\displaystyle\int x\sin x\,dx=-x\cos x+\sin x$, so $V=2\pi\Big[-x\cos x+\sin x\Big]_0^{\pi}=2\pi\cdot\pi=2\pi^2\approx19.739$.
7 The curves meet at $x=0,1$, and $\sqrt x\ge x$ on $[0,1]$. $V=\displaystyle\int_0^12\pi x\left(\sqrt x-x\right)dx=2\pi\left(\frac25-\frac13\right)=\frac{2\pi}{15}\approx0.419$.
5 For $0\le y\le1$ the strip runs from $x=y^2$ to $x=1$: $V=\displaystyle\int_0^12\pi y\left(1-y^2\right)dy=2\pi\left(\frac12-\frac14\right)=\frac\pi2$. Disks: $\pi\displaystyle\int_0^1\left(\sqrt x\right)^2dx=\frac\pi2$ โ.
6 Here $x=y^{1/3}$, $0\le y\le8$; the strip runs from $x=0$ to $x=y^{1/3}$: $V=\displaystyle\int_0^82\pi y\cdot y^{1/3}dy=2\pi\cdot\frac37y^{7/3}\Big|_0^8=\frac{6\pi}{7}\cdot128=\frac{768\pi}{7}\approx344.68$. Washers: $\pi\displaystyle\int_0^2\left(64-x^6\right)dx=\pi\left(128-\frac{128}{7}\right)=\frac{768\pi}{7}$ โ.
8 For $0\le y\le1$ the strip runs from $x=y$ to $x=\sqrt y$; radius $1-y$: $V=\displaystyle\int_0^12\pi(1-y)\left(\sqrt y-y\right)dy=2\pi\left(\frac23-\frac12-\frac25+\frac13\right)=\frac\pi5\approx0.628$. Washers: outer radius $1-x^2$, inner $1-x$: $\pi\displaystyle\int_0^1\left[\left(1-x^2\right)^2-(1-x)^2\right]dx=\pi\int_0^1\left(2x-3x^2+x^4\right)dx=\frac\pi5$ โ.
III
Choosing a method. In parts 1โ3 compute the volume both by shells and by disks/washers. All solids are revolved about the $y$-axis.
1 Region under $y=1-x^2$, $0\le x\le1$.
2 Region under $y=x^3$, $0\le x\le1$.
3 Region under $y=e^x$, $0\le x\le1$. (Washers need two integrals here โ why?)
4 Region under $y=x(x-1)^2$, $0\le x\le1$. Explain why shells are the better choice, then compute.
5 Region between $y=x^2$ and $y=4$ with $x\ge0$. Compute both ways.
6Find the error. To revolve the region under $y=x$, $0\le x\le2$, about the line $x=3$, a student writes $V=\displaystyle\int_0^22\pi x\cdot x\,dx=\frac{16\pi}{3}$. Explain the mistake and give the correct volume.
๐ก Show solutionsโถ
1 Shells: $\displaystyle\int_0^12\pi x\left(1-x^2\right)dx=2\pi\left(\frac12-\frac14\right)=\frac\pi2$. Disks: $x^2=1-y$, so $\pi\displaystyle\int_0^1(1-y)\,dy=\frac\pi2$ โ.
3 Shells: $\displaystyle\int_0^12\pi xe^x\,dx=2\pi\Big[(x-1)e^x\Big]_0^1=2\pi\approx6.283$. Washers: for $0\le y\le1$ a horizontal slice is a full disk of radius $1$; for $1\le y\le e$ it is a washer with outer radius $1$ and inner radius $\ln y$ โ the inner boundary changes at $y=1$, hence two integrals: $\pi\cdot1+\pi\displaystyle\int_1^e\left(1-(\ln y)^2\right)dy=\pi+\pi\big[(e-1)-(e-2)\big]=2\pi$ โ (using $\int(\ln y)^2dy=y(\ln y)^2-2y\ln y+2y$).
4 Washers would require solving the cubic $x(x-1)^2=y$ for $x$ (two branches). Shells: $\displaystyle\int_0^12\pi x\cdot x(x-1)^2dx=2\pi\int_0^1\left(x^4-2x^3+x^2\right)dx=2\pi\left(\frac15-\frac12+\frac13\right)=\frac{\pi}{15}\approx0.209$.
6 The radius must be the distance to the axis $x=3$, which is $3-x$, not $x$. Correct: $V=\displaystyle\int_0^22\pi(3-x)x\,dx=2\pi\left(6-\frac83\right)=\frac{20\pi}{3}\approx20.944$.
IV
Applications and proofs.
1Napkin ring. A hole of radius $6$ cm is drilled through the centre of a sphere of radius $10$ cm. Find the remaining volume and check it with the formula $\tfrac{\pi h^3}{6}$.
2Inner tube. The disk $(x-3)^2+y^2\le1$ (units: decimetres) is revolved about the $y$-axis to make a torus. Find its volume in litres. (Hint: after $u=x-3$, one integral is half the area of a unit disk.)
3Sand pile. A pile of sand has height $y=2e^{-x^2}$ metres at horizontal distance $x$ m from its centre. Find the volume of sand within $2$ m of the centre, and its mass if sand has density $1600$ kg/mยณ.
4 For the pile in part 3, find the total volume $\displaystyle\int_0^{\infty}2\pi x\cdot2e^{-x^2}dx$ (an improper integral โ Section 5.5).
5Proof. Use shells to show that a right circular cone of base radius $r$ and height $h$ has volume $\tfrac13\pi r^2h$. (Revolve the region under $y=h\left(1-\tfrac xr\right)$, $0\le x\le r$, about the $y$-axis.)
๐ก Show solutionsโถ
1 $V=\displaystyle\int_6^{10}2\pi x\cdot2\sqrt{100-x^2}\,dx$; with $w=100-x^2$: $V=2\pi\displaystyle\int_0^{64}w^{1/2}dw=\frac{4\pi}{3}\cdot512=\frac{2048\pi}{3}\approx2144.7\text{ cm}^3$. Height $h=2\sqrt{100-36}=16$ cm and $\tfrac{\pi\cdot16^3}{6}=\tfrac{2048\pi}{3}$ โ.
2 Strip at $x\in[2,4]$ has height $2\sqrt{1-(x-3)^2}$: $V=\displaystyle\int_2^42\pi x\cdot2\sqrt{1-(x-3)^2}\,dx$. With $u=x-3$: $V=4\pi\displaystyle\int_{-1}^{1}(u+3)\sqrt{1-u^2}\,du$. The $u\sqrt{1-u^2}$ part is odd, so it integrates to $0$; $\displaystyle\int_{-1}^1\sqrt{1-u^2}\,du=\frac\pi2$ (half a unit disk). So $V=4\pi\cdot3\cdot\frac\pi2=6\pi^2\approx59.2\text{ dm}^3=59.2$ L.
3 $V=\displaystyle\int_0^22\pi x\cdot2e^{-x^2}dx=2\pi\Big[-e^{-x^2}\Big]_0^2=2\pi\left(1-e^{-4}\right)\approx6.168\text{ m}^3$. Mass $\approx1600\times6.168\approx9869$ kg (about $9.9$ tonnes).
4 $\displaystyle\int_0^{\infty}4\pi xe^{-x^2}dx=\lim_{t\to\infty}2\pi\left(1-e^{-t^2}\right)=2\pi\approx6.283\text{ m}^3$. The sand beyond $2$ m adds only $2\pi e^{-4}\approx0.115$ mยณ.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. How much material is in a stockpile when all you have is a drone survey of its height profile?
Why it matters. Ports, mines and farms store coal, salt, gravel and grain in roughly circular piles. Operators in Rotterdam, Port Hedland and Sohar audit these piles with drone or laser surveys every month, and a 5% volume error can be worth hundreds of thousands of dollars.
What to do.
Sample data for an approximately circular gravel pile (height $h$ at radial distance $r$ from the peak): $r=0,5,10,15,20,25,30$ m; $h=12,\ 11.5,\ 10,\ 7.5,\ 4.5,\ 1.8,\ 0$ m.
Explain why $V=\displaystyle\int_0^{30}2\pi r\,h(r)\,dr$ โ each ring of ground of radius $r$ and width $dr$ supports a shell of height $h(r)$.
Estimate $V$ with Simpson's rule (Section 5.4) using the seven data points.
Compare with the volume of a cone of the same base radius and peak height, $\tfrac13\pi(30)^2(12)$. Which is larger, and what does that say about the pile's shape?
With a bulk density of $1.6$ t/mยณ, estimate the mass of gravel.
Tools: Calculator or spreadsheet; optionally a free elevation-profile tool to collect your own data.
Hand in: A half-page note with your table of $2\pi r h$ values, the Simpson estimate, the cone comparison and one source of error in the survey.
Going further. Real piles are not perfectly circular. Take profiles in four directions, compute the volume for each, and average them โ or look up how surveyors use a grid of heights instead.
Instructor note: Simpson gives $V\approx2\pi\times2066.7\approx12\,985$ mยณ (about $20\,800$ t); the cone gives $3600\pi\approx11\,310$ mยณ. The pile is "fuller" than a cone because its shoulders are rounded.
๐งช Self-check quiz
0/5 correct
Q1. A thin cylindrical shell has average radius $r$, height $h$ and thickness $\Delta r$. Its volume is
Q2. The region under $y=f(x)$ on $[1,3]$ is revolved about the line $x=5$. The shell radius at $x$ is
Q3. The region under $y=x$, $0\le x\le1$, is revolved about the $y$-axis. Its volume is
Q4. To revolve a region about the $x$-axis using shells, you integrate with respect to
Q5. For which solid is the shell method clearly easier than washers?