Back to the on-screen lesson ·
A strip parallel to the axis sweeps a tube; unrolled it is a slab of volume $2\pi r h\,dr$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to draw a strip parallel to the axis of revolution and read its radius and height off the picture, assemble the shell integrand $2\pi r h$ with the first power of each rather than a square, write a radius as a positive distance when the axis is not a coordinate axis, keep the limits in the variable of the thickness, choose between shells and slices by which description of the region is single-valued, and recognise a solid whose volume only the shell integral can express.
The last lesson cut a solid of revolution across its axis, into plates. There is a second way to cut it: along the axis, into nested tubes, like the rings of a tree.
Neither is more correct. They give the same volume for the same solid, as they must. What differs is which variable the integrand is written in — plates across a vertical axis are functions of $y$, tubes around it are functions of $x$ — and so which description of the region you need. When one description is awkward and the other is not, the choice is made for you.
A shell (or cylindrical shell) is a thin tube: a strip of the region, swept once round the axis of revolution.
Its radius is the distance from the axis to the strip, its height is the length of the strip, and its thickness is $dx$ or $dy$. Those three, and the $2\pi$ that turns a radius into a circumference, are the whole integrand.
The strip is drawn parallel to the axis of revolution — that is what distinguishes the method from slicing, where the cut is perpendicular.
The formula. A strip at distance $r$ from the axis, of height $h$ and thickness $dr$, sweeps a shell of volume $$dV = 2\pi r h\,dr,$$ so $$V = 2\pi\int r\,h\,dr.$$ For the region under $y = f(x)$ on $[a,b]$ revolved about the $y$-axis, that is $$V = 2\pi\int_a^b x\,f(x)\,dx.$$
Why. Cut the tube down its side and flatten it. It becomes a rectangular slab: length $2\pi r$, height $h$, thickness $dr$. The flattening is not exact for a tube of finite thickness — the outside is longer than the inside — but the error is of order $(dr)^2$ and vanishes in the limit, which is what the Riemann sum argument makes precise.
When the axis moves. The radius is the distance from the axis of revolution to the strip. About $x = c$ with the region to the right, $r = x - c$; with the region to the left, $r = c - x$. A radius is a distance and must be positive on the region; if yours is negative, it is written backwards.
Choosing between shells and washers. Cut in the variable the region's boundary is naturally a function of. If $y = f(x)$ is single-valued and its inverse is not, cut in $x$ — which means washers about a horizontal axis and shells about a vertical one. If a region needs to be split one way and not the other, take the way that does not.
Both always work. If both are available, both give the same answer, and computing it twice is an excellent check.
Another way: picture
A tin can with no top or bottom, cut down one side and flattened on the bench. What was a tube of radius $r$ and height $h$ is now a rectangle $2\pi r$ long and $h$ tall, as thin as the metal. Its volume did not change in the flattening, and $2\pi r h\,dr$ is that rectangle's volume written out.
Another way: steps
Revolve the region under $y = x^2$ on $[0, 2]$ about the $y$-axis.
By shells. A vertical strip at $x$ has radius $x$ and height $x^2$: $$V = 2\pi\int_0^2 x\cdot x^2\,dx = 2\pi\cdot\frac{16}{4} = 8\pi.$$
By washers. Cut horizontally. At height $y$, the plate runs from the curve $x = \sqrt y$ out to $x = 2$, so it is a washer with outer radius $2$ and inner radius $\sqrt y$, for $0 \le y \le 4$: $$V = \pi\int_0^4\left(4 - y\right)dy = \pi\left(16 - 8\right) = 8\pi.$$
Same number, as it must be. Here both were easy because $y = x^2$ inverts cleanly on $[0,2]$. Replace the curve by $y = x^3 + x$ and the inverse has no formula at all — the washer integral cannot be written down, and the shell integral is still $2\pi\int_0^2 x(x^3 + x)dx$.
That asymmetry is the practical reason shells exist: the method that cuts in $x$ never needs to solve for $x$.
Using $f(x)^2$. That is the washer integrand. A shell has $x\,f(x)$, first power of each — the radius multiplies the height rather than replacing it.
Forgetting the $2\pi$. It is the circumference, and without it you have computed an area.
Drawing the strip the wrong way. Parallel to the axis for shells, perpendicular for slices. A strip perpendicular to the axis does not sweep a tube.
A negative radius. About $x = 5$ with the region on $[0,3]$, the radius is $5 - x$, not $x - 5$. Check that it is positive somewhere in the region.
Limits in the wrong variable. Shells about the $y$-axis integrate $dx$, so the limits are $x$-values, even though the axis is the $y$-axis. This is the step the method's name works hardest against.
Assuming shells are for vertical axes. They are for whichever direction the region is a function in. Revolving a region described by $x = g(y)$ about the $x$-axis also uses shells.
Shells are usually met second and feel like a more advanced method, to be used when the examiner insists. The formula $2\pi\int x f(x)\,dx$ is then memorised alongside $\pi\int f(x)^2 dx$ and the two get confused, because they are being held as formulas rather than as pictures.
They are not variants of one technique. They are two different cuts of the same solid, and each produces exactly the integrand its picture describes. A plate's volume is face area times thickness. A tube's volume is circumference times height times thickness. Neither has to be remembered by anyone who draws the slice.
The reason the distinction matters is practical and one-sided. A cut in $x$ needs the region described as a function of $x$; a cut in $y$ needs it as a function of $y$. Solving $y = f(x)$ for $x$ is often hard and sometimes impossible in closed form — while revolving about the $y$-axis is completely ordinary. So there are solids whose volume only the shell integral can express, and none where a shell is available and a washer would have been better than merely equal.
The habit worth building: draw the strip before choosing anything. Parallel to the axis gives tubes, perpendicular gives plates, and the picture tells you which description of the region you will need.
Revolve the region under $y = x^3 + x$ on $[0, 1]$ about the $y$-axis. A strip at $x$ has radius $x$ and height $x^3 + x$.
Strip parallel to the axis.
$V = 2\pi\displaystyle\int_0^1 x\left(x^3 + x\right)dx = 2\pi\int_0^1\left(x^4 + x^2\right)dx = 2\pi\left(\dfrac15 + \dfrac13\right) = \dfrac{16\pi}{15}$.
Radius times height, then integrate.
The washer version needs $x$ as a function of $y$, which means solving a cubic in $x$ for every $y$. It exists — the function is increasing, so the inverse is a genuine function — but it has no usable formula, so the integral cannot be written. The volume is perfectly ordinary; only one of the two methods can express it.
Revolve the region under $y = x^2$ on $[0,2]$ about the line $x = 3$. The strip at $x$ is to the left of the axis.
Mark the axis and note which side the region is on.
Radius $= 3 - x$, which is positive throughout $[0, 2]$, and it should be; height $= x^2$.
A distance, written the way round that makes it one.
$V = 2\pi\displaystyle\int_0^2 (3 - x)x^2\,dx = 2\pi\left(8 - 4\right) = 8\pi$. Only the radius changed from the $y$-axis case; the height and the limits did not.
Strip parallel to the axis: vertical, at $x$, radius $x$, height $\sqrt x$.
Radius from the axis, height along the strip.
$V = 2\pi\displaystyle\int_0^4 x\sqrt{x}\,dx = 2\pi\int_0^4 x^{3/2}dx$.
Radius times height, first power of each.
$= 2\pi\cdot\dfrac{2}{5}\cdot 4^{5/2} = \dfrac{128\pi}{5}$. The washer version — outer radius $4$, inner $y^2$, for $0 \le y \le 2$ — gives the same number, and is worth doing once.
The region under $y = 3x$ from $x = 0$ to $x = 3$ is revolved about the $y$-axis. The volume is $k\pi$. Find $k$.
Answer:
The region under $y = 4x - x^{2}$ above the $x$-axis is revolved about the $y$-axis. Shells or washers?
Put the steps of finding the volume when the region under $y = 7x$ is revolved about the $y$-axis into the order you do them.
Number the steps in order (write the number in the box):
The region under $y = 6x - x^{2}$ is revolved about the $y$-axis. Fill in the numbers for the shell at $x = 5$.
| Value | |
|---|---|
| The radius of the shell | 5 |
| The height of the shell | |
| Radius times height |
A vertical strip sits at $x$ in the region $0 \le x \le 3$. Match each axis of revolution to the radius of the shell it sweeps.
| radius $= x$ | radius $= x + 3$ | radius $= 3 - x$ | |
|---|---|---|---|
| About the $y$-axis | |||
| About the line $x = -3$ | |||
| About the line $x = 3$ |
The region between $y = 7x - x^{2}$ and the $x$-axis is revolved about the $y$-axis. The volume is $k\pi$. Find $k$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The region under $y = 4x$ from $x = 0$ to $x = 5$ is revolved about the $y$-axis. The volume is $k\pi$. Find $k$.
Answer:
You can cut a solid of revolution either way and say which cut the region's own description favours. Say in your own words why the shell integrand has a first power where the washer integrand has a square. Next: the length of a curve, which needs no revolution at all.
10. Your turn: the region under $y = \sqrt{x}$ on $[0, 4]$, revolved about the $y$-axis, step 3