Unit 5, lesson 66

Volumes by Cylindrical Shells

How to find the volume of a solid of revolution by adding thin cylindrical shells, how to read the radius and height of a shell, and when shells are easier than washers.

  • Theory part 131
  • Practice part 132
  • Applications of Integration

Rotate the region under y=2x2−x3y = 2x^2 - x^3 about the yy-axis and try washers. You would need to solve a cubic for xx in terms of yy, which is painful. Instead, cut the region into thin vertical strips. Each strip, spun around the axis, sweeps out a thin hollow tube, like one layer of an onion or a paper towel roll. Adding those tubes gives the volume with no inverse functions needed.

Key terms

Cylindrical wall

Definition A cylindrical shell is the solid between two coaxial circular cylinders of radii r1<r2r_1 < r_2 and common height hh. Its volume is

V=πr22h−πr12h=2π⋅r1+r22⋅h⋅(r2−r1).V = \pi r_2^2 h - \pi r_1^2 h = 2\pi \cdot \frac{r_1 + r_2}{2} \cdot h \cdot (r_2 - r_1).

With average radius rˉ=(r1+r2)/2\bar r = (r_1 + r_2)/2 and thickness Δr=r2−r1\Delta r = r_2 - r_1, this is exactly

V=2πrˉ h Δr.V = 2\pi \bar r \, h \, \Delta r.

In plain words Cut the wall of a tube and unroll it. You get a nearly flat slab whose length is the circumference 2πrˉ2\pi\bar r, whose height is hh and whose thickness is Δr\Delta r.

Example With r1=2r_1 = 2, r2=3r_2 = 3 and h=5h = 5, the difference of cylinders gives π(9−4)⋅5=25π\pi(9 - 4) \cdot 5 = 25\pi, and the shell formula gives 2π⋅2.5⋅5⋅1=25π2\pi \cdot 2.5 \cdot 5 \cdot 1 = 25\pi.

Shell radius, height and thickness

Definition Let ff be continuous and nonnegative on [a,b][a, b] with 0≤a<b0 \le a < b. The solid obtained by rotating the region {(x,y)∣a≤x≤b, 0≤y≤f(x)}\{(x, y) \mid a \le x \le b, \ 0 \le y \le f(x)\} about the yy-axis has volume

V=∫ab2πx f(x) dx.V = \int_a^b 2\pi x \, f(x) \, dx.

It is the limit, as the widths Δxi\Delta x_i shrink to 00, of the sums of shell volumes

∑i=1n2πxˉif(xˉi) Δxi,\sum_{i=1}^n 2\pi \bar x_i f(\bar x_i) \, \Delta x_i,

where xˉi\bar x_i is the midpoint of the iith subinterval. The shell radius is xx, its distance from the axis. The shell height is f(x)f(x), or f(x)−g(x)f(x) - g(x) for a region between g≤fg \le f. The thickness is Δx\Delta x, which becomes dxdx in the integral.

In plain words For each vertical strip, ask how far it is from the axis (radius), how tall it is (height) and how wide it is (thickness). Multiply circumference by height by thickness and integrate.

Example Rotate the region under y=2x2−x3y = 2x^2 - x^3 on [0,2][0, 2] about the yy-axis. The radius is xx and the height is 2x2−x32x^2 - x^3, so

V=2π∫02(2x3−x4) dx=2π(8−325)=16π5.V = 2\pi\int_0^2 (2x^3 - x^4) \, dx = 2\pi\left(8 - \frac{32}{5}\right) = \frac{16\pi}{5}.

Axis of rotation

Definition For rotation about a vertical line x=kx = k of a region a≤x≤ba \le x \le b, g(x)≤y≤f(x)g(x) \le y \le f(x) that lies on one side of the line, the shell radius is ∣x−k∣\lvert x - k \rvert and

V=∫ab2π∣x−k∣(f(x)−g(x))dx.V = \int_a^b 2\pi \lvert x - k \rvert \left(f(x) - g(x)\right) dx.

For rotation about a horizontal line y=ky = k, the shells are horizontal. Use slices in yy, with radius ∣y−k∣\lvert y - k \rvert and height equal to the length of the horizontal strip, and integrate with respect to yy.

In plain words The radius is always the distance from the strip to the axis. Shells run parallel to the axis, so a vertical axis means vertical strips and dxdx, and a horizontal axis means horizontal strips and dydy.

Example Rotate the region under y=x2y = x^2 on [0,1][0, 1] about the line x=2x = 2. The radius is 2−x2 - x and the height is x2x^2, so

V=2π∫01(2−x)x2 dx=2π(23−14)=5π6.V = 2\pi\int_0^1 (2 - x) x^2 \, dx = 2\pi\left(\frac{2}{3} - \frac{1}{4}\right) = \frac{5\pi}{6}.

Choosing shells or washers

Definition For a given solid of revolution, slices perpendicular to the axis give disks or washers, and strips parallel to the axis give shells. Both methods give the same volume when they apply. The usual choice is the method whose strips are bounded by single formulas across the whole region and whose integrand is easier, often the one that avoids solving for the inverse function.

In plain words Washers stack across the axis, shells nest around it. Pick the one that needs fewer pieces and simpler algebra.

Example Rotate the region between y=xy = x and y=x2y = x^2 on [0,1][0, 1] about the yy-axis. With shells, radius xx and height x−x2x - x^2,

V=2π∫01x(x−x2) dx=2π(13−14)=π6.V = 2\pi\int_0^1 x(x - x^2) \, dx = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right) = \frac{\pi}{6}.

With washers in yy, the outer radius is y\sqrt{y} and the inner radius is yy, so

V=π∫01(y−y2) dy=π6V = \pi\int_0^1 (y - y^2) \, dy = \frac{\pi}{6}

as well.

Volume of a torus

Definition Let 0<r≤R0 < r \le R. The torus obtained by rotating the disk (x−R)2+y2≤r2(x - R)^2 + y^2 \le r^2 about the yy-axis has volume

V=2π2Rr2.V = 2\pi^2 R r^2.

With shells, the strip at xx has radius xx and height 2r2−(x−R)22\sqrt{r^2 - (x - R)^2}, so

V=∫R−rR+r4πxr2−(x−R)2 dx.V = \int_{R - r}^{R + r} 4\pi x \sqrt{r^2 - (x - R)^2} \, dx.

Substituting u=x−Ru = x - R splits this into two integrals,

V=4π∫−rrur2−u2 du+4πR∫−rrr2−u2 du=0+4πR⋅πr22.V = 4\pi\int_{-r}^r u\sqrt{r^2 - u^2} \, du + 4\pi R \int_{-r}^r \sqrt{r^2 - u^2} \, du = 0 + 4\pi R \cdot \frac{\pi r^2}{2}.

The first is zero because its integrand is odd, and the second is a half disk of area πr2/2\pi r^2/2.

In plain words A doughnut’s volume is the area of its circular cross section, πr2\pi r^2, times the distance 2πR2\pi R that the center of that circle travels.

Example A ring with R=3R = 3 cm and r=1r = 1 cm has volume 2π2⋅3⋅1=6π2≈59.222\pi^2 \cdot 3 \cdot 1 = 6\pi^2 \approx 59.22 cubic centimeters.

Avoiding double counting

Definition The shell formula for rotation about the yy-axis assumes the region lies on one side of the axis, here 0≤a0 \le a. If a region lies on both sides of the axis, the parts on each side sweep out solids that overlap, and the solid of revolution is the one swept by the union of the region’s right part and the mirror image of its left part. Its volume is at most the sum of the two separate volumes, and less whenever those two solids overlap, because adding them counts the overlap twice.

In plain words Before integrating, check that the region does not cross the axis. If it does, reflect everything to one side and keep only the outline that sticks out the farthest.

Example Rotate the region under y=1−x2y = 1 - x^2 on [−1,1][-1, 1] about the yy-axis. The left half is the mirror image of the right half, so it sweeps the same solid. The correct volume uses only [0,1][0, 1].

V=2π∫01x(1−x2) dx=2π(12−14)=π2.V = 2\pi\int_0^1 x(1 - x^2) \, dx = 2\pi\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{\pi}{2}.

The careless integral over the whole interval counts the same solid twice.

∫−112π∣x∣(1−x2) dx=π\int_{-1}^1 2\pi \lvert x \rvert (1 - x^2) \, dx = \pi

Check yourself

  1. Rotate the region under y=xy = \sqrt{x} on [0,4][0, 4] about the yy-axis. Find the volume.
Show the answer

The radius is xx and the height is x\sqrt{x}, so

V=2π∫04x⋅x1/2 dx=2π⋅25⋅32=128π5.V = 2\pi\int_0^4 x \cdot x^{1/2} \, dx = 2\pi \cdot \frac{2}{5} \cdot 32 = \frac{128\pi}{5}.
  1. The triangle with vertices (0,0)(0, 0), (1,0)(1, 0) and (1,1)(1, 1) is rotated about the xx-axis. Use shells to find the volume.
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Horizontal strips at height yy run from x=yx = y to x=1x = 1, so the radius is yy and the length is 1−y1 - y. Then

V=2π∫01y(1−y) dy=π3,V = 2\pi\int_0^1 y(1 - y) \, dy = \frac{\pi}{3},

the volume of a cone with radius 11 and height 11.

  1. Set up, without evaluating, the shell integral for rotating the region under y=e−xy = e^{-x} on [0,2][0, 2] about the line x=−1x = -1.
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The radius is x+1x + 1 and the height is e−xe^{-x}, so

V=∫022π(x+1)e−x dx.V = \int_0^2 2\pi (x + 1) e^{-x} \, dx.

How this lesson connects

It also uses cylindrical geometry.