Guides / Integrals

Volumes of revolution: disks, washers, and shells, and which one to use

Spin a flat region around a line and it sweeps out a solid. Its volume is an integral of thin slices, and there are only three kinds of slice. Most mistakes are about the radius: squared in the wrong order, measured from the wrong line, or written in the wrong variable.

Why it works

Cut the solid into thin slices perpendicular to the axis. Each slice is a flat disk, or a disk with a hole in it, and its volume is its area times its thickness. A disk of radius R has area πR². A washer is a big disk with a small disk taken out, so its area is πR² − πr². The integral adds up the slices.

Shells slice the other way. A thin strip of the region, parallel to the axis, sweeps out a thin cylindrical tube when it spins. Cut the tube open and flatten it: it's a slab 2π · radius long, as tall as the strip, and as thick as the strip is wide. Shells are the better choice when disks would need the curves solved for the other variable, or would split into pieces.

Example 1: disks

The region under y = x² from x = 0 to x = 2 is rotated about the x-axis. Find the volume.

1. R = x²The slice at x is a disk reaching from the axis up to the curve. The region touches the axis, so there's no hole.
2. V = ∫₀² π(x²)² dx = π∫₀² x⁴ dxπR², integrated across the region.
3. V = π · 325 = 32π5 ≈ 20.11[x⁵/5] from 0 to 2.

Example 2: washers

The region between y = 2x and y = x² is rotated about the x-axis. Find the volume.

1. 2x = x² at x = 0 and x = 2Where the curves meet. These are the limits.
2. R = 2x, r = x²At x = 1 the line is at 2 and the parabola at 1, so the line is farther from the axis.
3. V = π∫₀² (4x² − x⁴) dxπ(R² − r²): (2x)² = 4x² and (x²)² = x⁴.
4. V = π(323 − 325) = 64π15 ≈ 13.40160/15 − 96/15 = 64/15.

Example 3: shells

The same region, between y = 2x and y = x², is rotated about the y-axis. Find the volume.

1. height = 2x − x², radius = xA vertical strip at x runs from the parabola up to the line, and it's x away from the y-axis.
2. V = ∫₀² 2πx(2x − x²) dx = 2π∫₀² (2x² − x³) dxThe axis is vertical, so shells are integrated in x, with the same limits as before.
3. = 2π(163 − 4)∫ 2x² = 16/3 and ∫ x³ = 4, from 0 to 2.
4. V = 8π3 ≈ 8.38Washers in y agree: x runs from y/2 to √y for 0 ≤ y ≤ 4, and π∫₀⁴ (y − y²/4) dy = π(8 − 16/3) = 8π/3.

Example 4: an axis above the region

The same region is rotated about the line y = 5. Find the volume.

1. distance from y = 5 to a point at height y is 5 − yThe whole region is below the axis (its highest point is (2, 4)).
2. R = 5 − x², r = 5 − 2xThe parabola is the lower curve, so it's now the farther one from the axis. The line is the nearer one.
3. V = π∫₀² ((5 − x²)² − (5 − 2x)²) dx = π∫₀² (x⁴ − 14x² + 20x) dx(25 − 10x² + x⁴) − (25 − 20x + 4x²).
4. V = π(325 − 1123 + 40) = 136π15 ≈ 28.48(96 − 560 + 600)/15 = 136/15.

Common mistakes, and the exact line they happen on

1. Subtracting the radii before squaring

The washers in Example 2.

1. R = 2x, r = x²Right so far.
2. V = π∫₀² (2x − x²)² dx = 16π15 ≈ 3.35This is the mistake. π(R − r)² is the area of a disk as wide as the gap, not of a washer. A washer is a big disk minus a small one, π(R² − r²), and the volume is 64π/15 ≈ 13.40, four times as much. (a − b)² isn't a² − b².

2. Taking the top curve as the outer radius when the axis is above

The rotation about y = 5 in Example 4.

1. R = 5 − 2x, r = 5 − x², so V = −136π15This is the mistake. The line is the top curve, so it was used for the outer radius out of habit. With the axis above the region, the top curve is the one closer to the axis. A negative volume is the giveaway: measure both distances from y = 5, and the larger one, 5 − x², is R.

3. x-limits on a y-integral

The washer check in Example 3, integrated in y.

1. R = √y, r = y2Right so far: the radii are written in y.
2. V = π∫₀² (y − y²4) dy = 4π3This is the mistake. 0 to 2 is the range of x. The slices are stacked along the y-axis, from y = 0 up to y = 4, the height of the point (2, 4). With those limits the volume is 8π/3, matching the shells. The variable after d decides the limits and the radii.

Practice

1. The region under y = x³ from 0 to 1 is rotated about the x-axis. Find the volume.

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   Disks of radius x³: π∫₀¹ x⁶ dx = π/7 ≈ 0.449.

2. The region under y = e^x from 0 to 1 is rotated about the x-axis. Find the volume.

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   π∫₀¹ e^2x dx = π(e² − 1)/2 ≈ 10.04.

3. The region between y = √x and y = x is rotated about the x-axis. Find the volume.

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   They meet at 0 and 1, and √x is on top. π∫₀¹ (x − x²) dx = π/6 ≈ 0.524.

4. The region under y = 1/x from x = 1 to x = 3 is rotated about the y-axis. Find the volume with shells.

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   ∫₁³ 2πx · (1/x) dx = ∫₁³ 2π dx = 4π ≈ 12.57.

5. Use shells to find the volume when the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 is rotated about the y-axis, and check it against the formula for a cylinder.

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   ∫₀² 2πx · 3 dx = 3π · 4 = 12π. The solid is a cylinder of radius 2 and height 3, and π · 2² · 3 = 12π.