Volumes of Revolution: Disks, Washers and Shells
Set up a volume as an integral of cross-sections, and choose between slicing perpendicular to the axis and using cylindrical shells based on which integral is easier.
What a learner can do afterwards
- Write the integral for a solid generated by rotating a region about the x-axis
- Use washers when the solid has a hole and identify both radii
- Compare shell and disk or washer setups for the same solid, explaining which needs less rewriting or fewer pieces
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A volume integral adds cross-sectional area times slice thickness. Rotate 0 <= y <= f(x), a <= x <= b, about the x-axis, with f continuous and nonnegative. Vertical strips are perpendicular to the axis and make disks. Their radius is f(x), so V=π∫_a^b [f(x)]² dx. A thin slice contributes about π[f(x)]²Δx. If f(x)=c throughout, equal disks form a cylinder of radius c and length b-a.
Rotate 0 <= y <= √(2x), 0 <= x <= 3, about the x-axis. Squaring the radius gives 2x, so V=π∫_0^3 2x dx. An antiderivative of x is x²/2, so an antiderivative of 2x is x². Subtract endpoint values: V=π(3²-0²)=9π cubic units.
A perpendicular strip that misses the axis makes a washer. Its outer radius R and inner radius r satisfy 0 <= r <= R. Subtract circle areas: A=π(R²-r²). A disk has r=0. Square each radius before subtracting; (R-r)²=R²-2Rr+r² is a different expression. For constant radii, volume is washer area times the interval length. With R=6 and r=2, the area is π(36-4)=32π.
Rotate the region between y=3x and y=x², for 0 <= x <= 3, about the x-axis. Both curves are nonnegative, and 3x>=x² on this interval. Thus R=3x and r=x². The volume setup is V=π∫_0^3 [(3x)²-(x²)²] dx=π∫_0^3 (9x²-x⁴) dx. Each radius is a distance from the axis, not the gap between the curves.
Shells come from strips parallel to the axis. For a region on one side of the axis, a thin shell contributes about 2πrhΔt, where Δt is strip thickness. This is circumference times strip length times thickness. Add the contributions and take a limit. About the y-axis, vertical strips with x>=0 use r=x, h=top-bottom and V=2π∫ rh dx. About the x-axis, horizontal strips with y>=0 use r=y, h=right-left and V=2π∫ rh dy.
Rotate 0 <= y <= x³, 0 <= x <= 2, about the y-axis. Shells give V=2π∫_0^2 x⁴ dx=64π/5. Horizontal strips run from x=y^(1/3) to x=2, so they make washers: V=π∫_0^8 (4-y^(2/3)) dy=64π/5. Both methods describe the same solid. Shells keep x as the variable and avoid solving the curve for x; washers remain valid.
Rotate the triangle bounded by y=x, y=4-x and y=0 about the x-axis. Vertical disks need a split at x=2 because the upper boundary changes there. Horizontal shells need one integral. Solve the lines as x=y and x=4-y. For 0 <= y <= 2, radius is y and strip length is (4-y)-y=4-2y. Thus V=2π∫_0^2 y(4-2y) dy.
Radii are nonnegative distances from the axis. About y=k, the distance from height y is |y-k|; about x=c, it is |x-c|. For a perpendicular strip on one side of the axis, compare its endpoint distances to choose R and r. Below the x-axis, y=-5 is farther away than y=-2, so R=5 and r=2. If a perpendicular strip crosses the axis, its slice is a full disk with radius equal to the greatest distance reached.
Measure radii from the axis. Perpendicular slices use π(R²-r²); parallel shells use 2πrh. Keep the limits and formulas in the same integration variable.
Lesson, practice questions and flashcards written for this dataset.
Additional readings, practice questions and flashcards written for this dataset. Linked videos by Khan Academy.
More to read
What a volume integral adds
Start with a cross-section of the solid. Its area tells you how much material lies in one thin slice. Multiply that area by the slice thickness to estimate the slice's volume. Add the estimates, then take a limit as the slices become thinner.
For rotation about the x-axis, use vertical strips to make perpendicular slices. A segment from the axis to a nonnegative curve gives a disk. A segment that stays away from the axis gives a washer with a circular hole. The areas are πr² for a disk and π(R²-r²) for a washer.
The integration limits locate the first and last slices. Vertical strips use x-limits and dx. Horizontal strips use y-limits and dy. After integrating, subtract the antiderivative's value at the lower limit from its value at the upper limit.
A constant cross-section gives a useful check. A cylinder of radius 4 and length 1.5 has area 16π in every perpendicular section. Its volume is 16π·1.5=24π cubic units.
Comparing two descriptions of a solid
A region can give different kinds of slices depending on the strip direction. Perpendicular strips produce disks or washers. Parallel strips produce cylindrical shells.
For a shell, multiply circumference 2πr by strip length h and thickness. Vertical strips about the y-axis use a vertical length for h. Horizontal strips about the x-axis use a horizontal length, found by subtracting the left x-boundary from the right x-boundary.
Write both complete setups before choosing between them. Check the radius, any inner radius, the strip length and the limits. Shells can use a curve written as y=f(x) directly when perpendicular slices would require solving for x. Horizontal shells can also avoid splitting an integral where a vertical strip changes its upper boundary.
Neither method removes the need to describe the region accurately. Horizontal strips may leave a hole even when some part of the region touches the rotation axis. When two valid setups cover the same solid without overlap, their volumes agree. Compare the work needed to form and evaluate the integrals.
Checking the radii
A radius measures distance from the rotation axis. Its value cannot be negative. About the x-axis, a boundary at y=-6 is 6 units away. About y=2, a boundary at y=5 is 3 units away. Write these as |y| and |y-2| when the boundary varies.
For a perpendicular strip that stays on one side of the axis, compare the distances to its endpoints. The larger distance is the outer radius R, and the smaller is the inner radius r. If the strip crosses the axis, there is no central hole: use a full disk reaching the farthest endpoint.
Washer area subtracts two circle areas. For radii 8 and 5, it is π(64-25)=39π. The difference of the radii is 3, but π·3² would be the area of a different circle.
Finally check the differential and limits together. A radius written in terms of y belongs in a dy integral with y-limits. If a boundary changes within the interval, describe the pieces separately before adding their volumes.
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