Difficulty: Intermediate | Prerequisites: Basic integration, area between curves
TL;DR
Volumes of revolution are found by rotating a flat region around an axis and computing the resulting 3D solid's volume with an integral. Two main methods exist: the disk/washer method (integrate cross-sectional areas perpendicular to the axis) and the shell method (integrate cylindrical shells parallel to the axis). Choosing the right method depends on which axis you revolve around and which variable makes the integrand simpler.
Solid of revolution
A three-dimensional shape created by rotating a two-dimensional region around a line (the axis of revolution). Think of it as spinning a flat cutout on a skewer and looking at the shape it traces out.
Axis of revolution
The line around which the region is rotated. Common choices are the x-axis, the y-axis, or a horizontal/vertical line such as y = 2 or x = 1. In simple terms, this is the "skewer" your region spins around.
Disk method
Used when the region touches the axis of revolution, producing solid circular cross-sections. The volume integral stacks thin disks of radius r(x) (or r(y)): V = integral of pi [r]^2 dx (or dy). Think of it as slicing the solid into thin coins and adding up their volumes.
Washer method
Used when the region does not touch the axis of revolution, so each cross-section is a ring (washer) with an outer radius R and an inner radius r. V = integral of pi (R^2 - r^2) dx (or dy). In simple terms, you are stacking rings instead of solid coins, subtracting the hole in the middle.
Shell method
Instead of slicing perpendicular to the axis, you peel off thin cylindrical shells parallel to the axis. V = integral of 2*pi * (shell radius) * (shell height) dx (or dy). Think of it as unrolling nested tubes, like the layers of a rolled-up poster.
Approximating rectangle
A small rectangle (width delta x or delta y) drawn inside the region to represent one "slice." Its orientation tells you which method you are setting up: perpendicular to the axis of revolution for disk/washer, parallel to it for shells.
Use disk/washer when your approximating rectangle is drawn perpendicular to the axis of revolution. If the region touches the axis, every cross-section is a full disk. If there is a gap between the region and the axis, every cross-section is a washer (a disk with a hole).
Identify the axis of revolution and the bounding curves.
Draw the approximating rectangle perpendicular to the axis.
If the axis is horizontal (e.g. y = 0, y = 2), integrate with respect to x. If vertical (e.g. x = 0, x = 1), integrate with respect to y.
Express the outer radius R and inner radius r as functions of the integration variable.
Determine the limits of integration from the intersection points of the bounding curves.
Region bounded by x = 0, y = sqrt(x), and y = 2, revolved about y = 2.
The approximating rectangle is vertical (width dx), perpendicular to the horizontal line y = 2. Since the region sits below y = 2 and above y = sqrt(x), the solid has no hole (the region touches the axis at y = 2), so this is a disk, not a washer.
Radius: r = 2 - sqrt(x) (distance from y = 2 down to the curve y = sqrt(x))
Limits: x runs from 0 to 4 (since y = sqrt(x) meets y = 2 at x = 4)
Integral: V = integral from 0 to 4 of pi (2 - sqrt(x))^2 dx
Result: V = 8*pi / 3
Region in the second quadrant bounded above by y = 9 - x^2, below by the x-axis, and on the right by the y-axis, revolved about x = 1.
The approximating rectangle is horizontal (height dy), perpendicular to the vertical line x = 1. Because the axis x = 1 does not touch the region (the region is in the second quadrant, to the left of x = 0), you get a washer.
Solving y = 9 - x^2 for x: x = -sqrt(9 - y) (taking the negative root for the second quadrant)
Outer radius: R = 1 - (-sqrt(9 - y)) = 1 + sqrt(9 - y)
Inner radius: r = 1 - 0 = 1 (distance from x = 1 to the y-axis)
Limits: y runs from 0 to 9
Integral: V = integral from 0 to 9 of pi [(1 + sqrt(9 - y))^2 - 1^2] dy
Result: V = 153*pi / 2
Use shells when your approximating rectangle is drawn parallel to the axis of revolution. This is often the easier choice when disk/washer would force you to solve for x in terms of y (or vice versa) in an awkward way.
Identify the axis of revolution and the bounding curves.
Draw the approximating rectangle parallel to the axis.
If the axis is vertical (e.g. the y-axis), integrate with respect to x. If horizontal, integrate with respect to y.
Shell radius = distance from the rectangle to the axis of revolution.
Shell height = length of the rectangle (top curve minus bottom curve, or right curve minus left curve).
V = integral of 2*pi * (radius) * (height) dx (or dy).
Region bounded by y = 5x^2 and y = 5*sqrt(x), revolved about the y-axis.
The curves intersect where 5x^2 = 5*sqrt(x), giving x = 0 and x = 1. The approximating rectangle is vertical (width dx), parallel to the y-axis.
Shell radius: x (distance from the rectangle to the y-axis)
Shell height: 5*sqrt(x) - 5x^2 (the upper curve minus the lower curve)
Limits: x from 0 to 1
Integral: V = integral from 0 to 1 of 2pi * x * (5sqrt(x) - 5x^2) dx
Result: V = 3*pi / 2
Region bounded by y = x^2 - 4, y = 3x, x = 0, for x >= 0, revolved about the y-axis.
The curves y = x^2 - 4 and y = 3x intersect where x^2 - 4 = 3x, i.e. x^2 - 3x - 4 = 0, giving (x - 4)(x + 1) = 0. Since x >= 0, the intersection is at x = 4. The approximating rectangle is vertical (width dx), parallel to the y-axis.
Shell radius: x
Shell height: 3x - (x^2 - 4) = 3x - x^2 + 4 (upper curve minus lower curve in the region)
Limits: x from 0 to 4
Integral: V = integral from 0 to 4 of 2*pi * x * (3x - x^2 + 4) dx
Result: V = 64*pi
Disk method (no hole)
V = integral of pi [r(x)]^2 dx, or pi [r(y)]^2 dy
Washer method (with hole)
V = integral of pi [R(x)^2 - r(x)^2] dx, or pi [R(y)^2 - r(y)^2] dy
where R = outer radius, r = inner radius.
Shell method
V = integral of 2*pi * (shell radius) * (shell height) dx, or dy
Choosing the variable of integration
Horizontal axis of revolution (e.g. x-axis, y = 2): disk/washer integrates in dx; shells integrate in dy.
Vertical axis of revolution (e.g. y-axis, x = 1): disk/washer integrates in dy; shells integrate in dx.
Students often confuse which radius is R (outer) and which is r (inner) in the washer method. The outer radius is always the one farther from the axis of revolution. Draw the axis, then measure outward.
When revolving about a line that is not an axis (e.g. y = 2 or x = 1), students sometimes forget to adjust the radius. The radius is the distance from the curve to the axis of revolution, not the value of the function itself. For y = 2 with the curve y = sqrt(x), the radius is 2 - sqrt(x), not sqrt(x).
Students mix up which variable to integrate with respect to. For disk/washer, the rectangle is perpendicular to the axis, so: horizontal axis means dx, vertical axis means dy. For shells it is the opposite.
A common error in shell problems is writing the height as a single function rather than the difference of two functions. The shell height is (top curve - bottom curve) or (right curve - left curve), not just one curve.
Volumes of revolution are among the most heavily tested topics on Calculus II exams. Expect at least one disk/washer and one shell problem.
You will almost always be told which method to use. Read the prompt carefully: "disk/washer method" vs. "shell method" is not optional.
Sketching the region and the approximating rectangle is often worth explicit marks. Even when it is not graded, the sketch prevents radius errors.
Showing the setup of the integral (with pi or 2*pi, the correct radii, and the correct limits) typically earns more partial credit than the final numerical answer.
Revolving about a non-standard axis (y = 2, x = 1, x = -3) is a favourite exam twist. Practise adjusting your radii for these.
True or false: In the disk method, the approximating rectangle is drawn parallel to the axis of revolution. (False, it is perpendicular.)
Fill in the blank: The washer method subtracts ____ from ____ inside the integral. (r^2 from R^2, i.e. the inner radius squared from the outer radius squared.)
True or false: When using the shell method about the y-axis, you integrate with respect to y. (False, you integrate with respect to x.)
Fill in the blank: The shell method formula includes a factor of ____ in front of the integral. (2*pi.)
True or false: If the region touches the axis of revolution, you use the washer method. (False, you use the disk method; washers are for when there is a gap.)
Q: The region bounded by y = x^2 and y = 4 is revolved about the x-axis. Using the disk/washer method, what is the integral for the volume?
A: The curves intersect at x = -2 and x = 2. The outer radius is R = 4, the inner radius is r = x^2. V = integral from -2 to 2 of pi (16 - x^4) dx.
Q: The region bounded by y = sqrt(x), y = 0, and x = 4 is revolved about the y-axis. Set up the volume integral using the shell method.
A: Shell radius = x, shell height = sqrt(x). V = integral from 0 to 4 of 2pi * x * sqrt(x) dx = integral from 0 to 4 of 2pi * x^(3/2) dx.
Q: When revolving the region bounded by y = x and y = x^2 about the line y = -1, what are the outer and inner radii for the washer method?
A: Outer radius R = x - (-1) = x + 1 (distance from y = x to y = -1). Inner radius r = x^2 - (-1) = x^2 + 1 (distance from y = x^2 to y = -1).
Q: A region is revolved about the y-axis. You are told to use the shell method. What variable do you integrate with respect to, and why?
A: You integrate with respect to x. In the shell method, the rectangle is parallel to the axis of revolution. With a vertical axis, the rectangle is vertical (width dx), so the variable of integration is x.
Volumes of revolution build directly on the concept of area between curves (the "height" or "radius" in each formula comes from the same curve-minus-curve logic). Arc length, covered separately, uses a similar integral structure but measures distance along a curve rather than volume. Work problems (also in this course) use the same slice-and-integrate strategy: divide a physical quantity into thin pieces, approximate each piece, sum, and take the limit.
Related Terms / Search Tags
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