Source: MasterMathMentor Ch. 37, Advanced Calculus I (UCF)
Tags: volumes of revolution, disk method, washer method, solids of revolution, cross-sectional area, Riemann sum, definite integral, rotation about axis, dx vs dy integration
Difficulty: Intermediate Prerequisites: Definite integrals, basic integration techniques, area between curves, graphing functions. You should be comfortable setting up and evaluating integrals before tackling these problems.
This topic takes the definite integral and extends it from finding flat areas to computing three-dimensional volumes. When you rotate a region in the xy-plane around a line (an axis of revolution), you get a solid whose volume can be calculated by slicing it into thin disks or washers. This is one of the major practical payoffs of integration and appears heavily on AP Calculus exams. You should already be comfortable with setting up definite integrals for area problems, because the logic here is identical: slice, approximate, sum, take the limit.
Rotating a region around a line creates a 3D solid. Slice that solid perpendicular to the axis of rotation to get circles (disks) or rings (washers). Integrate the cross-sectional area along the axis to find the volume.
Disk
A solid circular cross-section obtained by slicing a solid of revolution perpendicular to the axis of rotation. Think of it as a very thin coin or cylinder. It arises when the region touches the axis of rotation, so there is no hole in the middle.
Washer
A cross-section shaped like a ring (annulus), with an outer radius R and an inner radius r. Think of it as a disk with a hole punched through the centre. It arises when the region does not touch the axis of rotation, creating a gap.
Outer radius (R)
The distance from the axis of rotation to the farthest boundary of the region. This is always the larger radius in a washer problem.
Inner radius (r)
The distance from the axis of rotation to the nearest boundary of the region. This is the radius of the hole in a washer.
Axis of revolution
The line about which the region is rotated. Can be the x-axis, y-axis, or any horizontal or vertical line.
Solid of revolution
The 3D object formed when a 2D region is rotated 360 degrees about an axis.
If there is only one curve between the axis of rotation and the boundary of the region, it is a disk problem. You need one radius, R.
If there are two curves (the region floats away from the axis of rotation), it is a washer problem. You need both R (outer) and r (inner).
A good visual test: draw the axis of rotation and colour the boundaries of the region. If you cross only one coloured boundary going from the axis outward, it is a disk. If you cross two, it is a washer.
Rotation about a horizontal line (the x-axis or any line y = c): integrate with respect to x (dx problem). All expressions for R and r must be in terms of x.
Rotation about a vertical line (the y-axis or any line x = c): integrate with respect to y (dy problem). All expressions for R and r must be in terms of y.
This rule is non-negotiable. If you set up the wrong variable, the integral will be incorrect.
Label a representative point (x, y) on the relevant curve.
For a dx problem, the vertical distance from the axis to the curve gives R (or r). Since y is a vertical distance, express it in terms of x.
For a dy problem, the horizontal distance from the axis to the curve gives R (or r). Since x is a horizontal distance, express it in terms of y.
When rotating about a line that is not the axis itself (e.g. y = 4 instead of y = 0), the radius is the distance from that line to the curve, not just y or x. Calculate it as a difference.
Disk: the cross-sectional area is pi R^2, so the volume is the integral of pi R^2 with respect to the appropriate variable over the relevant interval.
Washer: the cross-sectional area is pi (R^2 - r^2), so the volume is the integral of pi (R^2 - r^2) with respect to the appropriate variable.
Limits of integration come from the region's boundaries along the axis you are integrating.
Disk formula (horizontal rotation, dx):
V = pi * integral from a to b of [R(x)]^2 dx
Disk formula (vertical rotation, dy):
V = pi * integral from c to d of [R(y)]^2 dy
Washer formula (horizontal rotation, dx):
V = pi * integral from a to b of ([R(x)]^2 - [r(x)]^2) dx
Washer formula (vertical rotation, dy):
V = pi * integral from c to d of ([R(y)]^2 - [r(y)]^2) dy
This single region is rotated about four different axes to show how the setup changes each time.
a) Rotation about the x-axis (disk, dx) R = y = x^2. Limits: x from 0 to 2. V = pi integral from 0 to 2 of (x^2)^2 dx = pi integral of x^4 dx = 32pi/5.
b) Rotation about y = 4 (washer, dx) R = 4 (distance from y = 4 down to y = 0). r = 4 - y = 4 - x^2. V = pi * integral from 0 to 2 of [16 - (4 - x^2)^2] dx = 224pi/15.
c) Rotation about the y-axis (washer, dy) R = 2 (the vertical line x = 2). r = x = sqrt(y). Limits: y from 0 to 4. V = pi * integral from 0 to 4 of [4 - y] dy = 8pi.
d) Rotation about x = 2 (disk, dy) R = 2 - x = 2 - sqrt(y). Limits: y from 0 to 4. V = pi * integral from 0 to 4 of (2 - sqrt(y))^2 dy = 8pi/3.
Not all volume problems involve rotation. Some give you a base region and tell you the cross-sections perpendicular to an axis are squares, equilateral triangles, isosceles right triangles, or semicircles.
Find the length of the side (or diameter) of the cross-section at a general point. For a base bounded by a circle x^2 + y^2 = r^2, the half-width at position x is y = sqrt(r^2 - x^2), so the full width is 2y.
Compute the area of the cross-sectional shape in terms of that length.
Integrate the area along the axis.
Square: area = s^2 = 4y^2
Equilateral triangle: area = (sqrt(3)/4) s^2 = y^2 sqrt(3)
Isosceles right triangle (legs = s, hypotenuse along the base): area = (1/2) s (s/2) = y^2. (Note: when the base of the triangle is 2y and the equal legs meet at a right angle, the height equals y.)
Semicircle: area = (1/2) pi (s/2)^2 = (pi/2) * y^2
y^2 = 16 - x^2, so area = 4(16 - x^2). Integrate from -4 to 4 (or double the integral from 0 to 4 by symmetry): V = 8 * integral from 0 to 4 of (16 - x^2) dx = 1024/3.
These problems appear at the end of the chapter and on exams. They connect calculus back to formulas you already know.
Sphere: rotate the upper semicircle y = sqrt(r^2 - x^2) about the x-axis. The disk method gives V = (4/3) pi r^3.
Cone: a line from the origin to the point (h, r) is y = (r/h)x. Rotate about the x-axis: V = (1/3) pi r^2 * h.
Paraboloid: rotate y = ax^2 about the y-axis. The resulting volume is exactly half the circumscribed cylinder.
Volumes of revolution are how engineers calculate the capacity of tanks, silos, rocket nose cones, and turbine housings. The fuel-tank problem in the homework (C-130 wing tank modelled by y = (1/500) x^2 sqrt(18 - x)) is a direct example: you revolve the profile about the x-axis and multiply by a unit conversion to get gallons. The submarine problem uses known circular cross-sections to find displacement tonnage.
Students often forget to convert R and r into the correct variable. If it is a dx problem, everything must be in terms of x, even if the radius looks like a y-distance. Solve for x or y as needed.
Squaring a binomial radius is a frequent algebra error. (4 - x^2)^2 is not 16 - x^4. Expand carefully.
Students confuse which curve gives R and which gives r. R is always the one farther from the axis of rotation, r is the one closer. Sketch the picture first.
When rotating about a line like y = 6 instead of the x-axis, students sometimes use y = 6 as R and the curve as r when the entire region sits below y = 6. Both R and r should be measured as distances from y = 6 downward.
⚠️ Disk vs washer identification is the very first step and the most common source of wrong setups.
⚠️ Expect at least one problem where the axis of rotation is not the x-axis or y-axis (e.g. y = 4, x = 3). These shift the radius expressions.
⚠️ Cross-section problems (squares, triangles, semicircles) appear regularly on AP free-response questions.
⚠️ You may be asked to prove the volume formula for a sphere or cone using integration. Memorise the setup.
True or false: if a region is rotated about a vertical line, you always integrate with respect to y.
Fill in the blank: the area of a washer is pi * ( ______ - ______ ).
True or false: for the region between y = x and y = 0, rotated about the x-axis, it is a washer problem.
If cross-sections perpendicular to the x-axis are equilateral triangles with side length s, the area of each cross-section is ______.
True or false: when rotating about y = 5, the outer radius R can never equal 5.
Answers: 1) True. 2) R^2, r^2. 3) False (it is a disk, the region touches the axis). 4) (sqrt(3)/4) * s^2. 5) False (R = 5 when the inner boundary sits on the x-axis, i.e. y = 0).
Q: The region bounded by y = 2x, y = 0, and x = 3 is rotated about the x-axis. Set up and evaluate the volume integral.
A: Disk problem. R = y = 2x. V = pi integral from 0 to 3 of (2x)^2 dx = pi integral of 4x^2 dx from 0 to 3 = pi * [4x^3/3] from 0 to 3 = 36pi.
Q: The same region is rotated about the line y = 6. Identify R and r, and set up the integral.
A: Washer problem. R = 6 (distance from y = 6 to y = 0). r = 6 - 2x. V = pi * integral from 0 to 3 of [36 - (6 - 2x)^2] dx.
Q: The region bounded by y = x^2 and y = cube root of x is rotated about the x-axis. What are R and r?
A: The curves intersect at (0,0) and (1,1). Between these, cube root of x is above x^2. So R = x^(1/3) and r = x^2.
Q: A solid has base bounded by x^2 + y^2 = 16 and cross-sections perpendicular to the x-axis are semicircles. Set up the volume integral.
A: Half-width = y = sqrt(16 - x^2). Diameter = 2y, so radius of semicircle = y. Area = (pi/2) y^2 = (pi/2)(16 - x^2). V = (pi/2) integral from -4 to 4 of (16 - x^2) dx, or equivalently pi * integral from 0 to 4 of (16 - x^2) dx.
This connects directly to the area-between-curves work from earlier chapters, since the same region definitions and intersection-finding techniques carry over. It also sets the stage for the shell method (often covered next) and for physics applications such as computing moments of inertia and centroids using similar slicing arguments.
volume of revolution, disk method calculus, washer method calculus, solids of revolution, rotation about x-axis, rotation about y-axis, cross-sectional volume, known cross-sections, square cross-section volume, equilateral triangle cross-section, semicircle cross-section, AP Calculus volume problems, Riemann sum to integral, pi R squared, volume integral setup, dx vs dy volume