Difficulty: Intermediate | Prerequisites: Calc I (definite integrals, basic antiderivatives, u-substitution), familiarity with area between curves.
In Calc I you learned to find the area under a curve by slicing it into thin rectangles and summing. Volumes of solids extend that idea into three dimensions: slice a solid into thin cross-sections, compute the area of each slice, then integrate. This is one of the first major applications of the definite integral beyond flat area, and it comes up in nearly every Calc II exam. You should already be comfortable setting up definite integrals, performing u-substitution, and finding the region bounded by two curves.
To find the volume of a solid, slice it perpendicular to an axis, write down the cross-sectional area as a function of position, and integrate. When the solid is formed by rotating a region around an axis, each cross-section is a disk (one boundary) or a washer (two boundaries), and the area formulas follow directly from the geometry of circles.
Solid of revolution
A three-dimensional shape created by rotating a two-dimensional region around a straight line (the axis of rotation). Think of it as spinning a flat cutout on a skewer and looking at the shape it sweeps out.
Cross-sectional area, A(x) or A(y)
The area of the flat slice you see when you cut the solid perpendicular to the axis of integration. In simple terms, this is the "face" of each thin piece you are adding up.
Disk method
A technique for finding volumes of revolution when the region has one boundary curve and the axis of rotation. Each cross-section is a filled circle (a disk). Think of it as stacking coins of varying radius.
Washer method
A technique for volumes of revolution when the region sits between two boundary curves, both away from the axis of rotation. Each cross-section is a ring (a washer), so you subtract the inner circle's area from the outer circle's area.
Outer radius, R(x)
The distance from the axis of rotation to the farther boundary curve. This is the radius of the larger circle in a washer cross-section.
Inner radius, r(x)
The distance from the axis of rotation to the closer boundary curve. This is the radius of the hole in the washer.
The foundational idea is the same one behind Riemann sums for area: cut, approximate, sum, take the limit.
Vertical slicing (integrating with respect to x)
Slice the solid perpendicular to the x-axis.
Each slice has cross-sectional area A(x) and infinitesimal thickness dx.
Volume formula:
V = \int_a^b A(x)\,dxHorizontal slicing (integrating with respect to y)
Slice the solid perpendicular to the y-axis.
Each slice has cross-sectional area A(y) and thickness dy.
You need the boundary curves written as x = g(y), so if you are given y = f(x), invert it.
Volume formula:
V = \int_c^d A(y)\,dyChoosing the axis of integration comes down to which variable gives you a simpler expression for A. If the region's boundaries are easier to describe as functions of x, integrate in x. If they are easier as functions of y, integrate in y.
Use the disk method when the region touches the axis of rotation along one edge. Each cross-section is a solid circle.
Rotation about the x-axis
The region sits between y = f(x) and the x-axis, from x = a to x = b.
Each disk has radius r = f(x) and area A(x) = pi [f(x)]^2.
Volume formula:
V = \pi \int_a^b [f(x)]^2\,dxRotation about the y-axis
Rewrite boundaries as x = g(y), with y ranging from c to d.
Each disk has radius r = g(y).
Volume formula:
V = \pi \int_c^d [g(y)]^2\,dyFind the volume of the solid obtained by rotating y = (1/2)x + 1, from x = 0 to x = 4, about the x-axis.
Set up the integral:
V = \pi \int_0^4 \left(\tfrac{1}{2}x + 1\right)^2 dxUse the substitution u = (1/2)x + 1, so du = (1/2)dx, giving 2 du = dx. When x = 0, u = 1; when x = 4, u = 3.
V = 2\pi \int_1^3 u^2\,du = 2\pi \cdot \frac{u^3}{3}\bigg|_1^3 = 2\pi\left(\frac{27}{3} - \frac{1}{3}\right) = 2\pi \cdot \frac{26}{3} = \frac{52\pi}{3}Find the volume of the solid formed by rotating the region between y = x^(3/2) and y = x^(1/3), from x = 0 to x = 1, about the x-axis. (Here y = x^(1/3) is the outer curve and y = x^(3/2) is the inner curve on this interval, so the region does not touch the axis. However, since both curves pass through the origin and (1,1), the notes treat this by subtracting disk areas.)
V = \pi \int_0^1 \left[(x^{1/3})^2 - (x^{3/2})^2\right] dx = \pi \int_0^1 (x^{2/3} - x^3)\,dx= \pi \left[\frac{x^{5/3}}{5/3} - \frac{x^4}{4}\right]_0^1 = \pi\left(\frac{3}{5} - \frac{1}{4}\right) = \pi \cdot \frac{7}{20} = \frac{7\pi}{20}Use the washer method when the region does not touch the axis of rotation. Each cross-section is a ring with an outer radius R and an inner radius r.
The area of each washer is A = pi(R^2 - r^2).
Volume formula:
V = \pi \int_a^b \left[R(x)^2 - r(x)^2\right] dx = \pi \int_a^b \left[f(x)^2 - g(x)^2\right] dxwhere f(x) is the curve farther from the axis (outer radius) and g(x) is the curve closer to the axis (inner radius).
When the axis of rotation is a horizontal line y = k (not the x-axis), the radii are distances from that line to each curve:
Outer radius R = |(farther curve) - k|
Inner radius r = |(closer curve) - k|
The same logic applies for vertical axes x = k.
Consider the region bounded by y = x^2 + 1 and y = 1, from x = 0 to x = 2. Find the volume when this region is rotated about the x-axis.
Here f(x) = x^2 + 1 (farther from the axis) and g(x) = 1 (closer).
V = \pi \int_0^2 \left[(x^2+1)^2 - 1^2\right] dx = \pi \int_0^2 (x^4 + 2x^2)\,dx= \pi \left[\frac{x^5}{5} + \frac{2x^3}{3}\right]_0^2 = \pi\left(\frac{32}{5} + \frac{16}{3}\right) = \frac{176\pi}{15}Determine the volume obtained by rotating the region bounded by y = 2 sqrt(x - 1), y = x - 1, and x = -1, about the line x = -1.
Rewrite the curves as functions of y. From y = 2 sqrt(x - 1): x = y^2/4 + 1. From y = x - 1: x = y + 1. The region runs from y = 0 to y = 4.
Since we are rotating about the vertical line x = -1, each radius is the horizontal distance from x = -1 to the curve:
Outer radius: R = (y + 1) - (-1) = y + 2
Inner radius: r = (y^2/4 + 1) - (-1) = y^2/4 + 2
V = \pi \int_0^4 \left[(y+2)^2 - \left(\frac{y^2}{4}+2\right)^2\right] dy= \pi \int_0^4 \left(4y - \frac{y^4}{16}\right) dy = \pi \left[2y^2 - \frac{y^5}{80}\right]_0^4 = \frac{96\pi}{5}Students often forget to square the radius before integrating. The disk formula is pi [f(x)]^2, not pi f(x). You are computing the area of a circle, which requires squaring.
Students sometimes confuse which curve is outer and which is inner. The outer radius is always the curve farther from the axis of rotation, regardless of which function has the larger y-value in the original problem. Sketch the region and the axis first.
When the axis of rotation is shifted (e.g. y = -1 or x = 3), students often use the raw function values as radii. The radius is the distance from the curve to the axis, not the function value itself. Always subtract (or add) to account for the shift.
Students occasionally set up a washer integral when a disk integral is sufficient. If the region touches the axis of rotation, there is no hole, so there is no inner radius to subtract.
Disk and washer problems appear on nearly every Calc II midterm. Expect at least one volume-of-revolution question.
Professors frequently test whether you can set up the integral correctly, not just evaluate it. Show your reasoning for which method you chose, which curve is R and which is r, and the limits of integration.
Problems that rotate about a line other than the x-axis or y-axis (e.g. y = 4, x = -1) are a favourite way to test deeper understanding. Practise these specifically.
This material is the foundation for the cylindrical shells method (coming next), so gaps here compound quickly.
True or false: The disk method requires the region to touch the axis of rotation.
True.
Fill in the blank: The cross-sectional area of a washer with outer radius R and inner radius r is A = ______.
pi(R^2 - r^2).
True or false: When rotating about y = 3 instead of y = 0, you can use the same radius formula as for the x-axis.
False. You must adjust each radius to measure the distance from the curve to y = 3.
Fill in the blank: To set up a volume integral with respect to y, you need the boundary curves written as x = ______.
g(y), i.e. solved for x in terms of y.
Q: The region bounded by y = x^2 + 1 and y = 1, from x = 0 to x = 2, is rotated about the x-axis. Set up the volume integral and evaluate.
A: V = pi integral from 0 to 2 of [(x^2 + 1)^2 - 1^2] dx = pi integral of (x^4 + 2x^2) dx = pi [x^5/5 + 2x^3/3] from 0 to 2 = 176pi/15.
Q: The region bounded by y = 2 sqrt(x - 1), y = x - 1, and x = -1 is rotated about the line x = -1. Which method do you use, and what are the outer and inner radii?
A: Use the washer method, integrating with respect to y. Outer radius R = y + 2 (from x = y + 1 to x = -1). Inner radius r = y^2/4 + 2 (from x = y^2/4 + 1 to x = -1).
Q: Why does the disk method formula include pi but the general slicing formula V = integral of A(x) dx does not?
A: The general formula has no pi because A(x) can be any cross-sectional shape. In the disk method, every cross-section is a circle with area pi r^2, so pi appears when you substitute that specific area formula.
Q: A region bounded by y = sqrt(x) and y = 0, from x = 0 to x = 4, is rotated about the x-axis. Compute the volume.
A: V = pi integral from 0 to 4 of [sqrt(x)]^2 dx = pi integral of x dx = pi [x^2/2] from 0 to 4 = 8pi.
This material connects directly to the cylindrical shells method, which provides an alternative way to compute volumes of revolution and is sometimes simpler when the boundaries are easier to express in the "other" variable. Together, disks/washers and shells give you two tools for any rotation problem.
The general slicing idea (integrate a cross-sectional area) also extends to solids with known cross-sections that are not circles, such as squares, triangles, or semicircles. That topic is covered in the companion notes.
Volumes of revolution also connect to arc length and surface area of revolution, which appear later in Calc II and use a similar "slice and integrate" strategy.
Volumes of solids, volume of revolution, disk method, washer method, solids of revolution, cross-sectional area, axis of rotation, rotating a region, Calc II volumes, M408D, integral calculus volumes, slicing method, pi r squared, outer radius, inner radius, volume integral setup, u-substitution in volumes, UT Austin Calc 2