Difficulty: Intermediate | Prerequisites: Basic integration techniques, Fundamental Theorem of Calculus, graphing functions
Big picture: This material sits at the heart of Integral Calculus, right after you have learned how to evaluate definite integrals. You are now using integration as a tool to measure geometric quantities: first the area trapped between two curves, then the volume of a three-dimensional solid formed by spinning a region around an axis or by stacking known cross-sections. If you can set up the correct integral (choosing the right variable, the right method, and the right limits), the evaluation is usually straightforward calculus you already know.
You are learning to set up integrals that measure two things: the area trapped between curves (by integrating the "top minus bottom" or "right minus left" function), and the volume of a solid created by revolving a region around an axis or by stacking cross-sections. The main decision is which method to use: disks/washers when you slice perpendicular to the axis of rotation, shells when you slice parallel to it, and known cross-sections when the problem gives you the shape of each slice directly.
Area between curves
The net area of the region enclosed by two functions over an interval, computed by integrating the absolute difference of the functions. In simple terms, you subtract the lower curve from the upper curve and integrate.
Disk method
A technique for finding the volume of a solid of revolution by summing thin circular disks perpendicular to the axis of rotation. Think of it as stacking coins of varying radius along the axis.
Washer method
An extension of the disk method used when the region being rotated does not touch the axis of rotation, producing a hollow core. Each cross-section is a ring (washer) with an outer radius and an inner radius.
Shell method (cylindrical shells)
A technique for computing volumes of revolution by summing thin cylindrical shells parallel to the axis of rotation. Think of it as peeling layers off a cylinder, one shell at a time.
Cross-sectional volume (slicing method)
A general method where you know the shape of every cross-section perpendicular to a given axis, so volume equals the integral of the cross-sectional area along that axis.
Frustum
The portion of a cone (or other solid) lying between two parallel planes cutting through it. In simple terms, it is a cone with its tip sliced off.
Setting up the integral
When integrating with respect to x: Area = integral from a to b of [f(x) - g(x)] dx, where f(x) is the top curve and g(x) is the bottom curve on the interval [a, b].
When integrating with respect to y: Area = integral from c to d of [f(y) - g(y)] dy, where f(y) is the right curve and g(y) is the left curve.
Always find intersection points first. These are your limits of integration.
If the curves cross within the interval, split into sub-intervals so you always subtract the lower from the upper.
Worked example (HW Q1): f(x) = x^2 - 2x, g(x) = x
Find intersections: x^2 - 2x = x gives x^2 - 3x = 0, so x(x - 3) = 0, meaning x = 0 and x = 3.
On [0, 3], check which is on top: at x = 1, f(1) = -1 and g(1) = 1, so g is above f.
Area = integral from 0 to 3 of [x - (x^2 - 2x)] dx = integral from 0 to 3 of (3x - x^2) dx.
Evaluating: [3x^2/2 - x^3/3] from 0 to 3 = (27/2 - 9) - 0 = 27/2 - 18/2 = 9/2.
Answer: 9/2 sq. units.
Integrating with respect to y (HW Q2)
When curves are given as x = f(y) and x = g(y), it can be more natural to integrate with respect to y. The diagram in Q2 shows regions A1, A2, A3 in a rectangle with corners (1,1) and (4,3).
A1 sits to the right of g and below the rectangle boundary. A1 = area of the rectangle portion to the right of g, which is 8 - integral from 1 to 3 of g(y) dy. (Statement I is correct.)
A2 lies between the two curves. A2 = integral from 1 to 3 of [f(y) - g(y)] dy. (Statement II is correct.)
A3 sits to the left of f and above the rectangle boundary. A3 = integral from 1 to 3 of f(y) dy - 2. (Statement III is correct.)
All three statements are correct without further restrictions on f and g. Answer: all of them.
Disk method, rotating about the x-axis
Slice the solid perpendicular to the x-axis. Each slice is a disk of radius R(x) equal to the distance from the curve to the axis.
V = pi times integral from a to b of [R(x)]^2 dx.
Use this when the region touches the axis of rotation (no gap between the region and the axis).
Worked example (HW Q3): Frustum from y = x + 1
Region bounded by y = x + 1, y = 0, x = 0, x = 3, rotated about the x-axis.
Each disk has radius R(x) = x + 1.
V = pi times integral from 0 to 3 of (x + 1)^2 dx = pi times integral from 0 to 3 of (x^2 + 2x + 1) dx.
Evaluating: pi times [x^3/3 + x^2 + x] from 0 to 3 = pi times (9 + 9 + 3) = 21pi.
Answer: V = 21pi cu. units.
Worked example (HW Q4): y = 3/x rotated about the x-axis
Region bounded by y = 3/x, x = 2, x = 5, y = 0.
R(x) = 3/x.
V = pi times integral from 2 to 5 of (3/x)^2 dx = pi times integral from 2 to 5 of 9/x^2 dx = pi times [-9/x] from 2 to 5.
Evaluating: pi times (-9/5 + 9/2) = pi times (9/2 - 9/5) = pi times (45/10 - 18/10) = 27pi/10.
Answer: V = 27pi/10.
Disk method, rotating about the y-axis
If you rotate about the y-axis, express x as a function of y and integrate with respect to y.
V = pi times integral from c to d of [R(y)]^2 dy.
Worked example (HW Q5): y = x^2 rotated about the y-axis
Region bounded by y = x^2, x = 0, y = 16. Since y = x^2, we have x = sqrt(y), so R(y) = sqrt(y).
V = pi times integral from 0 to 16 of y dy = pi times [y^2/2] from 0 to 16 = pi times 128 = 128pi.
Answer: V = 128pi cu. units.
Washer method
When the region does not touch the axis of rotation, each slice is a washer (ring) with outer radius R and inner radius r.
V = pi times integral of [R^2 - r^2] dx (or dy).
Worked example (HW Q6): y = 2 - sqrt(x), x = 1, y = 2 about the x-axis
The region is bounded above by y = 2 and below by y = 2 - sqrt(x), from x = 0 (where 2 - sqrt(x) = 2, so x = 0, but the boundary x = 1 applies) to x = 1 (given). Wait, let us re-examine: y = 2 - sqrt(x) equals 2 when x = 0, and y = 2 - 1 = 1 when x = 1. The region is enclosed by y = 2 - sqrt(x), x = 1, and y = 2.
Outer radius R = 2 (from x-axis to y = 2). Inner radius r = 2 - sqrt(x).
V = pi times integral from 0 to 1 of [4 - (2 - sqrt(x))^2] dx = pi times integral from 0 to 1 of [4 - 4 + 4sqrt(x) - x] dx = pi times integral from 0 to 1 of [4sqrt(x) - x] dx.
Evaluating: pi times [8x^(3/2)/3 - x^2/2] from 0 to 1 = pi times (8/3 - 1/2) = pi times (16/6 - 3/6) = 13pi/6.
Answer: volume = 13pi/6 cu. units.
Cone volume (HW Q7): height h = 1, base radius r = 2
The cone's profile is a line from (0, 0) to (1, 2) in the xy-plane, giving R(x) = 2x. Rotating about the x-axis:
V = pi times integral from 0 to 1 of (2x)^2 dx = pi times integral from 0 to 1 of 4x^2 dx = pi times [4x^3/3] from 0 to 1 = 4pi/3.
This matches V = (1/3) pi r^2 h = (1/3) pi (4)(1) = 4pi/3.
Answer: volume = 4pi/3 cu. units.
Shell method setup
When slicing parallel to the axis of rotation, each slice is a thin cylindrical shell.
V = 2pi times integral of (radius)(height) dx (or dy).
The radius is the distance from the slice to the axis of rotation. The height is the length of the slice.
Shell method is often easier when the axis of rotation is vertical and the functions are given as y = f(x), because you avoid having to solve for x.
Worked example (HW Q8): f(x) = x, g(x) = x^3 about the line x = -3
Region A is in the first quadrant, bounded by y = x and y = x^3. These intersect at x = 0 and x = 1.
The axis of rotation is x + 3 = 0, i.e. x = -3.
Using shells (integrating with respect to x): the shell radius is x - (-3) = x + 3. The shell height is x - x^3 (since x > x^3 on (0,1)).
V = 2pi times integral from 0 to 1 of (x + 3)(x - x^3) dx.
Expanding: (x + 3)(x - x^3) = x^2 - x^4 + 3x - 3x^3.
Integrating: [x^3/3 - x^5/5 + 3x^2/2 - 3x^4/4] from 0 to 1 = 1/3 - 1/5 + 3/2 - 3/4.
Finding common denominator (60): 20/60 - 12/60 + 90/60 - 45/60 = 53/60.
V = 2pi times 53/60 = 53pi/30.
Answer: volume = 53pi/30.
Cross-sectional (slicing) method
Used when you know the cross-sectional shape at each point along an axis.
V = integral from a to b of A(y) dy (or A(x) dx), where A is the area of the cross-section.
Worked example (HW Q9-Q10): Semicircular cross-sections
The base is the circle x^2 + y^2 = 16. Cross-sections perpendicular to the y-axis are semicircles.
At height y, the base chord runs from x = -sqrt(16 - y^2) to x = sqrt(16 - y^2), so the chord length (diameter of the semicircle) is 2sqrt(16 - y^2). The semicircle has radius sqrt(16 - y^2).
Area of a semicircle with radius R is pi R^2 / 2, so A(y) = pi(16 - y^2)/2.
V = integral from -4 to 4 of pi(16 - y^2)/2 dy.
Evaluating: (pi/2) times [16y - y^3/3] from -4 to 4 = (pi/2) times [(64 - 64/3) - (-64 + 64/3)] = (pi/2) times [128 - 128/3] = (pi/2) times (256/3) = 128pi/3.
Answer: V = 128pi/3 cu. units.
\text{Area between curves (x):} \quad A = \int_a^b \bigl[f(x) - g(x)\bigr]\, dx\text{Area between curves (y):} \quad A = \int_c^d \bigl[f(y) - g(y)\bigr]\, dy\text{Disk method:} \quad V = \pi \int_a^b \bigl[R(x)\bigr]^2\, dx\text{Washer method:} \quad V = \pi \int_a^b \bigl[R(x)^2 - r(x)^2\bigr]\, dx\text{Shell method:} \quad V = 2\pi \int_a^b (\text{radius})(\text{height})\, dx\text{Cross-sectional volume:} \quad V = \int_a^b A(x)\, dxStudents often forget to square the radius in the disk/washer formula. The volume integral involves pi R^2, not pi R. The squaring comes from the area of a circle.
Students sometimes set up the washer method with R - r instead of R^2 - r^2. You are subtracting areas of circles, not radii.
When the axis of rotation is not the x-axis or y-axis (e.g. x = -3 in Q8), students often use the raw function value as the radius. The radius is the distance from the slice to the axis, which means you must adjust: if rotating about x = -3, the shell radius from a point at x is x + 3, not x.
Students frequently confuse when to use disks/washers versus shells. The decision depends on which variable you integrate with respect to, and whether your slices are perpendicular or parallel to the axis of rotation.
Setting up the integral correctly (choosing the method, identifying radii, finding limits) is worth more marks than evaluating it. Exams test setup more than arithmetic.
The cone volume formula V = (1/3) pi r^2 h is a standard result you should be able to derive by integration. Expect at least one problem asking you to verify a known geometric formula.
Rotating about a line other than a coordinate axis (like x = -3) appears regularly on exams and trips up students who have only practised with the standard axes.
Cross-sectional volume problems (semicircles, squares, equilateral triangles as cross-sections) are a favourite exam topic because they test whether you understand the slicing idea beyond just rotation.
True or false: In the washer method, you subtract the inner radius from the outer radius before squaring. _____ (False. You square each radius separately, then subtract: pi(R^2 - r^2).)
Fill in the blank: The shell method formula is V = _____ times the integral of (radius)(height). (2pi)
True or false: When rotating about the line x = -3, the shell radius for a vertical strip at position x is x + 3. _____ (True.)
Fill in the blank: To find the area between two curves, you always integrate _____ minus _____. (top function minus bottom function, or right minus left if integrating with respect to y)
True or false: The disk method and the shell method will always give the same answer for the same solid. _____ (True, when set up correctly, both methods compute the same volume.)
Q: Find the area enclosed by y = x^2 and y = 2x.
A: Intersections at x = 0 and x = 2. Area = integral from 0 to 2 of (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2 = 4 - 8/3 = 4/3.
Q: Set up (do not evaluate) the integral for the volume obtained by rotating the region bounded by y = sqrt(x), y = 0, and x = 4 about the x-axis.
A: Using the disk method, V = pi times integral from 0 to 4 of (sqrt(x))^2 dx = pi times integral from 0 to 4 of x dx.
Q: A solid has a circular base x^2 + y^2 = 9, and every cross-section perpendicular to the x-axis is a square. Set up the volume integral.
A: At position x, the base chord has half-length sqrt(9 - x^2), so side length = 2sqrt(9 - x^2). Area = [2sqrt(9 - x^2)]^2 = 4(9 - x^2). V = integral from -3 to 3 of 4(9 - x^2) dx.
Q: The region between y = x and y = x^3 in the first quadrant is rotated about the y-axis. Would you choose shells or washers, and why?
A: Shells are simpler here. With shells (integrating in x), the radius is x and the height is x - x^3, giving V = 2pi times integral from 0 to 1 of x(x - x^3) dx. The washer method would require expressing both curves as functions of y and splitting the integral, which is more work.
Q: The region bounded by y = 1/x, x = 1, x = 4, and y = 0 is rotated about the x-axis. Find the volume.
A: V = pi times integral from 1 to 4 of (1/x)^2 dx = pi times integral from 1 to 4 of 1/x^2 dx = pi times [-1/x] from 1 to 4 = pi times (-1/4 + 1) = 3pi/4.
This material connects directly to work and energy in physics, where the integral of a force over a distance gives work done, and to moments of inertia, which use similar integral setups with different integrands. Later in the course, you will extend these ideas to arc length and surface area of revolution, which use the same slicing logic but with different formulas for the integrand.
Related Terms / Search Tags
area between curves, bounded area, enclosed region, volume of revolution, solid of revolution, disk method, washer method, cylindrical shells, shell method, cross-sectional area, slicing method, frustum, cone volume derivation, rotating about a line, axis of rotation, inner radius, outer radius, semicircular cross-section, integral calculus applications, Riemann sums for volume