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Area Between Curves and Volumes of Revolution

Area and volume problems are where integration stops being an algebra exercise and starts being a modelling skill. The integrals themselves are usually easy. What separates a correct answer from a wrong one is the setup: which curve is on top, where the region actually begins and ends, whether to slice vertically or horizontally, and which method matches the axis of revolution.

This guide walks through that setup logic in order, with worked numbers you can check. The theme throughout is the one behind every application of the integral: slice the region into pieces so thin that each is a shape you already know, then add them up.

One Idea: Slice, Approximate, Integrate

Every formula in this guide is the same three-step move. Cut the region or solid into thin slices. Approximate each slice by an elementary shape whose measurement you know, a rectangle for area, a disk or washer or cylindrical shell for volume. Then add the slices with an integral, which is the limit of that sum as the slices become infinitely thin.

Because the slice shape determines the integrand, the productive first question is not which formula do I use but what does one slice look like. If you can describe a typical slice in words, including its thickness, you can write the integral.

Area Between Two Curves

For a region bounded above by one curve and below by another over an interval, a vertical slice is a thin rectangle whose height is the top function minus the bottom function and whose width is dx. Integrating that height across the interval gives the area. The single most common error is subtracting in the wrong order, which flips the sign; if your area comes out negative, you almost certainly reversed the subtraction rather than made an arithmetic slip.

Take the region enclosed by the line y equals x and the parabola y equals x squared. Setting the two expressions equal gives x equals x squared, so x times the quantity x minus one is zero, and the curves meet at x equals zero and x equals one. Between those points the line sits above the parabola, so the area is the integral from zero to one of x minus x squared. The antiderivative is x squared over two minus x cubed over three, giving one half minus one third, which is one sixth.

A = \int_a^b \left[f(x) - g(x)\right] dx, \quad f(x) \ge g(x) \text{ on } [a,b]

Area between curves: integrate top minus bottom across the interval.

Find the Intersections, Then Split Where the Curves Swap

The limits of integration are usually not handed to you. Solve for the intersection points algebraically rather than reading them off a sketch, because a sketch will not tell you whether two curves meet at exactly two points or three. And once you have the intersections, check which curve is on top on each subinterval, because the answer can change from one piece to the next.

The curves y equals x cubed and y equals x show why this matters. Setting them equal gives x cubed equals x, so x times the quantity x squared minus one is zero, and there are three intersections: negative one, zero, and one. On the interval from negative one to zero the cubic is above the line, and on the interval from zero to one the line is above the cubic. So the enclosed area is the integral from negative one to zero of x cubed minus x, plus the integral from zero to one of x minus x cubed. Each piece evaluates to one quarter, for a total area of one half. Integrating top minus bottom in a single sweep with one fixed ordering would have produced zero, since the two halves would cancel.

Integrating in y Instead of x

Some regions are described far more naturally by horizontal slices. If a region is bounded on the right by one curve and on the left by another, and those roles stay fixed as you move up the region, slice horizontally: a typical slice has width equal to the right function minus the left function, expressed in terms of y, and thickness dy. This often replaces a two-piece x-integral with a single clean y-integral.

Consider the region bounded by the parabola x equals y squared and the line x equals y plus two. Setting them equal gives y squared minus y minus two equals zero, which factors as the quantity y minus two times the quantity y plus one, so the curves meet at y equals negative one and y equals two. Between those values the line lies to the right of the parabola. The area is the integral from negative one to two of y plus two minus y squared, whose antiderivative is y squared over two plus two y minus y cubed over three. Evaluating gives ten thirds at the top and negative seven sixths at the bottom, and the difference is nine halves.

The practical rule: slice in whichever direction keeps one boundary pair valid across the whole region.

Volumes by Disks and Washers

Revolve a region about an axis and slice perpendicular to that axis. Each slice becomes a circular disk when the region touches the axis, with volume equal to pi times the radius squared times the thickness. Integrating gives pi times the integral of the radius squared. For instance, revolving the region under the square root of x from zero to four about the x-axis makes the radius the square root of x, so the volume is pi times the integral of x from zero to four, which is pi times eight, or eight pi.

When there is a gap between the region and the axis, the slice is a washer: a disk with a hole. Subtract the inner radius squared from the outer radius squared before integrating, and be careful that this is the difference of the squares, never the square of the difference. Revolving the region between y equals x and y equals x squared from zero to one about the x-axis gives an outer radius of x and an inner radius of x squared, so the volume is pi times the integral of x squared minus x to the fourth. The antiderivative is x cubed over three minus x to the fifth over five, giving pi times one third minus one fifth, which is two pi over fifteen.

Watch the axis: when the region is revolved about a horizontal line other than the x-axis, each radius is a distance to that line rather than the curve's own height.

V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right) dx

Washer method: outer radius squared minus inner radius squared.

Volumes by Cylindrical Shells

Shells slice parallel to the axis of revolution instead of perpendicular to it. A thin vertical strip revolved about a vertical axis sweeps out a hollow cylinder, and unrolling it gives a thin rectangular sheet whose area is the circumference times the height. That is where the formula comes from: two pi times the radius times the height, integrated over the thickness variable.

Revolve the region under y equals x squared from x equals zero to x equals two about the y-axis. The shell radius is x and the shell height is x squared, so the volume is two pi times the integral of x cubed from zero to two, which is two pi times four, or eight pi. As a check, the same solid can be done with washers in y: the outer radius is the constant two and the inner radius is the square root of y, so the volume is pi times the integral from zero to four of four minus y, which is pi times sixteen minus eight, again eight pi.

That agreement is the useful lesson. Disks and shells are two bookkeeping schemes for the same solid, so when both work you may pick the easier integral. Prefer shells when revolving a region given as a function of x about a vertical axis.

V = 2\pi \int_a^b (\text{radius})(\text{height})\, dx

Shell method: circumference times height, integrated across the thickness.

A Setup Checklist

Before integrating anything, settle the geometry. These questions, answered in order, prevent nearly every setup error in this family of problems.

Sketch the region and find every intersection algebraically, not by eye.
Decide the slice direction: vertical slices give a dx integral, horizontal slices a dy integral.
Check whether the boundary curves swap roles inside the region; if they do, split the integral there.
For volumes, identify the axis of revolution and write each radius as a distance to that axis.
Perpendicular slices give disks or washers; parallel slices give shells.
Use the difference of squared radii for washers, never the square of the difference.

Frequently asked questions

How do I know which curve is on top?

Test a single point strictly between two consecutive intersections and compare the two function values there. The ordering cannot change without an intersection, so one test point settles each subinterval.

When should I integrate with respect to y instead of x?

Integrate in y when the region has a fixed left curve and right curve but its top or bottom boundary changes partway across. Horizontal slices then need one integral where vertical slices would need several.

Do I use disks, washers, or shells?

Slice perpendicular to the axis of revolution for disks or washers, and use a washer whenever a gap separates the region from the axis. Slice parallel to the axis for shells, which are usually easier when the axis is vertical and the curves are given as functions of x.

Why is the washer integrand the difference of squares?

The cross-section is a large disk with a smaller disk removed, so its area is pi times the outer radius squared minus pi times the inner radius squared. Squaring the difference of the radii would measure a different, meaningless region.