Area Between Curves and Volumes of Revolution

Use slices, distances, and a worked example to choose and build the right definite integral.

To find an area between curves, integrate the distance between their boundaries. To find a volume of revolution, rotate the region and integrate either washer areas or thin cylindrical shells. The essential decisions are the slice direction, the integration bounds, and the distances measured from the rotation axis—not simply which formula to memorize. [1], [2], [3]

In this article

1. The central idea: identify what each slice contributes
2. Area means a distance between boundaries
3. Worked example: the region between y = x and y = x²
4. Rotate the region: discs and washers
5. Shells offer a different way to slice the same solid
6. Choose the setup and check the geometry

1. The central idea: identify what each slice contributes

Area: use upper minus lower for vertical strips, or right minus left for horizontal strips. [1]
Washers: slices perpendicular to the rotation axis contribute π(R² − r²) times their thickness. [3], [2]
Shells: slices parallel to the rotation axis contribute circumference × height × thickness. [3]

These methods connect geometry to a definite integral. Begin with a thin piece of the region, identify its dimensions, and then accumulate its contribution across the appropriate interval. For planar area, that piece is a narrow rectangle. For a rotated region, it becomes a washer, disc, or cylindrical shell. The integral represents the limit of adding increasingly thin pieces. Drawing one representative slice is therefore more useful than choosing a formula from memory: it reveals what the integrand must measure. [1], [3]

2. Area means a distance between boundaries

For vertical strips, let u(x) be the upper boundary and l(x) the lower boundary throughout a ≤ x ≤ b. The strip height is u(x) − l(x), giving A = ∫ₐᵇ[u(x) − l(x)] dx. For horizontal strips, write the boundaries as x = R(y) and x = L(y). Their horizontal separation gives A = ∫꜀ᵈ[R(y) − L(y)] dy. Choose the direction that describes the region most simply. The letters dx and dy specify both the strip thickness and the variable used for the limits. [1]

Before integrating, identify the entire bounded region. Intersections can supply its endpoints, but additional boundaries—such as a vertical line or the y-axis—may supply them instead. Also check whether the ordering of the curves changes within the interval. If it does, split the calculation where the ordering changes and use upper minus lower on each part. This prevents signed contributions from cancelling. A sketch helps establish the geometry; solving the boundary equations supplies exact endpoints rather than estimates read from the drawing. [1]

3. Worked example: the region between y = x and y = x²

Consider this illustrative example, calculated here using the area rule: find the region enclosed by y = x and y = x². The intersections satisfy x² = x, so x(x − 1) = 0 and the endpoints are x = 0 and x = 1. Throughout this interval, x ≥ x², making the line the upper boundary and the parabola the lower boundary. A vertical strip therefore has height x − x². No interval splitting is needed because that ordering holds across the whole enclosed region. [1]

The resulting calculation is A = ∫₀¹(x − x²) dx = [x²/2 − x³/3]₀¹ = 1/6 square units. Horizontal strips provide an independent setup: solving for x gives x = y for the line and x = √y for the nonnegative branch of the parabola. For 0 ≤ y ≤ 1, the parabola is the right boundary. Thus A = ∫₀¹(√y − y) dy = 2/3 − 1/2 = 1/6. Both calculations describe the same region, but subtract boundaries in different directions. [1]

4. Rotate the region: discs and washers

When a slice is perpendicular to the rotation axis, rotating it creates a circular cross-section. If the slice reaches the axis, the cross-section is a disc. If a gap separates it from the axis, the cross-section is a washer: an outer disc with an inner disc removed. Its area is π(R² − r²), where R and r are distances from the axis to the outer and inner edges. Integrating these areas along the solid gives its volume; setting r = 0 recovers the disc formula. [2], [3]

Rotate our example about the x-axis. The vertical strips are perpendicular to that axis, so they produce washers. Since both curves are nonnegative on [0, 1], the outer radius is R = x and the inner radius is r = x². Therefore V = π∫₀¹[x² − (x²)²] dx = π[x³/3 − x⁵/5]₀¹ = 2π/15 cubic units. Notice that we subtract the squared radii, not square their difference: π(x − x²)² would describe a different cross-sectional area. [1], [3], [2]

5. Shells offer a different way to slice the same solid

A strip parallel to the rotation axis produces a thin cylindrical shell. Its contribution is circumference × height × thickness, leading to the shell integrand 2πρh, where ρ is the shell radius and h is its height. For a region to the right of the y-axis, rotated about that axis using vertical strips, ρ = x and h is upper minus lower. Shells can preserve equations written as y-functions when washers would require rewriting them as x-functions. The choice is geometric first and algebraic second. [3]

Rotate our example about the y-axis instead. Vertical strips now produce shells with radius x and height x − x². Thus V = 2π∫₀¹x(x − x²) dx = 2π[ x³/3 − x⁴/4 ]₀¹ = π/6 cubic units. Check the same solid with horizontal washers: their outer radius is √y and inner radius is y, so V = π∫₀¹(y − y²) dy = π/6. These methods agree because they describe the same rotation. The x-axis result differs because changing the axis changes the solid. [3]

6. Choose the setup and check the geometry

Use the following comparison to connect the desired quantity to a representative slice. It is a setup guide, not a ranking: the most convenient method depends on how the boundaries are expressed and where the axis lies. For a volume problem, draw the axis before assigning radii. Then compare a perpendicular slice with a parallel slice and choose the one whose dimensions are easiest to express. If the geometry changes across the interval, handle the resulting pieces separately rather than forcing one expression to describe them all. [1], [2], [3]

A shifted axis makes the distance rule particularly important. For example, when a boundary lies below y = 4, its distance from that axis is 4 minus its y-coordinate—not simply its function value. The cited ring example uses an inner radius 4 − x and an outer radius 4 − (x² − 2x). The farther boundary supplies the outer radius, even when that boundary is the lower curve. Before integrating, check that both radii are nonnegative and R ≥ r; this checks the geometry rather than merely the algebra. [2], [3]

Quantity or method	Slice geometry	Integral contribution	Main check
Area with dx	Vertical strip	(upper − lower) dx	Correct ordering across the interval [1]
Area with dy	Horizontal strip	(right − left) dy	Boundaries expressed as functions of y [1]
Washer or disc volume	Perpendicular to rotation axis	π(R² − r²) × thickness	Radii measured from the actual axis; r = 0 for a disc [3], [2]
Shell volume	Parallel to rotation axis	2π × radius × height × thickness	Radius is distance to the axis; height spans the region [3]

Frequently asked questions

Why can’t I always subtract the functions in their original order?

Geometric area uses the distance between boundaries, so vertical strips require upper minus lower. Reversing the order produces negative contributions. If the upper curve changes within the interval, split the integral at those changes and order each piece correctly. The names or order in which the functions were introduced do not determine their geometric positions. [1]

When should I integrate with respect to y?

Use dy when horizontal strips give a convenient description of the region. For area, subtract the left boundary from the right boundary. For washers about a vertical axis, horizontal strips are perpendicular to that axis. In either case, express the needed boundaries as functions of y and use lower and upper y-values as the limits. [1], [3]

Why is a washer’s area not π(R − r)²?

A washer is an outer disc with an inner disc removed. Subtracting their areas gives πR² − πr². The difference R − r measures the ring’s radial thickness; treating that thickness as a disc radius does not describe the washer. Write the two radii separately before squaring them to keep this distinction clear. [3]

Should shells and washers always give the same answer?

They should agree when both setups correctly describe the same region rotated about the same axis. They accumulate the volume using different pieces. In the worked example, y-axis shells and y-axis washers both give π/6. Comparing results for different axes is not the same check: the x-axis rotation gives 2π/15 because it creates a different solid. [2], [3]

What changes when the rotation axis is shifted?

The radius expressions change because every radius must be measured from the actual axis. A function value measures a coordinate, not automatically that distance. For a boundary below y = 4, the distance to the axis is 4 minus its y-coordinate. Determine which boundary is nearer and which is farther before assigning the inner and outer radii. [2]

Conclusion

The reliable sequence is region → slice → dimensions → integral. For area, measure the gap between boundaries. For revolution, decide whether the slice creates a washer or a shell, then measure every radius from the rotation axis. The example between y = x and y = x² shows the distinction: its area is 1/6, while its rotation produces 2π/15 about the x-axis and π/6 about the y-axis. Correct geometry makes the integration meaningful; evaluating an integral comes only after that setup is established. [1], [2], [3]

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