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Integrate horizontally to compute enclosed areas when curves are naturally expressed as functions of y.
The idea of computing areas bounded by curves is one of the oldest motivations for what eventually became integral calculus. Ancient Greek mathematicians, most notably Archimedes, devised the method of exhaustion to approximate areas enclosed by parabolas and circles by inscribing sequences of polygons whose areas converged to the desired value. While these early techniques were ingenious, they were limited to specific geometric shapes and lacked a general algebraic framework for handling arbitrary curves. The conceptual leap from vertical slicing—integrating with respect to x—to horizontal slicing with respect to y became natural only after Leibniz and Newton established the formal machinery of the integral in the late seventeenth century, enabling mathematicians to choose the variable of integration that best suits a given geometric configuration.
When two curves are more naturally described as functions of y—for instance, a sideways parabola x = y² and a vertical line x = 4—slicing vertically with respect to x would require splitting the region or solving for multiple branches. The central question this lesson addresses is: how can we set up and evaluate a single integral that captures the entire enclosed area by integrating with respect to y?
Before diving into calculations, it is essential to establish the foundational ideas that govern horizontal integration. When we integrate with respect to y, we think of each infinitesimal strip as a horizontal rectangle of width dy and length equal to the horizontal distance between the right curve and the left curve. The following principles outline the framework for setting up such integrals correctly.
The diagram below illustrates the core geometric idea. Two curves, x = f(y) and x = g(y), enclose a region in the xy-plane. Rather than slicing the region into vertical strips (which might require splitting the region or dealing with multi-valued functions), we slice it into horizontal strips of thickness dy. Each strip extends from the left boundary g(y) to the right boundary f(y), so its length is f(y) − g(y). Summing these infinitesimal rectangular areas from y = c to y = d yields the total enclosed area.
The formal setup for computing the area between two curves expressed as functions of y mirrors the familiar dx-integration formula but with the roles of the axes swapped. Where vertical integration uses "top minus bottom" with respect to x, horizontal integration uses "right minus left" with respect to y. The derivation begins with a Riemann sum of horizontal rectangles and passes to a limit, yielding the definite integral below.
To derive this, partition the interval [c, d] into n sub-intervals each of width Δy = (d − c)/n. In the i-th sub-interval, choose a sample point yi*. The area of the i-th horizontal rectangle is [f(yi*) − g(yi*)] Δy. Summing and taking the limit as n → ∞ yields the integral.
To solidify the technique, let us walk through the setup procedure with a concrete geometric example. Consider the region enclosed by the curves x = y² (a rightward-opening parabola) and x = y + 2 (a line). Both curves are already expressed as functions of y, making horizontal integration the natural choice. The diagram below shows the region and the representative horizontal strip.
Let us compute the area of the region enclosed by x = y² and x = y + 2, the same region discussed in Section 5. This worked example will carry the calculation from setup through evaluation, showing every algebraic detail.
A natural question arises: when should you integrate with respect to y rather than x? The answer depends on the geometry of the region and the algebraic form of the bounding curves. The table below summarizes the key differences and considerations, helping you choose the more efficient approach for a given problem on the AP exam.
| Criterion | dx-Integration (Vertical Strips) | dy-Integration (Horizontal Strips) |
|---|---|---|
| Integrand | f(x) − g(x) (top minus bottom) | f(y) − g(y) (right minus left) |
| Limits | x-coordinates of intersections | y-coordinates of intersections |
| Best When | Curves are naturally y = f(x); region has consistent top/bottom | Curves are naturally x = f(y); region has consistent right/left |
| Common Pitfall | Top/bottom curves switch, requiring split integrals | Forgetting to solve for x in terms of y before integrating |
| Typical Curves | y = x², y = sin x, y = eˣ | x = y², x = √y, x = eʸ |
The technique of integrating with respect to y is not an isolated skill—it forms a conceptual bridge to several advanced topics that appear later in AP Calculus BC and in multivariable calculus. Understanding horizontal slicing deeply prepares you for the more general idea of choosing an appropriate variable (or coordinate system) for integration.
| This Lesson | Advanced Extension |
|---|---|
| Horizontal strips: ∫ [f(y) − g(y)] dy | Volumes by the shell method: V = 2π ∫ y [f(y) − g(y)] dy (shells are generated by rotating horizontal strips about the x-axis) |
| Choosing dy vs. dx for area | Choosing the order of integration in double integrals: ∫∫ dA can be evaluated as ∫∫ dx dy or ∫∫ dy dx depending on region geometry |
| Solving y = f(x) for x to get x = g(y) | Parametric and polar area: A = ½ ∫ r² dθ uses the natural parameter θ, just as we use y when it is the natural parameter |
| Right minus left across a single interval in y | Cross-sectional area method for volumes: A(y) integrated along y yields volume when cross-sections are perpendicular to the y-axis |
In particular, the shell method for computing volumes of revolution is a direct descendant of horizontal-strip integration. When you rotate a horizontal strip about the x-axis, the strip sweeps out a cylindrical shell whose volume element is 2πy × [f(y) − g(y)] × dy. Mastering the area-between-curves setup in this lesson therefore gives you a head start on one of the most important volume techniques in BC Calculus.
When a region in the xy-plane is bounded by curves more naturally expressed as functions of y, integrating with respect to y allows you to compute the enclosed area using a single integral of the form A = ∫ from c to d [f(y) − g(y)] dy, where f(y) is the right boundary and g(y) is the left boundary. The limits of integration are the y-coordinates of the intersection points, found by setting f(y) = g(y).
The key strategy is to sketch the region, identify which curve lies to the right at a test y-value, and confirm that the right-minus-left relationship holds throughout the interval. If the curves cross, split the integral at the crossing y-value. This technique avoids the multi-integral complications of dx-integration for sideways-opening curves, and it directly extends to the shell method for volumes of revolution and to iterated double integrals in multivariable calculus.
Keep learning with more lessons from the same subject.