Area Between Curves (y-functions) — Finding the Area Between Curves Expressed as Functions of y - Calculus 2

SECTION 1

Historical Context & Motivation

The problem of computing the area enclosed between curves has been a central preoccupation of mathematics since antiquity. Ancient Greek geometers such as Archimedes devised ingenious methods to compute areas bounded by parabolas and other conic sections, using the method of exhaustion — a precursor to modern integration. However, these classical approaches were restricted to specific geometric shapes and lacked a unifying algebraic framework that could handle arbitrary curves.

The development of analytic geometry by René Descartes and Pierre de Fermat in the seventeenth century allowed curves to be described by equations, opening the door to algebraic methods for area computation. When Newton and Leibniz independently formulated the Fundamental Theorem of Calculus, they established that integration — the inverse of differentiation — provides the definitive tool for computing areas under and between curves. Critically, their framework was flexible enough to accommodate integration with respect to any variable, not just x.

~250 BC • Archimedes' Method of Exhaustion

Archimedes computed the area of parabolic segments by inscribing successively finer polygons, establishing the foundational idea that areas can be approximated by summing infinitely many thin slices.

1637 • Descartes' Analytic Geometry

The publication of La Géométrie unified algebra and geometry, enabling curves to be expressed as equations in x and y — a prerequisite for formulating area problems as integrals.

1665–1676 • Newton & Leibniz Develop Calculus

The independent discoveries of the Fundamental Theorem of Calculus provided a systematic method to evaluate areas via antiderivatives, enabling integration with respect to either x or y.

18th–19th c. • Formalization and Riemann's Integral

Euler, Cauchy, and Riemann rigorously defined the integral, solidifying the theoretical underpinning for computing areas between arbitrary curves in both dx and dy formulations.

While most introductory calculus courses emphasize integrating with respect to x, many geometric configurations are far more naturally described when the curves are expressed as functions of y. Consider, for instance, the region between x = y² and x = 2y + 3. Setting up this area as a dx-integral would require splitting the region and solving for inverse functions, while a dy-integral yields a single, elegant expression. This section of the course addresses the key question: when and how should we integrate with respect to y to compute the area between two curves?

SECTION 2

Core Principles & Definitions

Before diving into computations, it is essential to internalize the foundational ideas that govern integration with respect to y. The approach mirrors the standard dx-method, but the orientation of the approximating rectangles changes from vertical to horizontal strips. Understanding this geometric shift is the key to correctly setting up and evaluating these integrals.

Horizontal Rectangles

When integrating with respect to y, the region is partitioned into thin horizontal strips of height Δy. Each strip has a width equal to the difference x_right(y) − x_left(y), where 'right' and 'left' refer to positions on the x-axis.

Right Minus Left

Analogous to the 'top minus bottom' rule for dx-integrals, the integrand in a dy-integral is always the rightmost curve minus the leftmost curve. This guarantees a non-negative integrand over the interval of integration.

Limits in y

The bounds of integration are the y-coordinates where the two curves intersect (or the y-values defining the region of interest). These are found by setting the two x-expressions equal and solving for y.

When to Choose dy over dx

Integrate with respect to y whenever the curves are more naturally expressed as x = f(y), or when a dx-setup would require splitting the integral across multiple subintervals due to the geometry of the region.

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KEY TAKEAWAY

Think of the dy-method like scanning a document horizontally rather than vertically. If you are measuring the width of a river from bank to bank, you would naturally draw horizontal lines across the river (measuring the distance from the left bank to the right bank at each elevation). Each horizontal line is a single strip, and the total area is obtained by stacking all such strips from the lowest to the highest point. Similarly, when curves are given as functions of y, the natural 'scanning direction' is horizontal, and each strip measures right minus left.

SECTION 3

Visual Explanation

The diagram below illustrates the core geometric idea behind integrating with respect to y. Two curves are expressed as functions of y, and the shaded region between them is partitioned into horizontal rectangular strips. Each strip has a height of Δy and a width of x_right(y) − x_left(y). The total area is obtained by summing these strips from y = c to y = d.

The shaded region between x = g(y) (left, violet) and x = f(y) (right, cyan) is partitioned into horizontal strips. Each strip has width f(y) − g(y) and height Δy. Summing from y = c to y = d via integration yields the total enclosed area.

Notice the fundamental geometric logic at work: for a fixed value of y, the horizontal distance from the left curve to the right curve gives the width of that particular strip. Multiplying by the infinitesimal height dy and integrating over the full range of y-values accumulates the total area. This is the exact analog of the dx-method, with the roles of x and y interchanged.

SECTION 4

Mathematical Framework

The formal setup for computing the area between two curves expressed as functions of y closely parallels the dx-formulation. Suppose two curves are given by x = f(y) and x = g(y), where f(y) ≥ g(y) on the interval [c, d]. Then the area of the region enclosed between the curves and the horizontal lines y = c and y = d is given by the following integral.

AREA BETWEEN CURVES (dy-FORM)

A = ∫[c to d] [f(y) − g(y)] dy

where f(y) is the rightmost curve, g(y) is the leftmost curve, and c and d are the y-coordinates at which the curves intersect (or the y-values bounding the region).

In many problems, the limits of integration c and d are determined by finding the intersection points of the two curves. Setting f(y) = g(y) and solving for y provides these critical values. The integrand f(y) − g(y) must remain non-negative throughout [c, d]; if the curves cross within the interval, the integral must be split into subintervals where the identity of the right and left curves is consistent.

RIEMANN SUM MOTIVATION

A ≈ Σ [f(yᵢ) − g(yᵢ)] Δy → A = lim (n→∞) Σ [f(yᵢ) − g(yᵢ)] Δy

The integral is the limit of a Riemann sum over n horizontal strips, each of height Δy = (d − c)/n and width f(yᵢ) − g(yᵢ) at sample point yᵢ.

When the region is bounded by curves that serve as functions of y but not of x, the dy-formulation is not merely convenient — it is necessary to avoid multi-part integrals. A classic example is the region between x = y² (a leftward-opening parabola in the standard x-y plane) and a line such as x = y + 2. Solving for y in terms of x for the parabola yields two branches (y = √x and y = −√x), forcing a dx-setup to split. In contrast, the dy-approach uses the single expressions x = y² and x = y + 2 directly.

COMPARISON: dx vs dy FORMULATION

dx: A = ∫[a to b] [f_top(x) − f_bottom(x)] dx dy: A = ∫[c to d] [f_right(y) − f_left(y)] dy

Both formulations compute the same area. The choice depends on which variable produces a simpler integrand and avoids splitting the integral.

SECTION 5

When to Integrate with Respect to y — A Decision Guide

Choosing between integrating with respect to x or y is a strategic decision that can dramatically simplify (or complicate) a problem. The following diagram and discussion present a systematic decision framework. In general, you should prefer the dy-method whenever the region's left and right boundaries are naturally described as single-valued functions of y, or whenever a dx-setup would force you to split the integral into multiple pieces.

Decision flowchart for choosing between dx and dy integration. When curves are naturally functions of y, or when a dx-setup requires splitting, the dy-method is preferred. The bottom panels show a classic example where x = y² and x = y + 2 are most naturally integrated with respect to y.

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Common Pitfall

When setting up a dy-integral, students frequently confuse 'right' and 'left.' Always sketch the region and identify which curve has the larger x-value for a given y in the interval of integration. The integrand is x_right − x_left. Getting this backwards yields a negative area.

Indicator	Use dx	Use dy
Curves given as	y = f(x), y = g(x)	x = f(y), x = g(y)
Strip orientation	Vertical (width dx)	Horizontal (height dy)
Integrand sign rule	Top − Bottom	Right − Left
Limits of integration	x-values of intersection	y-values of intersection
Requires splitting when	Top/bottom curves switch	Right/left curves switch

Comparison of dx and dy integration approaches

SECTION 6

Worked Example

Let us compute the area of the region enclosed between the curves x = y² and x = y + 2. This is a classic problem that showcases the elegance of integrating with respect to y.

Area Between x = y² and x = y + 2

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Step 1 — Identify the Curves and Sketch

The curve x = y² is a parabola opening to the right with vertex at the origin. The curve x = y + 2 is a straight line with slope 1 (when viewed as a function of y, the line has x increasing at the same rate as y). From a quick sketch, the line lies to the right of the parabola in the region of interest.
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Step 2 — Find the Intersection Points

Set the two expressions equal: y² = y + 2. Rearranging gives y² − y − 2 = 0, which factors as (y − 2)(y + 1) = 0.

y = −1 and y = 2. These are the limits of integration: c = −1, d = 2.
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Step 3 — Determine Right and Left

For a sample value between −1 and 2, say y = 0: f(0) = 0 + 2 = 2 (line) and g(0) = 0² = 0 (parabola). Since 2 > 0, the line x = y + 2 is the right curve and x = y² is the left curve throughout the interval [−1, 2].
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Step 4 — Set Up the Integral

A = ∫ from −1 to 2 of [(y + 2) − y²] dy = ∫ from −1 to 2 of (y + 2 − y²) dy.
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Step 5 — Evaluate the Integral

Find the antiderivative: (y²/2) + 2y − (y³/3). Evaluate at the bounds: At y = 2: (4/2) + 4 − (8/3) = 2 + 4 − 8/3 = 6 − 8/3 = 10/3. At y = −1: (1/2) + (−2) − (−1/3) = 1/2 − 2 + 1/3 = −7/6. A = 10/3 − (−7/6) = 10/3 + 7/6 = 20/6 + 7/6 = 27/6.

A = 27/6 = 9/2 = 4.5 square units
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Step 6 — Verify (Optional: Compare with dx)

A dx-approach would require solving x = y² as y = ±√x and x = y + 2 as y = x − 2, then splitting the integral at x = 1 (where y = −1 meets the parabola at (1, −1)) and x = 4 (where y = 2 meets x = y + 2 at (4, 2)). The dx-setup requires two integrals: ∫₀¹ [√x − (−√x)] dx + ∫₁⁴ [(x − 2) − (−√x)] dx, which is considerably more work. The dy-method gave us the answer in a single integral.

SECTION 7

Strengths, Limitations & Comparisons

The dy-integration technique is a powerful addition to the integration toolkit, but like every method it has specific contexts where it shines and others where alternative approaches are preferable. Understanding these trade-offs helps you select the most efficient strategy for any given problem.

Aspect	Strengths of dy-Integration	Limitations / Caveats
Avoiding split integrals	Often reduces multi-part dx-integrals to a single dy-integral, saving significant computation.	If the right/left relationship switches in y, the dy-integral itself must be split.
Natural representation	Ideal when curves are already given as x = f(y), avoiding the need to find inverse functions.	When curves are naturally y = f(x), converting to x = g(y) may be algebraically difficult or impossible in closed form.
Geometric intuition	Horizontal strips align naturally with the width of the region, often making the setup more intuitive.	Students accustomed to dx-methods may initially struggle with the 'right minus left' orientation.
Integration difficulty	The resulting integrand may be simpler and easier to antidifferentiate.	Occasionally the dy-integrand is harder to evaluate than the corresponding dx-integrand — always compare before committing.

Strengths and limitations of integrating with respect to y

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KEY TAKEAWAY

Choosing between dx and dy integration is analogous to choosing a coordinate system in physics: the answer doesn't change, but the right choice of coordinates simplifies the work dramatically. Just as polar coordinates simplify problems with circular symmetry, dy-integration simplifies area problems where curves are naturally expressed as functions of y. The sign of a good mathematician is recognizing which framework suits the geometry of the region before diving into computation.

SECTION 8

Connections to Advanced Theory

The technique of integrating with respect to y is not merely a calculational convenience — it connects to deeper structures in multivariable calculus, differential geometry, and applied mathematics. Understanding these connections enriches your conceptual grasp and prepares you for more advanced coursework.

This Course (Calc 2)	Advanced Extension
Area = ∫[c to d] [f(y) − g(y)] dy as a single integral	In Calc 3, the area becomes ∬ dA over a region R, computed as an iterated integral. Choosing dy first corresponds exactly to horizontal slicing.
Choosing dx vs dy for simplicity	Generalizes to choosing the order of integration (dy dx vs dx dy) in double integrals, governed by Fubini's Theorem.
Volume by shells: V = 2π ∫ y·[f(y) − g(y)] dy	The shell method for volumes of revolution uses the same dy-framework, with horizontal strips rotated about the x-axis.
Area between explicit curves x = f(y)	In differential geometry, areas between parametric or implicitly defined curves require line integrals and Green's Theorem: A = ½ ∮ (x dy − y dx).

How dy-integration connects to more advanced topics

In particular, the shell method for computing volumes of revolution — which you will likely encounter later in this course — builds directly on dy-integration. When a region bounded by curves x = f(y) and x = g(y) is revolved about the x-axis, each horizontal strip at height y generates a cylindrical shell with radius y and height f(y) − g(y). The total volume is V = 2π ∫ y·[f(y) − g(y)] dy. Mastering the dy area setup now provides a seamless bridge to these more complex applications.

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Looking Ahead

In multivariable calculus, when you encounter a double integral ∬ f(x,y) dA over a non-rectangular region, you will need to decide whether to integrate dx dy or dy dx. The skills you are developing now — sketching regions, determining which curve is 'outer' vs. 'inner,' and identifying where curves intersect — transfer directly to setting up iterated integrals in Calc 3.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain geometrically why integrating with respect to y uses the rule 'right minus left' rather than 'top minus bottom.' How does the orientation of the approximating rectangles change when switching from a dx-integral to a dy-integral?

Show answer

When integrating with respect to x, we use vertical strips of width dx whose height is determined by the vertical distance between curves — hence 'top minus bottom.' When integrating with respect to y, we instead use horizontal strips of height dy, and the relevant dimension of each strip is its horizontal extent — the distance from the leftmost curve to the rightmost curve. Since moving rightward increases x, the width is x_right − x_left, giving the 'right minus left' rule. The rectangles rotate 90° from the dx-setup, and correspondingly, the role of the two coordinates is exchanged: the limits become y-values, and the integrand measures a horizontal distance.

Question 2: BASIC CALCULATION

Find the area enclosed between the curves x = y and x = y² − 2 by integrating with respect to y.

Show answer

First, find the intersection points by setting y = y² − 2, which gives y² − y − 2 = 0, factoring as (y − 2)(y + 1) = 0, so y = −1 and y = 2. For y between −1 and 2, test y = 0: x = 0 (line) and x = −2 (parabola), so the line x = y is to the right. The integral is A = ∫ from −1 to 2 of [y − (y² − 2)] dy = ∫ from −1 to 2 of (y + 2 − y²) dy. The antiderivative is y²/2 + 2y − y³/3. Evaluating: at y = 2, we get 2 + 4 − 8/3 = 10/3; at y = −1, we get 1/2 − 2 + 1/3 = −7/6. Thus A = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2.

Question 3: INTERMEDIATE

Find the area of the region bounded by x = 3 − y² and x = y² − 1 using integration with respect to y.

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Set 3 − y² = y² − 1, so 4 = 2y², giving y² = 2 and y = ±√2. For y in [−√2, √2], test y = 0: x = 3 (first parabola) vs. x = −1 (second parabola), so x = 3 − y² is the right curve. The integral is A = ∫ from −√2 to √2 of [(3 − y²) − (y² − 1)] dy = ∫ from −√2 to √2 of (4 − 2y²) dy. By symmetry (the integrand is even), this equals 2 ∫ from 0 to √2 of (4 − 2y²) dy = 2[4y − 2y³/3] from 0 to √2 = 2[4√2 − 2(2√2)/3] = 2[4√2 − 4√2/3] = 2 × (8√2/3) = 16√2/3 ≈ 7.54 square units.

Question 4: APPLIED

A cross-section of a river channel is modeled by the curves x = y² (left bank) and x = 8 − y² (right bank), with y measured in meters from the center depth. Find the cross-sectional area of the channel. Then, if water flows at an average velocity of 1.5 m/s, estimate the volumetric flow rate in m³/s.

Show answer

Setting y² = 8 − y² gives 2y² = 8, so y = ±2. The right bank is x = 8 − y² and the left bank is x = y² for y ∈ [−2, 2]. The cross-sectional area is A = ∫ from −2 to 2 of [(8 − y²) − y²] dy = ∫ from −2 to 2 of (8 − 2y²) dy. Using symmetry: A = 2 ∫ from 0 to 2 of (8 − 2y²) dy = 2[8y − 2y³/3] from 0 to 2 = 2[16 − 16/3] = 2 × 32/3 = 64/3 ≈ 21.33 m². The volumetric flow rate is Q = A × v = (64/3)(1.5) = 32 m³/s. This problem illustrates how dy-integration naturally models physical cross-sections where the horizontal extent varies with vertical position.

Question 5: CRITICAL THINKING

Consider the region R bounded by x = y³ − y and x = 0. (a) Sketch the region carefully, noting that the cubic crosses x = 0 at y = −1, 0, and 1. (b) Set up — but do not combine into a single integral — the dy-integral(s) needed to compute the total area of R. (c) Explain why a single integral ∫ from −1 to 1 of (y³ − y) dy would give the wrong answer, and state what value it does yield.

Show answer

(a) The curve x = y³ − y factors as x = y(y² − 1) = y(y − 1)(y + 1), crossing x = 0 at y = −1, 0, 1. For y ∈ (−1, 0), test y = −0.5: x = −0.125 + 0.5 = 0.375 > 0, so the cubic is to the right of x = 0. For y ∈ (0, 1), test y = 0.5: x = 0.125 − 0.5 = −0.375 < 0, so the cubic is to the left of x = 0. (b) The total area requires two integrals: A = ∫ from −1 to 0 of [(y³ − y) − 0] dy + ∫ from 0 to 1 of [0 − (y³ − y)] dy = ∫ from −1 to 0 of (y³ − y) dy + ∫ from 0 to 1 of (y − y³) dy. Each evaluates to 1/4, so A = 1/2. (c) A single integral ∫ from −1 to 1 of (y³ − y) dy = 0 because y³ − y is an odd function, and the positive and negative contributions cancel. This illustrates the critical principle that signed area ≠ geometric area when the integrand changes sign — one must split at the zeros and take absolute values (or use |f(y) − g(y)|).

SUMMARY

Summary

When curves are expressed as functions of y, the area between them is computed using horizontal rectangular strips of height dy and width equal to right curve minus left curve. The fundamental formula is A = ∫ from c to d of [f(y) − g(y)] dy, where c and d are the y-coordinates of intersection, f(y) is the rightmost curve, and g(y) is the leftmost curve. This method is preferred whenever curves are more naturally described as x = f(y), or whenever a dx-setup would require splitting the integral into multiple parts.

The key procedural steps are: (1) express both boundaries as x = f(y), (2) find intersection points by solving f(y) = g(y) to determine the limits of integration in y, (3) identify which curve is to the right for each subinterval, and (4) integrate. If the right-left relationship reverses within the interval, the integral must be split at the crossover points. This technique connects directly to the shell method for volumes of revolution and to iterated double integrals in multivariable calculus, making it a foundational skill for advanced coursework.