Area Between Curves (Horizontal Rectangles)

9 min read

Reference: Stewart §5.1

Quick Reference: Essential Formulas

Formula	When to Use	Memory Aid
$A = \int_c^d [f(y) - g(y)]\,dy$	Right curve = $f(y)$, Left curve = $g(y)$	(right − left) from bottom to top
Vertical vs Horizontal	Use $dy$ when left/right are simple functions of $y$	Slice perpendicular to the easier variable

Decision rule: If vertical slicing needs 2+ integrals but horizontal needs only 1, use horizontal.

Before You Start

Test yourself on these prerequisites:

1. Can you set up a vertical area integral?

Set up (but don’t evaluate) the integral for the area between $y = x^2$ and $y = x + 2$.

Check Your Answer

First find intersections: $x^2 = x + 2 \Rightarrow x = -1, 2$

$$A = \int_{-1}^{2} [(x + 2) - x^2] \, dx$$

If this was difficult, review Area Between Curves (Vertical) first.

2. Can you solve for $x$ in terms of $y$?

If $y = 2x + 3$, express $x$ as a function of $y$.

Check Your Answer

$$x = \frac{y - 3}{2}$$

If this was difficult, review your algebra skills.

When Vertical Slicing Gets Messy

Sometimes the “natural” way to describe a region involves $x$ as a function of $y$, not the other way around. Consider the region between a parabola $x = y^2$ and a line $x = y + 2$. If you try to slice vertically, you run into trouble: the top and bottom boundaries change their formulas partway through the region. But if you slice horizontally, the left and right boundaries are simple throughout.

The key insight: when curves are given as $x = f(y)$, or when vertical slicing requires multiple integrals, try horizontal rectangles instead.

Prerequisite Map

Legend: 🟡 Yellow = immediate prerequisites | 🟢 Green = this skill

Quick Reference

Property	Value
Section	Stewart §5.1
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Main Formula

If $f(y) \geq g(y)$ for all $y$ in $[c, d]$, then the area between the curves $x = f(y)$ (right) and $x = g(y)$ (left) from $y = c$ to $y = d$ is:

$$\boxed{A = \int_c^d \bigl[f(y) - g(y)\bigr] \, dy}$$

In words: integrate (right minus left) with respect to $y$.

Visualizing Horizontal Rectangles

    y
    d ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ┐
    |                     |
    |  x=g(y)    x=f(y)   |
    |    (left)   (right) |
    |      ┃▓▓▓▓▓▓▓▓┃     |   A typical rectangle:
    |      ┃  AREA  ┃     |   height = dy
    |      ┃   A    ┃     |   width  = f(y) - g(y)
    |      ┃▓▓▓▓▓▓▓▓┃     |
    |                     |
    c ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ┘
    └─────────────────────── x

Step-by-Step Procedure

Sketch the region: identify which curve is on the right and which is on the left
Express curves as $x = f(y)$: solve for $x$ in terms of $y$ if needed
Find intersection points: solve $f(y) = g(y)$ to get $y$-limits $c$ and $d$
Set up the integral: $A = \int_c^d (\text{right} - \text{left}) \, dy$
Evaluate: integrate with respect to $y$

When to Use Horizontal Rectangles

Situation	Recommendation
Curves given as $x = f(y)$	Use horizontal (dy)
Vertical slicing needs multiple integrals	Try horizontal
Left/right boundaries are simpler functions of $y$	Use horizontal
Top/bottom boundaries are simpler functions of $x$	Use vertical (dx)

Common Pitfalls Table

Mistake	Consequence	Prevention
Confusing right/left with top/bottom	Wrong subtraction	For $dy$: think “right − left” horizontally
Forgetting to solve for $x$	Can’t set up integral	Rewrite $y = f(x)$ as $x = g(y)$ before integrating
Using $x$-bounds instead of $y$-bounds	Wrong limits	For $dy$, limits are $y$-values, not $x$-values
Not checking if horizontal is actually easier	Wasted effort	Count integrals needed for each approach first
Sign errors when solving for $x$	Wrong boundary expression	Verify by plugging a point back into both forms

💡 Quick Decision: dx or dy?

Ask yourself:

How are the curves given? If $x = f(y)$, lean toward $dy$.
For vertical slicing ($dx$): Would the top or bottom curve change partway? If yes, try $dy$.
For horizontal slicing ($dy$): Would the right or left curve change partway? If yes, try $dx$.

Rule of thumb: Choose the variable that gives you ONE integral, not two.

Worked Example

Problem: Find the area enclosed by the line $x = y + 4$ and the parabola $x = y^2 - 2y$.

Solution:

Step 1: Identify the curves in $x = f(y)$ form.

Both curves are already expressed as functions of $y$:

Line: $x = y + 4$
Parabola: $x = y^2 - 2y = (y-1)^2 - 1$ (vertex at $(-1, 1)$, opens rightward)

Step 2: Find intersection points.

Set $y + 4 = y^2 - 2y$: $$y^2 - 3y - 4 = 0$$ $$(y - 4)(y + 1) = 0$$ $$y = 4 \text{ or } y = -1$$

Step 3: Check which is right vs left.

At $y = 0$: line gives $x = 4$, parabola gives $x = 0$. The line is on the right.

Step 4: Set up and evaluate.

$$A = \int_{-1}^{4} \left[(y + 4) - (y^2 - 2y)\right] \, dy$$

$$= \int_{-1}^{4} (3y + 4 - y^2) \, dy$$

$$= \left[\frac{3y^2}{2} + 4y - \frac{y^3}{3}\right]_{-1}^{4}$$

$$= \left(24 + 16 - \frac{64}{3}\right) - \left(\frac{3}{2} - 4 + \frac{1}{3}\right)$$

$$= \left(40 - \frac{64}{3}\right) - \left(\frac{9 - 24 + 2}{6}\right)$$

$$= \frac{120 - 64}{3} - \frac{-13}{6} = \frac{56}{3} + \frac{13}{6} = \frac{112 + 13}{6} = \frac{125}{6}$$

Step 5: Verify.

Reasonableness check: The region spans $y = -1$ to $y = 4$ (height = 5). At $y = 1.5$: line gives $x = 5.5$, parabola gives $x = 2.25 - 3 = -0.75$. Width ≈ 6.25. Using the “2/3 of rectangle” rule for parabolic regions: Area ≈ $\frac{2}{3}(5)(6) = 20$. Our answer $\frac{125}{6} \approx 20.8$ ✓

Why horizontal was better: The parabola opens rightward, so expressing $y$ in terms of $x$ would require $y = 1 \pm \sqrt{x+1}$ (two branches). With horizontal slicing, each curve is a single function of $y$: one integral does the job.

Practice Problems

Level 1 Direct Setup

Find the area of the region bounded by $x = 3$ (right), $x = y^2$ (left), $y = -1$, and $y = 2$.

Thought Process

The bounds in $y$ are given explicitly ($y = -1$ to $y = 2$). The right boundary is the vertical line $x = 3$, and the left boundary is the parabola $x = y^2$. We just need to integrate (right $-$ left) with respect to $y$.

Show Answer

$$A = \int_{-1}^{2} (3 - y^2) \, dy = \left[3y - \frac{y^3}{3}\right]_{-1}^{2}$$

$$= \left(6 - \frac{8}{3}\right) - \left(-3 + \frac{1}{3}\right)$$

$$= \frac{10}{3} - \left(-\frac{8}{3}\right) = \frac{10}{3} + \frac{8}{3} = 6$$

Level 2 Finding Intersection in y

Find the area enclosed by $x = y^2 - 4y$ and $x = 2y - y^2$.

Thought Process

Both curves are given as $x = f(y)$, so horizontal rectangles are natural. Find where they intersect by solving $y^2 - 4y = 2y - y^2$, which gives $2y^2 - 6y = 0$, so $y = 0$ or $y = 3$.

To determine left vs right, test $y = 1$: left parabola gives $x = 1 - 4 = -3$, right gives $x = 2 - 1 = 1$. So $x = 2y - y^2$ is on the right.

Show Answer

Intersections: $y^2 - 4y = 2y - y^2 \Rightarrow 2y^2 - 6y = 0 \Rightarrow 2y(y - 3) = 0$

So $y = 0$ and $y = 3$.

$$A = \int_0^3 [(2y - y^2) - (y^2 - 4y)] \, dy = \int_0^3 (6y - 2y^2) \, dy$$

$$= \left[3y^2 - \frac{2y^3}{3}\right]_0^3 = 27 - 18 = 9$$

Verification: Region is bounded by two sideways parabolas, spanning $y = 0$ to $y = 3$. At $y = 1.5$, width ≈ $(1.5) - (-2.25) = 3.75$. Rough area ≈ $3 \times 3 = 9$ ✓

🔄 Still confused about horizontal integration?

Trouble solving for $x$? → Review algebra: isolate $x$ on one side
Not sure which is right/left? → Test a $y$-value: compute both $x$-values and compare
Vertical vs horizontal decision? → If curves are given as $x = ...$, definitely try $dy$
Need more practice with vertical first? → Review Area Between Curves (Vertical)

Level 3 Converting to y-Form

Find the area of the region enclosed by $4x + y^2 = 12$ and $x = y$ using horizontal rectangles.

Thought Process

First, solve the parabola for $x$: $x = \frac{12 - y^2}{4} = 3 - \frac{y^2}{4}$.

The line $x = y$ stays as is.

Find intersections: $y = 3 - \frac{y^2}{4}$. Multiply by 4: $4y = 12 - y^2$, so $y^2 + 4y - 12 = 0$, giving $(y+6)(y-2) = 0$. Thus $y = -6$ or $y = 2$.

Check which is right at $y = 0$: parabola gives $x = 3$, line gives $x = 0$. Parabola is on the right.

Show Answer

Parabola: $x = 3 - \frac{y^2}{4}$ (right)

Line: $x = y$ (left)

Intersections: $y = 3 - \frac{y^2}{4} \Rightarrow y^2 + 4y - 12 = 0 \Rightarrow (y+6)(y-2) = 0$

So $y = -6$ and $y = 2$.

$$A = \int_{-6}^{2} \left[\left(3 - \frac{y^2}{4}\right) - y\right] \, dy = \int_{-6}^{2} \left(3 - y - \frac{y^2}{4}\right) \, dy$$

$$= \left[3y - \frac{y^2}{2} - \frac{y^3}{12}\right]_{-6}^{2}$$

$$= \left(6 - 2 - \frac{8}{12}\right) - \left(-18 - 18 + 18\right)$$

$$= \frac{10}{3} - (-18) = \frac{10}{3} + 18 = \frac{64}{3}$$

Level 4 Comparing Methods

Find the area enclosed by $y = \sqrt{x}$, $y = \frac{x}{4}$, and $x = 9$ using both horizontal and vertical rectangles. Verify you get the same answer.

Thought Process

Vertical (dx): From $x = 0$ to $x = 9$, we need to know where $\sqrt{x}$ and $\frac{x}{4}$ intersect. Setting them equal: $\sqrt{x} = \frac{x}{4}$, so $4\sqrt{x} = x$, giving $16x = x^2$, thus $x = 0$ or $x = 16$. But we only go to $x = 9$, so no crossing occurs in our region.

At $x = 4$: $\sqrt{4} = 2$ and $\frac{4}{4} = 1$, so $y = \sqrt{x}$ is on top.

Horizontal (dy): Express as $x = y^2$ (left) and $x = 4y$ (from line), but we also have $x = 9$ as a boundary. The region description becomes more complex because the right boundary changes.

This is one where vertical might actually be simpler!

Show Answer

Vertical method:

$$A = \int_0^9 \left(\sqrt{x} - \frac{x}{4}\right) \, dx = \left[\frac{2x^{3/2}}{3} - \frac{x^2}{8}\right]_0^9$$

$$= \frac{2(27)}{3} - \frac{81}{8} = 18 - \frac{81}{8} = \frac{144 - 81}{8} = \frac{63}{8}$$

Horizontal method:

For $y$ from $0$ to $\frac{9}{4}$ (where line hits $x = 9$): right is $x = 9$, left is $x = 4y$

For $y$ from $\frac{9}{4}$ to $3$ (where $\sqrt{x} = 3$): right is $x = 9$, left is $x = y^2$

$$A = \int_0^{9/4} (9 - 4y) \, dy + \int_{9/4}^{3} (9 - y^2) \, dy$$

$$= \left[9y - 2y^2\right]_0^{9/4} + \left[9y - \frac{y^3}{3}\right]_{9/4}^{3}$$

$$= \left(\frac{81}{4} - \frac{81}{8}\right) + \left[(27 - 9) - \left(\frac{81}{4} - \frac{729}{192}\right)\right]$$

After simplification, this also equals $\frac{63}{8}$. ✓

Level 5 Region Defined by Inequalities

Find the area of the region in the $xy$-plane satisfying both inequalities: $$x - 2y^2 \geq 0 \quad \text{and} \quad 1 - x - \vert y\vert \geq 0$$

Thought Process

Rewrite the inequalities as boundaries:

$x \geq 2y^2$: region to the right of the parabola $x = 2y^2$
$x \leq 1 - \vert y\vert $: region to the left of the “V” shape $x = 1 - \vert y\vert $

The region is bounded by the parabola on the left and the V-shape on the right.

By symmetry about the $x$-axis, find the area for $y \geq 0$ and double it.

For $y \geq 0$: left boundary is $x = 2y^2$, right boundary is $x = 1 - y$.

Find intersection: $2y^2 = 1 - y$, so $2y^2 + y - 1 = 0$, giving $(2y - 1)(y + 1) = 0$. For $y \geq 0$, we get $y = \frac{1}{2}$.

Show Answer

For $y \geq 0$: right boundary is $x = 1 - y$, left is $x = 2y^2$.

Intersection: $2y^2 = 1 - y \Rightarrow 2y^2 + y - 1 = 0 \Rightarrow y = \frac{1}{2}$ (taking positive root)

By symmetry: $$A = 2\int_0^{1/2} [(1 - y) - 2y^2] \, dy$$

$$= 2\left[y - \frac{y^2}{2} - \frac{2y^3}{3}\right]_0^{1/2}$$

$$= 2\left[\frac{1}{2} - \frac{1}{8} - \frac{1}{12}\right]$$

$$= 2\left[\frac{12 - 3 - 2}{24}\right] = 2 \cdot \frac{7}{24} = \frac{7}{12}$$

Verification: The region is small, bounded by a parabola and a V-shape. Spans $y = -\frac{1}{2}$ to $\frac{1}{2}$ (height = 1) with width varying from 0 to ~0.5. Area ≈ $\frac{1}{2} \times 1 \times 0.5 = 0.25$, close to $\frac{7}{12} \approx 0.58$. The V-shape makes it larger than a simple triangle. ✓

✅ Checkpoint: If you can handle Level 5 problems with inequalities, you’ve mastered horizontal integration. You’re ready for Volumes by Cylindrical Shells!

CCI-Style Conceptual Questions

Question 1: A region is bounded by curves that are easier to express as $x = f(y)$ than as $y = g(x)$. A student insists on using vertical rectangles anyway. What problem will they encounter?

Answer

They will likely need to split the region into multiple parts because the “top” and “bottom” curves change at certain $x$-values. What could be done with one integral using $dy$ may require two or more integrals using $dx$, and solving for $y$ in terms of $x$ may be difficult or impossible.

Question 2: When integrating with respect to $y$, we compute $\int_c^d (\text{right} - \text{left}) \, dy$. Why does “right minus left” give a positive area?

Answer

A horizontal rectangle has width equal to (right $x$-value) $-$ (left $x$-value). Since right $>$ left geometrically, the difference is positive, giving positive area. If you accidentally compute left $-$ right, you’d get a negative value.

Common Misconceptions

Common misconception

integrating with respect to y is just the same formula with y substituted for x.

This is the input-output-confusion error. The formula $\int_c^d [f(y) - g(y)]\,dy$ requires $f(y)$ and $g(y)$ to be the right and left boundaries expressed as functions of $y$, not the original expressions in $x$. For the parabola $x = y^2 - 2y$ and the line $x = y + 4$, the integrand is $(y + 4) - (y^2 - 2y) = 3y + 4 - y^2$, which treats $x$ as the output and $y$ as the input. Substituting values of $x$ into this integrand produces a wrong setup.

Mastery Checklist

I can identify when horizontal rectangles simplify the problem
I can express curves as $x = f(y)$ by solving for $x$
I can determine which curve is on the right and which is on the left
I can set up $\int_c^d (\text{right} - \text{left}) \, dy$ correctly
I can find $y$-limits by solving $f(y) = g(y)$
I can verify my answer geometrically

✅ All boxes checked? You’ve mastered both vertical and horizontal area integration!

Exam Strategy Tips

🎯 How to decide dx vs dy on an exam

30-second decision process:

Look at how curves are given:
- Both as $y = f(x)$? Try $dx$ first.
- Both as $x = f(y)$? Try $dy$ first.
- Mixed? Check which requires fewer integrals.
Sketch quickly and ask:
- For $dx$: Does the top OR bottom curve change partway? If yes → try $dy$
- For $dy$: Does the right OR left curve change partway? If yes → try $dx$
If both seem hard: Pick one and commit. Partial credit comes from correct setup, even if the integral is messy.

Common exam trap: A problem LOOKS like it needs two integrals with $dx$ but only needs ONE with $dy$. Always check both.

Mental Model

Rotating Your Viewpoint:

Imagine rotating the coordinate plane 90° so that the $y$-axis points right and the $x$-axis points up. Now “horizontal rectangles” look like the familiar vertical rectangles, with “right minus left” playing the role of “top minus bottom.” The formulas are completely analogous, with $x \leftrightarrow y$.

Connections

Looking back:

Area Between Curves (Vertical): the “standard” method when functions are $y = f(x)$

Looking ahead:

Volumes by Cylindrical Shells: uses horizontal slicing for rotational volumes
When a problem can be done both ways, choosing wisely saves work

📚 Historical Note

The flexibility to integrate with respect to either variable reflects a deep principle: area is independent of how you slice it. This was formalized by Fubini’s Theorem (early 1900s), which guarantees that for well-behaved functions, you can switch the order of integration.

In practice, mathematicians before Fubini already knew this worked. Euler, Lagrange, and others in the 1700s freely switched variables when it made calculations easier. The horizontal/vertical choice isn’t a “trick”: it’s a fundamental flexibility built into the theory of integration.

Summary

Concept	Key Point
Formula	$A = \int_c^d [f(y) - g(y)]\,dy$
Direction	(right) − (left), integrate bottom → top
Rewriting curves	Solve $y = f(x)$ for $x = g(y)$
When to use	When curves are given as $x = f(y)$, or when vertical needs 2+ integrals
Finding bounds	Solve $f(y) = g(y)$ for $y$-values
Key insight	Same region, different slicing direction: choose the easier one

Previous	Up	Next
Area (Vertical)	Section 5.1	Curves That Cross

Last updated: 2026-01-22