Navigation: Wiki Home > Skills > Area Between Curves (Horizontal Rectangles)
| 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.
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 $y = x + 2$.
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 $y$.
$$x = \frac{y - 3}{2}$$
If this was difficult, review your algebra skills.
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.
Legend: 🟡 Yellow = immediate prerequisites | 🟢 Green = this skill
| Property | Value |
|---|---|
| Section | Stewart §5.1 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
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 $y$.
y
d ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ┐
| |
| x=g(y) x=f(y) |
| (left) (right) |
| ┃▓▓▓▓▓▓▓▓┃ | A typical rectangle:
| ┃ AREA ┃ | height = dy
| ┃ A ┃ | width = f(y) - g(y)
| ┃▓▓▓▓▓▓▓▓┃ |
| |
c ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ┘
└─────────────────────── x| 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) |
| 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 |
Ask yourself:
Rule of thumb: Choose the variable that gives you ONE integral, not two.
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 $y$:
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.
Find the area of the region bounded by $x = 3$ (right), $x = y^2$ (left), $y = -1$, and $y = 2$.
Find the area enclosed by $x = y^2 - 4y$ and $x = 2y - y^2$.
Find the area of the region enclosed by $4x + y^2 = 12$ and $x = y$ using horizontal rectangles.
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.
Find the area of the region in the $xy$-plane satisfying both inequalities: $$x - 2y^2 \geq 0 \quad \text{and} \quad 1 - x - \vert y\vert \geq 0$$
✅ Checkpoint: If you can handle Level 5 problems with inequalities, you've mastered horizontal integration. You're ready for Volumes by Cylindrical Shells!
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?
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?
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.
✅ All boxes checked? You've mastered both vertical and horizontal area integration!
30-second decision process:
Common exam trap: A problem LOOKS like it needs two integrals with $dx$ but only needs ONE with $dy$. Always check both.
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—just swap $x \leftrightarrow y$.
Looking back:
Looking ahead:
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.
| 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