Area Between Curves (Integrating with Respect to y)

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 = $f(y)$, Left = $g(y)$	(right − left) from bottom to top
Horizontal slicing	Curves given as $x = f(y)$, or vertical slicing needs 2+ integrals	Slice perpendicular to the EASIER variable

Comparison:

Variable	Direction	Bounds	Subtraction
$dx$	left → right	$x$-values	top − bottom
$dy$	bottom → top	$y$-values	right − left

Before You Start

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

If $y^2 = 4x + 8$, express $x$ as a function of $y$.

Check Your Answer

$$4x = y^2 - 8 \Rightarrow x = \frac{y^2 - 8}{4} = \frac{y^2}{4} - 2$$

If this was difficult, review basic algebra.

2. Can you set up a vertical area integral?

Set up (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).

When Horizontal Beats Vertical

Sometimes a region is much easier to describe using horizontal slices instead of vertical ones. If the left and right boundaries are functions of $y$, then integrating with respect to $y$ can turn a complicated multi-part problem into a single clean integral.

The key insight: there’s nothing special about $x$. The same “subtract the boundaries” logic works in any direction. Just slice perpendicular to the variable you’re integrating with respect to.

Prerequisite Map

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

Quick Reference

Property	Value
Chapter	Chapter 5: Applications of Integration
Section	§5.1 Areas Between Curves
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The y-Integration Formula

If $f(y) \geq g(y)$ for all $y$ in $[c, d]$, then the area of the region bounded by $x = f(y)$ (right), $x = g(y)$ (left), and the horizontal lines $y = c$ and $y = d$ is:

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

In words: Integrate (right minus left) from bottom to top.

Visualizing Horizontal Slices

        y
        |  d ___________________________
        |    |                         |
        |    |        SHADED           |
        |    |         AREA         x = f(y)
        |    |    <-- width -->      (right)
        |    |                         |
        |  c |_________________________|
        |    x = g(y)
        |    (left)
        |_________________________________ x

Each horizontal slice has:

Width: $f(y) - g(y)$ (right minus left)
Height: $\Delta y$
Area: $[f(y) - g(y)] \Delta y$

Rewriting Curves as Functions of y

To use this method, you need the boundary curves expressed as $x = (\text{something in } y)$.

Example: The parabola $y^2 = 4x + 8$ can be solved for $x$: $$x = \frac{y^2 - 8}{4} = \frac{y^2}{4} - 2$$

Example: The line $y = 2x + 3$ becomes $x = \frac{y - 3}{2}$.

When to Choose y-Integration

Use y-integration when:

The curves are given as $x = f(y)$ (like $x = y^2$)
The region’s left/right boundaries are single functions of $y$, but the top/bottom would require splitting
One curve “wraps around” another vertically

Classic sign: If using x-integration would require two (or more) separate integrals, check if y-integration gives just one.

Comparison of Approaches

Situation	Integrate with respect to:
Curves given as $y = f(x)$ with clear top/bottom	$x$
Curves given as $x = f(y)$ with clear right/left	$y$
Sideways parabolas ($x = y^2$, etc.)	$y$ often easier
Region with a corner that splits horizontally	Try $y$

Common Pitfalls Table

Mistake	Consequence	Prevention
Using $x$-limits instead of $y$-limits	Completely wrong bounds	For $dy$, limits are $y$-values; find where curves meet in $y$
Confusing right/left with top/bottom	Wrong subtraction	For $dy$: test a $y$-value, compare $x$-values
Forgetting to solve for $x$	Can’t write the integrand	Rewrite $y = f(x)$ as $x = g(y)$ before setting up
Not checking if $dy$ is actually easier	Wasted effort on harder form	Count integrals needed each way first
Solving for $x$ incorrectly	Wrong integrand	Verify by substituting a point back

💡 Quick Decision Framework: dx or dy?

Step 1: Look at how curves are given.

Both $y = f(x)$? → Lean toward $dx$
Both $x = f(y)$? → Lean toward $dy$
Mixed? → Continue to Step 2

Step 2: Sketch the region. Ask:

For $dx$: Do I need to split because top/bottom changes? If yes → try $dy$
For $dy$: Do I need to split because right/left changes? If yes → try $dx$

Step 3: Count integrals.

The method needing ONE integral is usually better than needing TWO

When equal: Choose the one with simpler antiderivatives.

Practice Problems

Level 1 Identify Right and Left

For the region bounded by $x = y^2$ and $x = 4$ from $y = -2$ to $y = 2$:

(a) Which curve is the right boundary? (b) Which curve is the left boundary? (c) Write (but don’t evaluate) the integral for the area using y-integration.

Thought Process

To find right vs. left, I check x-values at a point in the interval. At $y = 0$:

$x = y^2 = 0$
$x = 4$

Since $4 > 0$, the line $x = 4$ is to the right of the parabola $x = y^2$.

Show Answer

(a) Right boundary: $x = 4$ (the vertical line)

(b) Left boundary: $x = y^2$ (the parabola opening rightward)

Level 2 Direct y-Integration

Find the area of the region bounded by $x = y^2 - 2$ and $x = 1$ from $y = 0$ to $y = 1$.

Thought Process

At $y = 0$: $x = y^2 - 2 = -2$ and $x = 1$. Since $1 > -2$, the line $x = 1$ is on the right.

At $y = 1$: $x = y^2 - 2 = -1$ and $x = 1$. Since $1 > -1$, still consistent.

Right: $x = 1$, Left: $x = y^2 - 2$.

Show Answer

$$A = \int_0^1 [1 - (y^2 - 2)]\,dy = \int_0^1 (3 - y^2)\,dy$$

$$= \left[3y - \frac{y^3}{3}\right]_0^1 = 3 - \frac{1}{3} = \boxed{\frac{8}{3}}$$

Level 3 Sideways Parabola and Line

Find the area enclosed by the line $x = y + 3$ and the parabola $x = y^2 - y$ by integrating with respect to $y$.

Thought Process

Both curves are already given as $x = f(y)$, so y-integration is natural.

Step 1: Find intersection points by setting the x-values equal: $$y + 3 = y^2 - y$$ $$y^2 - 2y - 3 = 0$$ $$(y - 3)(y + 1) = 0$$ $$y = 3 \text{ or } y = -1$$

Step 2: Determine right vs. left. At $y = 0$:

Line: $x = 0 + 3 = 3$
Parabola: $x = 0 - 0 = 0$

Since $3 > 0$, the line is on the right.

Step 3: The parabola $x = y^2 - y = (y - \frac{1}{2})^2 - \frac{1}{4}$ has vertex at $(-\frac{1}{4}, \frac{1}{2})$ and opens rightward, which confirms the line is to the right of the parabola between the intersections.

Show Answer

Both curves in $x = f(y)$ form:

Line: $x = y + 3$
Parabola: $x = y^2 - y$

Find intersections: $y + 3 = y^2 - y \Rightarrow y^2 - 2y - 3 = 0 \Rightarrow (y-3)(y+1) = 0$

So $y = 3$ or $y = -1$.

At $y = 0$: line gives $x = 3$, parabola gives $x = 0$. Line is on right.

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

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

$$= \left(9 + 9 - 9\right) - \left(1 - 3 + \frac{1}{3}\right)$$

$$= 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27 + 5}{3} = \boxed{\frac{32}{3}}$$

Verification: Region spans $y = -1$ to $y = 3$ (height = 4). At $y = 1$: width = $(1+3) - (1-1) = 4 - 0 = 4$. At $y = 0$: width = $3 - 0 = 3$. Rough area ≈ $\frac{2}{3}(4)(4) \approx 10.7$. Our answer $\frac{32}{3} \approx 10.67$ ✓

🔄 Still confused about y-integration?

Can’t solve for $x$? → Practice algebra: isolate $x$ on one side
Not sure which is right/left? → Test a $y$-value, compute both $x$-values, compare
Getting wrong bounds? → For $dy$, you need $y$-limits, not $x$-limits
Need more practice with $dx$ first? → Review Area (x-integration)

Level 4 Choosing the Right Variable

Find the area of the region bounded by $y = \sqrt{x}$, $y = 0$, and $y = x - 2$.

Hint: Try both approaches and see which is simpler.

Thought Process

x-integration approach:

For $0 \leq x \leq 1$: top is $y = \sqrt{x}$, bottom is $y = 0$
For $1 \leq x \leq 4$: I need to find where $\sqrt{x}$ and $x - 2$ intersect...

Actually, let me find all intersection points first:

$\sqrt{x} = 0$ at $x = 0$
$\sqrt{x} = x - 2$: Let $u = \sqrt{x}$, so $u = u^2 - 2$, giving $u^2 - u - 2 = 0$, so $u = 2$ or $u = -1$. Since $u \geq 0$, we have $u = 2$, meaning $x = 4$.
$x - 2 = 0$ at $x = 2$

So the vertices are at $(0, 0)$, $(4, 2)$, and $(2, 0)$.

For x-integration, from $x = 0$ to $x = 2$: top is $\sqrt{x}$, bottom is $0$. From $x = 2$ to $x = 4$: top is $\sqrt{x}$, bottom is $x - 2$.

That’s two integrals!

y-integration approach: Rewrite: $y = \sqrt{x} \Rightarrow x = y^2$ and $y = x - 2 \Rightarrow x = y + 2$.

For $0 \leq y \leq 2$: right is $x = y + 2$, left is $x = y^2$.

That’s one integral! Much simpler.

Show Answer

Using x-integration (harder): $$A = \int_0^2 \sqrt{x}\,dx + \int_2^4 [\sqrt{x} - (x-2)]\,dx$$

This requires two separate integrals.

Using y-integration (easier):

Express as functions of $y$:

From $y = \sqrt{x}$: $x = y^2$ (left boundary)
From $y = x - 2$: $x = y + 2$ (right boundary)

The curves meet where $y^2 = y + 2$, giving $y = 2$ or $y = -1$. Since $y \geq 0$ (from $y = \sqrt{x}$), we use $y = 0$ to $y = 2$.

$$A = \int_0^2 [(y + 2) - y^2]\,dy = \int_0^2 (y + 2 - y^2)\,dy$$

$$= \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_0^2 = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \boxed{\frac{10}{3}}$$

Level 5 Comparing Methods

The region $R$ is bounded by $x = y^2 - 4y$ and $x = 2y - y^2$.

(a) Sketch the region and find the intersection points.

(b) Set up the area integral using y-integration and evaluate it.

Thought Process

For part (a): Set $y^2 - 4y = 2y - y^2$ to find intersections. $$2y^2 - 6y = 0 \Rightarrow 2y(y - 3) = 0 \Rightarrow y = 0 \text{ or } y = 3$$

At $y = 0$: both curves give $x = 0$, so one intersection is $(0, 0)$. At $y = 3$: $x = 9 - 12 = -3$ and $x = 6 - 9 = -3$, so intersection is $(-3, 3)$.

For part (b): I need to determine which is right/left. At $y = 1$:

$x = 1 - 4 = -3$ (curve 1)
$x = 2 - 1 = 1$ (curve 2)

So $x = 2y - y^2$ is on the right.

For part (c): Both curves are parabolas opening left/right with vertices at different x-values. Converting to x would give multi-valued functions requiring splitting.

Show Answer

(a) Intersection points:

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

So $y = 0$ or $y = 3$.

Intersection points: $(0, 0)$ and $(-3, 3)$.

(b) y-integration:

At $y = 1$: $x = 1 - 4 = -3$ and $x = 2 - 1 = 1$.

Right: $x = 2y - y^2$, Left: $x = y^2 - 4y$.

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

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

(c) Why x-integration is harder:

The curve $x = y^2 - 4y = (y-2)^2 - 4$ is a parabola opening right with vertex at $(-4, 2)$.

The curve $x = 2y - y^2 = -(y-1)^2 + 1$ is a parabola opening left with vertex at $(1, 1)$.

If we solved for $y$ in terms of $x$, each parabola would give two $y$-values for most $x$-values (upper and lower branches). We would need to:

Integrate from $x = -4$ to $x = -3$ between the two branches of the first parabola
Integrate from $x = -3$ to $x = 0$ with mixed boundaries
Integrate from $x = 0$ to $x = 1$ between branches of the second parabola

This would require at least 3 integrals, compared to just 1 with y-integration.

✅ Checkpoint: If you can solve Level 5 problems comparing both methods, you’ve mastered the art of choosing the right integration variable!

Common Misconceptions

Common misconception

when integrating with respect to $y$, the integration limits are still the $x$-values where the curves intersect.

This is the input-output-confusion error. When the integral is $\int_c^d [f(y) - g(y)]\,dy$, the bounds $c$ and $d$ must be $y$-values, found by solving $f(y) = g(y)$ for $y$. For the region enclosed by $x = y + 3$ and $x = y^2 - y$, the curves intersect at $y = -1$ and $y = 3$, so the integral runs from $y = -1$ to $y = 3$, not from the corresponding $x$-values of $-4$ and $6$. Using $x$-values as limits in a $dy$ integral produces an integral over the wrong range and the wrong answer.

Mastery Checklist

I can rewrite curves in the form $x = f(y)$ when needed
I can identify which curve is the right boundary and which is the left boundary
I can set up and evaluate $\int_c^d [f(y) - g(y)]\,dy$
I can recognize when y-integration is simpler than x-integration
I understand that the choice of variable is about convenience, not correctness
I can verify my answer geometrically

✅ All boxes checked? You’ve mastered area between curves with both variables! Ready for volumes.

Exam Strategy Tips

🎯 How to choose dx vs dy on exams

The 30-second decision:

Look at curve forms:
- Both $y = ...$? → Try $dx$ first
- Both $x = ...$? → Try $dy$ first
- Mixed? → Sketch and check both
Count integrals needed:
- One method needs 1 integral, other needs 2? → Use the 1-integral method
- Both need 1? → Choose whichever has easier algebra
Check antiderivatives:
- Will you need substitution? Which variable makes that cleaner?

The definitive test: If a sideways parabola ($x = y^2$, etc.) is involved, $dy$ is almost always easier.

Partial credit tip: If unsure, write BOTH setups and compute the easier one. Showing you know both methods earns points.

Mental Model

The “Turning Your Head Sideways” Picture:

Everything you know about x-integration works for y-integration: just rotate your mental image 90°.

“Top minus bottom” becomes “Right minus left”
“From left to right” becomes “From bottom to top”
Vertical rectangles become horizontal rectangles

Decision rule: Look at the region. If vertical slicing gives a clean setup, use x. If horizontal slicing is cleaner, use y.

Connections

Looking back:

This builds directly on x-integration: same logic, different orientation

Looking ahead:

The shell method for volumes (Ch5 §3) uses horizontal slices even more naturally
Arc length formulas can be set up with either variable

Real-world connections:

Finding the area of a lake given depth measurements at different elevations
Computing work done against a force that varies with height
Calculating hydrostatic pressure on a dam (integrate with respect to depth)

📚 Historical Note

The flexibility to integrate with respect to either variable is a consequence of Fubini’s Theorem (Guido Fubini, 1907), which guarantees that for continuous functions, the order of integration doesn’t affect the result.

This was intuitively understood much earlier. Newton and Leibniz both freely switched variables when convenient. The formal proof had to wait for the development of measure theory in the early 20th century.

The practical wisdom: mathematicians and scientists have always chosen variables to make calculations easier. There’s no “correct” variable, only more or less convenient ones. Learning to make this choice quickly is a mark of mathematical maturity.

Summary

Concept	Key Point
Formula	$A = \int_c^d [f(y) - g(y)]\,dy$
Direction	(right) − (left), integrate bottom → top
Bounds	$y$-values (NOT $x$-values)
Rewriting	Solve $y = f(x)$ for $x = g(y)$
When to use	Curves given as $x = f(y)$, or when $dx$ needs 2+ integrals
Key insight	Same area, different slicing: choose the easier method

Previous	Up	Next
Area (x-integration)	Section 5.1	Finding Intersection Points

Last updated: 2026-01-22