Unit 8.5 – Finding the Area Between Curves Expressed as Functions of y

AP® Calculus AB & BC | Integrating with Respect to y

Why This Matters: Sometimes it's easier (or necessary!) to integrate with respect to \(y\) instead of \(x\). When curves are expressed as functions of \(y\) (like \(x = f(y)\)), or when a region is more naturally described by horizontal slices, integrating with respect to \(y\) is the way to go! This technique appears on AP® exams and is essential for solving certain area problems that would be difficult or impossible using \(dx\). Master both methods!

🔄 Functions of y vs. Functions of x

THE FUNDAMENTAL SHIFT

Comparing Integration Methods
Aspect	Functions of x	Functions of y
Variable	Integrate with respect to \(x\)	Integrate with respect to \(y\)
Functions	\(y = f(x)\)	\(x = f(y)\)
Slices	Vertical rectangles	Horizontal rectangles
Limits	\(a \leq x \leq b\)	\(c \leq y \leq d\)
Formula	\(\int_a^b [\text{top} - \text{bottom}]\,dx\)	\(\int_c^d [\text{right} - \text{left}]\,dy\)

📐 The Fundamental Formula

Area Between Curves (Functions of y)

THE FORMULA:

\[ A = \int_c^d [\text{right} - \text{left}] \, dy \]

Or more formally:

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

Where:

\(f(y)\) = rightmost function (larger x-value)
\(g(y)\) = leftmost function (smaller x-value)
\([c, d]\) = interval in y (vertical bounds)
\(f(y) \geq g(y)\) on \([c, d]\)

📝 Key Insight: Just like with \(dx\), always subtract the LEFT function from the RIGHT function. Think: "Right minus Left" instead of "Top minus Bottom"

🤔 When to Use Functions of y

Use integration with respect to \(y\) when:

Functions are already given as \(x = f(y)\): Natural choice
Region bounded by vertical lines: Easier with horizontal slices
Avoid multiple integrals: Single integral with \(dy\) vs. splitting with \(dx\)
Function fails vertical line test: But passes horizontal line test
Problem specifically asks: "Integrate with respect to \(y\)"

💡 Pro Tip: Sometimes you can choose either method! Pick whichever makes the problem simpler—fewer integrals, easier algebra, simpler limits.

🔄 Converting Between x and y Functions

From \(y = f(x)\) to \(x = g(y)\):

Solve for \(x\) in terms of \(y\)!

Examples:

\(y = x^2\) → \(x = \pm\sqrt{y}\) (need to specify which branch!)
\(y = 2x + 1\) → \(x = \frac{y-1}{2}\)
\(y = e^x\) → \(x = \ln y\)
\(y = \sin x\) → \(x = \arcsin y\) (on appropriate domain)

📋 Step-by-Step Process

Complete Method for Functions of y

The 6-Step Approach:

Express curves as \(x = f(y)\): Solve for \(x\) in terms of \(y\)
Find intersection points: Set functions equal, solve for \(y\) values (\(c\) and \(d\))
Determine which is right/left: Test a \(y\)-value or analyze
Set up integral: \(\int_c^d [\text{right} - \text{left}]\,dy\)
Evaluate the integral: Find antiderivative with respect to \(y\)
State answer with units: Area = ___ square units

📖 Comprehensive Worked Examples

Example 1: Basic Functions of y

Problem: Find the area between \(x = y^2\) and \(x = y + 2\).

Solution:

Step 1: Functions already in form \(x = f(y)\)

\(x = y^2\) and \(x = y + 2\) ✓

Step 2: Find intersection points

\[ y^2 = y + 2 \]

\[ y^2 - y - 2 = 0 \]

\[ (y-2)(y+1) = 0 \]

So \(y = -1\) and \(y = 2\)

Step 3: Determine which is on right

Test \(y = 0\):

\(y^2 = 0\) and \(y + 2 = 2\)

So \(x = y + 2\) is on the right!

Step 4: Set up integral

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

Step 5: Evaluate

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

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

\[ = \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) \]

\[ = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20 + 7}{6} = \frac{9}{2} \]

ANSWER: Area = \(\frac{9}{2} = 4.5\) square units

Example 2: Converting from y = f(x)

Problem: Find the area bounded by \(y = x^2\), \(y = 0\), and \(x = 2\) using integration with respect to \(y\).

Step 1: Convert to functions of y

From \(y = x^2\): \(x = \sqrt{y}\) (taking positive root since \(x \geq 0\))

Vertical line: \(x = 2\)

Horizontal line: \(y = 0\) is the lower bound

Step 2: Find y-limits

Lower: \(y = 0\)

Upper: When \(x = 2\), \(y = 2^2 = 4\)

So integrate from \(y = 0\) to \(y = 4\)

Step 3: Set up integral

Right function: \(x = 2\)

Left function: \(x = \sqrt{y}\)

\[ A = \int_0^4 (2 - \sqrt{y}) \, dy \]

Step 4: Evaluate

\[ A = \left[2y - \frac{2y^{3/2}}{3}\right]_0^4 \]

\[ = 8 - \frac{2(8)}{3} = 8 - \frac{16}{3} = \frac{8}{3} \]

Example 3: Why dy is Better Than dx

Problem: Find the area bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\).

Method 1: With dx (straightforward)

\[ A = \int_0^4 \sqrt{x} \, dx = \frac{16}{3} \]

Method 2: With dy (to practice)

From \(y = \sqrt{x}\): \(x = y^2\)

Limits: \(y = 0\) to \(y = 2\) (when \(x = 4\))

Right: \(x = 4\), Left: \(x = y^2\)

\[ A = \int_0^2 (4 - y^2) \, dy = \left[4y - \frac{y^3}{3}\right]_0^2 = \frac{16}{3} \]

Note: Both methods work! Choose based on which is simpler.

Example 4: Region Between Two Parabolas

Problem: Find area between \(x = y^2 - 4\) and \(x = 3 - 2y^2\).

Find intersections:

\[ y^2 - 4 = 3 - 2y^2 \]

\[ 3y^2 = 7 \quad \Rightarrow \quad y = \pm\sqrt{\frac{7}{3}} \]

Determine position:

At \(y = 0\): \(x = -4\) and \(x = 3\)

So \(x = 3 - 2y^2\) is on the right

Set up and evaluate:

\[ A = \int_{-\sqrt{7/3}}^{\sqrt{7/3}} [(3-2y^2) - (y^2-4)] \, dy \]

\[ = \int_{-\sqrt{7/3}}^{\sqrt{7/3}} (7 - 3y^2) \, dy \]

By symmetry (even function):

\[ = 2\int_0^{\sqrt{7/3}} (7 - 3y^2) \, dy = 2\left[7y - y^3\right]_0^{\sqrt{7/3}} \]

⚖️ Choosing Between dx and dy

Decision Guide:

Use \(dx\) when:

Functions naturally expressed as \(y = f(x)\)
Region bounded by vertical lines \(x = a\) and \(x = b\)
Thinking of vertical slices makes sense

Use \(dy\) when:

Functions naturally expressed as \(x = f(y)\)
Region bounded by horizontal lines \(y = c\) and \(y = d\)
Using \(dx\) requires splitting into multiple integrals
Converting to \(x = f(y)\) is simpler than staying with \(y = f(x)\)

💡 Essential Tips & Strategies

✅ Success Strategies:

Sketch the region: Visualize horizontal slices
Right minus left: Just like top minus bottom with \(dx\)
Check your bounds: \(c\) and \(d\) are \(y\)-values
Solve for x carefully: Watch for ± when taking square roots
Test a point: To determine right/left position
Integrate with respect to y: Treat \(x\) as constant, \(y\) as variable
Don't mix variables: Keep everything in terms of \(y\)

🔥 Common Conversions:

\(y = x^2\) → \(x = \pm\sqrt{y}\) (specify branch!)
\(y = x^3\) → \(x = \sqrt[3]{y}\)
\(y = e^x\) → \(x = \ln y\)
\(y = \sin x\) → \(x = \arcsin y\)
\(y = mx + b\) → \(x = \frac{y-b}{m}\)

❌ Common Mistakes to Avoid

Mistake 1: Subtracting left from right (should be right - left)
Mistake 2: Using x-limits instead of y-limits
Mistake 3: Forgetting to solve for \(x\) in terms of \(y\)
Mistake 4: Taking wrong square root (positive vs negative)
Mistake 5: Integrating with respect to \(x\) when you mean \(dy\)
Mistake 6: Not converting ALL functions to form \(x = f(y)\)
Mistake 7: Wrong antiderivative (treating \(y\) as constant)
Mistake 8: Arithmetic errors in bounds evaluation
Mistake 9: Not checking which function is rightmost
Mistake 10: Confusing which method to use

📝 Practice Problems

Find the area using integration with respect to y:

Between \(x = y^2\) and \(x = 4\)
Between \(x = y\) and \(x = y^3\) from \(y = 0\) to \(y = 1\)
Between \(x = 2y\) and \(x = y^2 + 1\)
Region bounded by \(y = x\), \(y = 2\), and \(x = 0\)

Answers:

\(\frac{32}{3}\) square units
\(\frac{1}{4}\) square units
\(\frac{9}{2}\) square units
2 square units

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Show conversion: How you got from \(y = f(x)\) to \(x = g(y)\)
Correct integral setup: \(\int_c^d [\text{right} - \text{left}]\,dy\)
Y-limits clearly stated: Show how you found \(c\) and \(d\)
Show which is right/left: Test or reasoning
Integration work: Show antiderivative
Evaluate at bounds: Show substitution
Simplify answer: Exact or decimal as requested
Include units: "square units"

💯 Exam Strategy:

Read carefully: Does problem suggest or require \(dy\)?
Sketch region if possible
Express curves as \(x = f(y)\)
Find y-intersection points (limits)
Determine right/left functions
Write setup: \(\int_c^d [\text{right} - \text{left}]\,dy\)
Evaluate integral
Check: Is answer positive?

⚡ Quick Reference Guide

FUNCTIONS OF y ESSENTIALS

The Formula:

\[ A = \int_c^d [\text{right} - \text{left}] \, dy \]

Key Differences from dx:

Variables: \(x = f(y)\) instead of \(y = f(x)\)
Direction: Right - Left instead of Top - Bottom
Limits: \(y\)-values (\(c\) to \(d\)) instead of \(x\)-values
Slices: Horizontal instead of vertical

Remember:

Express as \(x = f(y)\)
Find y-intersection points
Determine right/left
Integrate: right - left
Use \(dy\) not \(dx\)!

Master Integration with Respect to y! The fundamental formula: Area = \(\int_c^d [\text{right} - \text{left}]\,dy\) where functions are expressed as \(x = f(y)\) and \(x = g(y)\). The key difference from \(dx\): use HORIZONTAL slices and subtract LEFT from RIGHT (instead of bottom from top). To convert: solve \(y = f(x)\) for \(x\) in terms of \(y\). Find limits \(c\) and \(d\) by solving for \(y\)-intersection points. Use \(dy\) when: (1) functions naturally given as \(x = f(y)\), (2) region bounded by horizontal lines, (3) avoids splitting integrals, (4) simpler than \(dx\) method. Common conversions: \(y = x^2 \to x = \pm\sqrt{y}\) (specify branch!), \(y = e^x \to x = \ln y\), \(y = mx+b \to x = \frac{y-b}{m}\). Process: convert to \(x = f(y)\), find y-limits, determine right/left, integrate. Both \(dx\) and \(dy\) methods give same area—choose based on simplicity! This appears regularly on AP® exams! 🎯✨