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Area Between Curves
Functions of \( y \)

Master the technique of integrating with respect to \( y \) to find enclosed areas when curves are expressed as functions of the vertical axis.

AP Calculus AB

Why Integrate with Respect to \( y \)?

Some regions are bounded by curves that are more naturally described as functions of \( y \) rather than \( x \). When curves open left/right (like horizontal parabolas), expressing them as \( x = f(y) \) and integrating along the \( y \)-axis gives us a single, clean integral instead of splitting into multiple integrals.

Horizontal vs. Vertical Slicing

Vertical slices (\(dx\)) x y Can be complex Horizontal slices (\(dy\)) x y Simpler setup
\(x=4\)
\(x = y^2\)

Right Minus Left: The Core Idea

When integrating with respect to \( y \), we think of each infinitesimal strip as a horizontal rectangle of width \( dy \). Its length is the difference between the right boundary and the left boundary at that \( y \)-value.

Right boundary
The curve with the larger \( x \)-value: \( x = f(y) \)
Left boundary
The curve with the smaller \( x \)-value: \( x = g(y) \)

The Formula

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

x y
\(g(y)\)
\(f(y)\)
 c d
\(dy\)
\(f(y) - g(y)\)

Integrating in \( x \) vs. \( y \)

With respect to \( x \) With respect to \( y \)
Slice direction Vertical strips Horizontal strips
Integrand \( f(x) - g(x) \) \( f(y) - g(y) \)
Meaning Top minus Bottom Right minus Left
Limits \( x \)-values at intersection \( y \)-values at intersection
Best when Curves are \( y = f(x) \) Curves are \( x = f(y) \)

How to Apply the Method

1
Sketch the region. Plot both curves and shade the enclosed area.
2
Express as \( x = f(y) \). Rewrite each curve so \( x \) is a function of \( y \).
3
Find limits in \( y \). Set the two \( x \)-expressions equal and solve for \( y \) to get \( c \) and \( d \).
4
Identify right and left. For a \( y \) between \( c \) and \( d \), which curve has the larger \( x \)?
5
Integrate: \( A = \int_{c}^{d}[\text{right} - \text{left}]\,dy \). Evaluate and simplify.

Example 1: \( x = y^2 \) and \( x = y + 2 \)

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

x y 0
\(x = y^2\)
\(x = y + 2\)
(1, -1)
(4, 2)
\(dy\)

Example 1: Solution

1
Intersections: Set \( y^2 = y + 2 \). Then \( y^2 - y - 2 = 0 \), so \( (y-2)(y+1) = 0 \). Limits: \( y = -1 \) to \( y = 2 \).
2
Right vs. Left: At \( y = 0 \): line gives \( x = 2 \), parabola gives \( x = 0 \). So \( x = y + 2 \) is right, \( x = y^2 \) is left.
3
Set up the integral:

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

Example 2: \( x = y^2 - 1 \) and \( x = 2y + 1 \)

Find the area of the region bounded by \( x = y^2 - 1 \) and \( x = 2y + 1 \).

x y
\(x = y^2 - 1\)
\(x = 2y + 1\)
(-0.46, -0.73)
(6.46, 2.73)

Example 2: Solution

1
Right vs. Left: At \( y = 1 \): line gives \( x = 3 \), parabola gives \( x = 0 \). So \( x = 2y+1 \) is right, \( x = y^2-1 \) is left.
2
Integrand: \( (2y + 1) - (y^2 - 1) = -y^2 + 2y + 2 \)

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

Example 3: Region with a Vertical Line

Find the area enclosed by \( x = y^2 \), the line \( x = 1 \), and the \( x \)-axis (in the first quadrant).

x y
\(x = y^2\)
\(x=1\)
(1, 1)
 1 1

Example 3: Solution

1
Limits: The curves meet at \( y = 0 \) (origin) and \( y = 1 \) (where \( y^2 = 1 \)).
2
Right: \( x = 1 \). Left: \( x = y^2 \).
3
Integrate:

$$ A = \int_{0}^{1} (1 - y^2)\,dy = \left[y - \frac{y^3}{3}\right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3} $$

When Should You Choose \( dy \)?

✓ Curves open left/right
Parabolas like \( x = y^2 \) or \( x = -y^2 + 4 \) are naturally functions of \( y \).
✓ Avoid splitting integrals
If integrating in \( x \) requires two or more separate integrals, try \( dy \) instead.
✓ Boundaries are \( y \)-based
Horizontal lines like \( y = 3 \) are constant limits in \( y \), making setup cleaner.

Common Mistakes

✕ Wrong subtraction order
Writing left minus right instead of right minus left. This gives a negative area.
✕ Using \( x \)-limits with \( dy \)
The limits of integration must be \( y \)-values when integrating with respect to \( y \).
✕ Forgetting to solve for \( x \)
Leaving curves as \( y = f(x) \) and integrating in \( y \). You must rewrite as \( x = g(y) \).

Quick Reference Summary

Step Action Key Detail
1 Sketch the region Identify which curves bound left/right
2 Solve for \( x = f(y) \) Both curves as functions of \( y \)
3 Find \( y \)-limits Set \( f(y) = g(y) \), solve for \( y \)
4 Test a midpoint Confirm right > left
5 Evaluate integral \( \int_c^d [\text{right} - \text{left}]\,dy \)

Worked practice: Section 9, Practice Problem 4 — Applied

Work through this original lesson problem, then reveal the solution and illustrations.

A water channel has a cross-section bounded on the left by the curve x = y² − 1 and on the right by the vertical line x = 3, for y ∈ [−2, 2]. The channel is partially filled so that the water surface is at y = 1. Set up and evaluate an integral with respect to y to find the cross-sectional area of the water in the channel (the region where −2 ≤ y ≤ 1 between the two boundaries). (a) Identify the limits of integration and the integrand. (b) Evaluate the integral to find the cross-sectional area. (c) If the channel extends 10 meters long, find the volume of water. (d) Explain why integrating with respect to y is preferable here.

Show answer and explanation

Part (a) — 1 point: The left boundary is x = y² − 1 and the right boundary is x = 3. The intersection at the bottom of the channel occurs where y² − 1 = 3, giving y = −2 (taking the negative root for the bottom). The water surface is at y = 1. So the limits are y = −2 to y = 1, and the integrand is 3 − (y² − 1) = 4 − y². Part (b) — 1 point: A = ∫ from −2 to 1 (4 − y²) dy = [4y − y³/3] from −2 to 1 = (4 − 1/3) − (−8 + 8/3) = (11/3) − (−16/3) = 27/3 = 9 square units. Part (c) — 1 point: Volume = cross-sectional area × length = 9 × 10 = 90 cubic meters (assuming consistent units). Part (d) — 1 point: Integrating with respect to y is preferable because the left boundary x = y² − 1 is a sideways parabola; solving for y in terms of x would produce two branches (y = ±√(x + 1)), requiring the region to be split. With dy-integration, the parabola is a single-valued function of y, so one integral suffices.

Water between x=y²−1 and x=3 for −2≤y≤1. A horizontal slice at y=0 has width 4; area is 9 m².
Horizontal water strips span from x=y²−1 to x=3; filling through y=1 gives 9 m² of cross-sectional area. The water face is an explanatory cutaway, not an unsupported real water wall.

Question 1 of 5

Which integral correctly gives the area between \( x = y^2 \) and \( x = 4 \)?

Question 2 of 5

When integrating with respect to \( y \), the integrand represents:

Question 3 of 5

Evaluate \( \displaystyle\int_{0}^{1}(1 - y^2)\,dy \).

Question 4 of 5

The area between \( x = y + 2 \) and \( x = y^2 \) from \( y = -1 \) to \( y = 2 \) equals:

Question 5 of 5

Which integral gives the area of the region bounded by the parabola \( x = y^2 \), the y-axis, and the line \( y = 2 \)?