When a region is easier to slice sideways, you integrate with respect to $y$ instead of $x$ . You rewrite curves as $x = f(y)$ , find where they meet by solving for $y$ , then integrate right curve minus left curve over those $y$ -bounds. For AP Calculus, choose horizontal slices when they avoid splitting the area into multiple $x$ -integrals.

Why This Matters for the AP Calculus Exam

Area between curves shows up across the integration applications in Unit 8, and the skill of choosing whether to integrate in $x$ or $y$ is exactly the kind of decision the AP Calculus exam wants you to make on your own. Some regions are messy with vertical slices because the top or bottom boundary switches functions, but they become clean single integrals when you slice horizontally.

This topic also builds directly into later Unit 8 work like the disc and washer methods, where deciding between horizontal and vertical orientation determines whether your "thickness" is $dx$ or $dy$ . Getting comfortable now with sideways regions pays off when you reach volumes of revolution.

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Key Takeaways

A horizontal slice has width equal to (rightmost $x$ ) minus (leftmost $x$ ), so the integrand is $x_{R}(y) - x_{L}(y)$ .
Integrate with respect to $y$ using $y$ -bounds: find intersection points by solving the equations for $y$ .
Rewrite each curve as $x$ in terms of $y$ before setting up the integral.
The integrand must be nonnegative over the interval, so subtract the left curve from the right curve (or take the absolute value of the difference).
Choose $y$ -slicing when a region is bounded left and right by curves, or when solving for $x = f(y)$ is simpler than splitting into multiple $x$ -integrals.

Area Between Curves Defined Using $y$

Most area-between-curves problems use functions of $x$ and vertical slices. If you want to review that setup, see Finding the Area Between Curves Expressed as Functions of x.

Here the region is split into thin horizontal rectangles instead. Each rectangle has a small height $dy$ and a width that goes from the leftmost curve to the rightmost curve. Adding up all those rectangles with an integral gives the total area. The whole process mirrors the $x$ -version, just rotated to think in terms of $y$ .

Setting Up the Integral

To find the area between two curves over a $y$ -interval $[c,d]$ , integrate with respect to $y$ . Write both curves as functions of $y$ , then take the rightmost curve minus the leftmost curve.

The area is given by:

A=\int_c^d \left| f(y)-g(y) \right| \, dy

Here $[c,d]$ is the interval on the $y$ -axis where the curves intersect. The absolute value keeps the integrand nonnegative, which means you are subtracting the left curve from the right curve.

When you integrate, make sure everything is in terms of $y$ . Visually, you are stacking thin rectangles up and down the $y$ -axis instead of left and right across the $x$ -axis.

Worked Examples

Example 1: Area Between Curves

Given $f(y) = y^2$ and $g(y) = y$ , find the area between the curves.

Start by finding intersection points. Set the functions equal: $y^2 = y$ . Solving gives $y = 0$ and $y = 1$ , so the interval of integration is $[0, 1]$ .

Over this interval, $y$ is greater than $y^2$ , so $g(y)$ is the rightmost curve. Set up the integral so the area comes out positive:

A = \int_{0}^{1} \left| (y^2 - y) \right| \,dy = \int_{0}^{1} (y - y^2) \,dy

Integrate with respect to $y$ :

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

Evaluate at the bounds:

A = \left( \frac{1}{2}(1)^2 - \frac{1}{3}(1)^3 \right) - \left( \frac{1}{2}(0)^2 - \frac{1}{3}(0)^3 \right)

A = \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{6}

The area between the curves over $[0, 1]$ is $\frac{1}{6}$ square units.

Example 2: Area Between Curves Using a Calculator

Find the area of the region above the $x$ -axis between $x = 2y - y^3$ and $x = -y$ using a calculator.

These curves are already written as $x$ in terms of $y$ , so $y$ -slicing is the natural choice. Graphing first helps you see which curve has the larger $x$ -values and where the curves intersect.

For the region above the $x$ -axis, the curve $x = 2y - y^3$ is to the right of $x = -y$ . Set up the integral as (right minus left):

\int_{0}^{1.73}\ \left(2y\ -\ y^{3}-\left(-y\right)\right)\ dy

Evaluating on a graphing calculator gives:

\int_{0}^{1.73}\ \left(2y\ -\ y^{3}-\left(-y\right)\right)\ dy = 2.2499873975

So the area is approximately 2.25 square units.

How to Use This on the AP Calculus Exam

MCQ

Watch for regions described or graphed with left and right boundaries. If the left or right edge is a single curve and the top or bottom switches, $y$ -slicing is usually cleaner.
Be ready to rewrite a curve like $y = x^2$ opening sideways as $x = \pm\sqrt{y}$ , or a line $y = mx + b$ as $x = \frac{y-b}{m}$ .

Free Response

Set up a correct integral expression with proper notation before evaluating. A clear expression like $A = \int_{c}^{d} (x_{R}(y) - x_{L}(y)) \, dy$ shows your reasoning and is important for clean exam work.
On calculator-active parts, you can find intersection points and evaluate the definite integral on the calculator, but still write the full integral setup.

Common Trap

Mixing variables: if you integrate $dy$ , every term in the integrand must be a function of $y$ , and your bounds must be $y$ -values.

Common Misconceptions

"I can always integrate with respect to $x$ ." You can sometimes, but it may force you to split into multiple integrals where the boundary changes. Slicing in $y$ can turn that into one clean integral.
"Right minus left is the same as top minus bottom." For horizontal slices, the width is rightmost $x$ minus leftmost $x$ . Subtracting in the wrong order gives a negative value, so use the absolute value or order the curves correctly.
"The bounds are still $x$ -values." When integrating in $y$ , the limits $c$ and $d$ are $y$ -coordinates of the intersection points, not $x$ -coordinates.
"I can leave equations as $y = f(x)$ ." You need each boundary written as $x$ in terms of $y$ before you set up a $dy$ integral.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
areas in the plane	Regions bounded by curves and axes in a coordinate system whose measurements can be determined using integration.
definite integral	The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.

Frequently Asked Questions

When should I integrate with respect to y for area between curves?

Integrate with respect to y when the region is easier to describe with right and left boundaries as functions of y, or when vertical slices would require splitting the region.

What is the formula for area between curves as functions of y?

Use the integral from bottom y-bound to top y-bound of right function minus left function with respect to y: area equals integral of x_right minus x_left dy.

How do I choose y-bounds for area between curves?

Find the lowest and highest y-values of the bounded region. These often come from intersection points or from the visible vertical extent of the region.

How do I know which function is right and which is left?

For a horizontal slice, the right boundary has the larger x-value and the left boundary has the smaller x-value at that y. The integrand should be right minus left.

What is the most common mistake in AP Calc 8.5?

The most common mistake is using top minus bottom while integrating with respect to y. For dy integrals, use right minus left, not upper minus lower.

How should I set up AP Calculus FRQ work for this topic?

State the y-bounds, identify the right and left functions, write the definite integral, and only then evaluate if the question asks for a numerical area.