8.5 Finding the Area Between Curves Expressed as Functions of y

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.

Choosing horizontal slices here carries straight into the disc and washer methods later in Unit 8, where the orientation you pick decides whether the thickness is $dx$ or $dy$. Getting comfortable with sideways regions now makes volumes of revolution easier to set up.

Setting Up Horizontal Slices With Right Minus Left

• 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.

Choosing Horizontal Slices and Writing the Setup

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

• Misconception: I can always integrate with respect to $x$.
  Actually: 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.
• Misconception: Right minus left is the same as top minus bottom.
  Actually: 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.
• Misconception: The bounds are still $x$-values.
  Actually: When integrating in $y$, the limits $c$ and $d$ are $y$-coordinates of the intersection points, not $x$-coordinates.
• Misconception: I can leave equations as $y = f(x)$.
  Actually: You need each boundary written as $x$ in terms of $y$ before you set up a $dy$ integral.

Related AP Calculus Guides

• Unit 8 Review: Applications of Integration
• 8.1 Finding the Average Value of a Function on an Interval
• 8.7 Volumes with Cross Sections: Squares and Rectangles
• 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals
• 8.4 Finding the Area Between Curves Expressed as Functions of x
• 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts