Finding the Area Between Curves Expressed as Functions of $y$ - AP Calculus BC Study Guide

The Core Idea: Finding the Area Between Curves Expressed as Functions of $y$

In calculus, we often find the area of a region by summing the areas of infinitesimally thin vertical rectangles, which leads to an integral with respect to $x$. However, some regions are defined by curves that are more easily expressed as functions of $y$, in the form $x=f(y)$. For these regions, it is often more convenient to calculate the area by summing infinitesimally thin horizontal rectangles.

This topic extends the concept of finding the area between curves to situations where integrating with respect to $y$ is more efficient. Instead of a "top minus bottom" approach over an interval of $x$-values, we use a "right minus left" approach over an interval of $y$-values. This requires reorienting our perspective to view $x$ as a function of $y$ and integrating along the $y$-axis. The key is choosing the variable of integration ($x$ or $y$) that simplifies the setup and calculation of the definite integral.

Key Formulas

The area of a region bounded by curves expressed as functions of $y$ is defined by a single definite integral.

The area $A$ of the region enclosed by the graphs of $x=f(y)$ and $x=g(y)$ between the horizontal lines $y=c$ and $y=d$ is given by:

$$A={\int }_c^d(f(y)-g(y))dy$$

Conditions and Components:

• Functions: The curves must be expressed as functions of $y$, i.e., in the form $x=\text{expression in }y$.

• Bounds: The limits of integration, $c$ and $d$, are the lower and upper $y$-values that bound the region. These are often the $y$-coordinates of the points where the curves intersect.

• Integrand: The expression $(f(y)-g(y))$ represents the width of a typical horizontal rectangle.

• Right vs. Left: For the formula to yield a positive area, the condition $f(y)\ge g(y)$ must hold for all $y$ in the interval $[c,d]$. This means that the graph of $x=f(y)$ must be to the right of the graph of $x=g(y)$ throughout the interval. The integrand is therefore conceptualized as "(right function) - (left function)".

Understanding the Change in Perspective

The decision to integrate with respect to $y$ rather than $x$ is a strategic one, driven by the goal of simplifying the problem.

Consider a region bounded by curves. If you were to integrate with respect to $x$, you would need to identify a single "top" function and a single "bottom" function over the entire interval of integration. If the function that is on top changes, or if a boundary is not a function of $x$ (like a vertical line), you would need to set up multiple integrals.

However, if that same region can be described by a single "right" function and a single "left" function over an entire interval of $y$-values, then one integral with respect to $y$ will suffice. This is often the case for regions bounded by curves that open horizontally, such as parabolas of the form $x=a{y}^2+by+c$. Choosing to integrate with respect to $y$ can turn a multi-integral problem into a single-integral problem, significantly reducing complexity.

To make this switch, you must be able to:

1. Express all bounding curves as functions of $y$ (i.e., solve for $x$ in terms of $y$).

2. Identify the limits of integration as $y$-values.

3. Consistently identify the "rightmost" and "leftmost" functions for the integrand.

Core Concepts & Rules

• Area Formula: The area between two curves $x=f(y)$ and $x=g(y)$ from $y=c$ to $y=d$, where $f(y)\ge g(y)$, is calculated by the definite integral ${\int }_c^d(f(y)-g(y))dy$. This represents integrating the "right curve" minus the "left curve".

• Finding Limits of Integration: If the vertical bounds of the region are not explicitly given, they are determined by finding the $y$-coordinates of the points of intersection of the bounding curves. This is done by setting the expressions for $x$ equal to each other and solving for $y$.

• Strategic Choice: Integrating with respect to $y$ is a method to be used when it is more convenient than integrating with respect to $x$. This convenience typically arises when the region's boundaries are more easily or consistently described by $x$ as a function of $y$.

Step-by-Step Example 1: Basic Application

Problem: Find the area of the region enclosed by the parabola $x=y^2-3$ and the line $x=y-1$.

Step 1: Find the points of intersection.

To find the limits of integration, set the two functions equal to each other and solve for $y$.

$$y^2-3=y-1$$

$$y^2-y-2=0$$

$$(y-2)(y+1)=0$$

The points of intersection occur at $y=-1$ and $y=2$. These will be our limits of integration, so $c=-1$ and $d=2$.

Step 2: Determine the "right" and "left" functions.

Choose a test value for $y$ within the interval $[-1,2]$, such as $y=0$.

• For the parabola: $x=(0{)}^2-3=-3$.

• For the line: $x=(0)-1=-1$.

Since $-1>-3$, the line $x=y-1$ is the "right" function, $f(y)$, and the parabola $x=y^2-3$ is the "left" function, $g(y)$, on the interval $[-1,2]$.

Step 3: Set up the definite integral.

Using the formula $A={\int }_c^d(f(y)-g(y))dy$:

$$A={\int }_{-1}^2\left((y-1)-(y^2-3)\right)dy$$

$$A={\int }_{-1}^2(-y^2+y+2)dy$$

Step 4: Evaluate the integral.

Find the antiderivative:

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

Apply the Fundamental Theorem of Calculus:

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

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

$$A=\left(-\frac{8}{3}+6\right)-\left(\frac{2}{6}+\frac{3}{6}-\frac{12}{6}\right)$$

$$A=\left(\frac{10}{3}\right)-\left(-\frac{7}{6}\right)$$

$$A=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$$

The area of the region is $\frac{9}{2}$ square units.

Step-by-Step Example 2: Exam-Style Application

Problem: Find the area of the region in the first quadrant bounded by the $y$-axis, the curve $y=\sqrt{x}$, and the line $y=x-2$.

Step 1: Analyze the region and choose the variable of integration.

A sketch of the region shows that if we integrate with respect to $x$, the "bottom" function changes. From $x=0$ to $x=2$, the bottom is the $x$-axis ($y=0$). From $x=2$ to $x=4$, the bottom is the line $y=x-2$. This would require two separate integrals.

Let's consider integrating with respect to $y$. The region has a consistent "right" boundary and "left" boundary. This suggests integrating with respect to $y$ will be more convenient.

Step 2: Express curves as functions of $y$.

• The $y$-axis is the line $x=0$.

• The curve $y=\sqrt{x}$ becomes $x=y^2$.

• The line $y=x-2$ becomes $x=y+2$.

Step 3: Find the limits of integration.

We need the $y$-values that bound the region. The region starts at the $x$-axis, so the lower bound is $y=0$. The upper bound is the $y$-coordinate of the intersection of $x=y^2$ and $x=y+2$.

$$y^2=y+2$$

$$y^2-y-2=0$$

$$(y-2)(y+1)=0$$

The intersection points are at $y=2$ and $y=-1$. Since the region is in the first quadrant, we are interested in the interval of $y$-values from $y=0$ to $y=2$. Thus, $c=0$ and $d=2$.

Step 4: Set up the definite integral.

Over the interval $[0,2]$, the "right" function is $x=y+2$ and the "left" function is $x=y^2$.

Wait, let's re-examine the region. The region is bounded by the $y$-axis ($x=0$), $x=y^2$, and $x=y+2$.

From $y=0$ to $y=2$, the right boundary is $x=y+2$ and the left boundary is the $y$-axis ($x=0$). The curve $x=y^2$ is also a boundary. Let's find all intersections.

• $x=y+2$ and $x=y^2$ intersect at $y=2$.

• $x=y+2$ and $x=0$ intersect at $y=-2$.

• $x=y^2$ and $x=0$ intersect at $y=0$.

The region is split at $y=1$.

• For $y\in [0,1]$, the right boundary is $x=y^2$ and the left boundary is $x=0$.

• For $y\in [1,2]$, the right boundary is $x=y+2$ and the left boundary is $x=y^2$.

This requires two integrals. Let's re-read the problem. "bounded by the $y$-axis, the curve $y=\sqrt{x}$, and the line $y=x-2$".

The intersection of $y=\sqrt{x}$ and $y=x-2$ is at $x=4,y=2$. The line $y=x-2$ hits the $y$-axis at $y=-2$. The curve $y=\sqrt{x}$ hits the $y$-axis at $y=0$.

The region is bounded on the left by the $y$-axis ($x=0$). The right boundary is composed of two different functions.

This problem is more complex than it seems. Let's try a clearer exam-style problem that highlights the convenience.

Revised Problem: Find the area of the region enclosed by the graphs of $y^2=x$ and $x-y=2$.

Step 1: Express curves as functions of $y$ and choose the variable.

We have $x=y^2$ and $x=y+2$. Since both are easily expressed in the form $x=f(y)$, integrating with respect to $y$ is a natural choice.

Step 2: Find the limits of integration.

Set the expressions for $x$ equal to find the intersection points.

$$y^2=y+2$$

$$y^2-y-2=0$$

$$(y-2)(y+1)=0$$

The curves intersect at $y=-1$ and $y=2$. These are our limits of integration.

Step 3: Determine the "right" and "left" functions.

Test a value in the interval $(-1,2)$, such as $y=0$.

• For $x=y^2$, $x=0$.

• For $x=y+2$, $x=2$.

Since $2>0$, the line $x=y+2$ is the right function and the parabola $x=y^2$ is the left function.

Step 4: Set up and evaluate the integral.

$$A={\int }_{-1}^2\left((y+2)-y^2\right)dy$$

$$A={\int }_{-1}^2(-y^2+y+2)dy$$

This is the same integral as in Example 1.

$$A={\left[-\frac{y^3}{3}+\frac{y^2}{2}+2y\right]}_{-1}^2=\frac{9}{2}$$

Note on convenience: To solve this by integrating with respect to $x$, we would have to express the curves as functions of $x$: $y=\pm \sqrt{x}$ and $y=x-2$. The region would require two integrals: one from $x=0$ to $x=1$ (bounded by $y=\sqrt{x}$ and $y=-\sqrt{x}$) and another from $x=1$ to $x=4$ (bounded by $y=\sqrt{x}$ and $y=x-2$). This is significantly more complex.

Using Your Calculator

While the setup of the integral is a crucial analytical skill, a graphing calculator can be used to verify intersections and evaluate the final definite integral.

To find the area between $x=f(y)$ and $x=g(y)$:

1. Graph the Functions: Since calculators graph functions of $x$, you can graph $Y_1=f(X)$ and $Y_2=g(X)$ and mentally treat the horizontal axis as the $y$-axis and the vertical axis as the $x$-axis.

2. Find Intersection Points: Use the calculator's intersection feature (2nd -> TRACE -> $5:intersect$) to find the intersection points of the two graphed curves. The "x-values" reported by the calculator will correspond to the $y$-values of intersection, which are your limits of integration, $c$ and $d$.

3. Evaluate the Integral: On the home screen, use the numerical integration function (e.g., fnInt on TI-84 or Math -> 9: fnInt(). The syntax is typically fnInt(function, variable, lower_bound, upper_bound).

   • For the example A = \int_{-1}^2 (-y^2 + y + 2) \,dy, you would enter:

     fnInt(-X^2 + X + 2, X, -1, 2)

   • The calculator will return the numerical value of the area, which is $4.5$.

Important: The calculator is a tool for evaluation. On the AP Exam, you must be able to write the correct definite integral expression to earn full credit on free-response questions.

AP Exam Quick Hit

Common Question Types

• Direct Calculation: You will be given two curves, typically already in the form $x=f(y)$ and $x=g(y)$, and asked to find the area of the region enclosed by them.

  • Example: "Find the area of the region enclosed by the graphs of $x=4-y^2$ and $x=y^2-4$."

• Choosing the Method: You will be given a region described by functions of $x$ (e.g., $y=\ln x$, $x=e$, $y=1$) where setting up the integral with respect to $y$ is significantly easier than with respect to $x$.

  • Example: "Find the area of the region bounded by the graphs of $y=\arctan (x)$, the line $x=1$, and the $x$-axis." (Integrating ${\int }_0^1\arctan (x)dx$ is difficult, but integrating ${\int }_0^{\pi /4}(1-\tan y)dy$ is simpler).

• Setup Only: On a free-response question, you may be asked only to write an integral expression for the area, without evaluating it.

  • Example: "Let $R$ be the region enclosed by the graphs of $x=e^y$ and $x=(y-2{)}^2$. Write, but do not evaluate, an integral expression that gives the area of $R$."

Common Mistakes

• Using $x$-bounds: After correctly solving for the $y$-coordinates of intersection, students mistakenly use the corresponding $x$-coordinates as the limits of integration. Remember, when integrating $dy$, the limits must be $y$-values.

• Incorrect Integrand Order: Setting up the integral as $(leftfunction)-(rightfunction)$. This will result in the negative of the correct area. Always confirm which function has the greater $x$-value on the interval by testing a point. Area must be a non-negative quantity.

• Mixing Variables: Writing an integral with a $dy$ differential but having the variable $x$ in the integrand, or vice-versa. The variable in the integrand must match the differential (e.g., $\int f(y)dy$).

• Algebraic Errors in Re-expression: When converting a function from $y=f(x)$ to $x=g(y)$, students may make algebraic mistakes. For example, incorrectly solving $y^2=x-4$ as $x=y^2-4$ instead of $x=y^2+4$.

• Failing to Consider Integration with respect to $y$: Some students will always default to integrating with respect to $x$, even when the problem is structured to be much simpler with respect to $y$. This can lead to unnecessarily complex or multiple integrals, wasting time and increasing the chance of error.