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

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

While we typically find the area of a region by summing the areas of infinitesimally thin vertical rectangles (integrating with respect to $x$), this is not always the most efficient method. Some regions are bounded by curves that are more easily expressed as functions of $y$, such as $x=y^2$. For these regions, it is advantageous to reorient our perspective and sum the areas of infinitesimally thin horizontal rectangles.

This topic introduces the method of finding the area between two curves by integrating with respect to $y$. The fundamental concept is to define the region's boundaries by a "right" curve, $x=f(y)$, and a "left" curve, $x=g(y)$, between two $y$-values, $y=c$ and $y=d$. The area is then the accumulation (the definite integral) of the lengths of these horizontal rectangles, which is the difference $(f(y)-g(y))$, over the interval of $y$-values. This approach can simplify problems that would otherwise require splitting the region into multiple subregions if approached with vertical rectangles.

Key Formulas

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

• Area Between Two Curves (as functions of $y$):

The area $A$ of the region enclosed by the graphs of $x=f(y)$ and $x=g(y)$ from $y=c$ to $y=d$ is given by the formula:

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

In this formula, it is required that $f(y)\ge g(y)$ for all $y$ in the interval $[c,d]$. Conceptually, this means we are integrating $(rightcurve-leftcurve)$ with respect to $y$.

Understanding the Method

The choice to integrate with respect to $y$ instead of $x$ is a strategic one, primarily made for efficiency. This method is often superior in two common scenarios:

1. When curves are not functions of $x$: If one of the bounding curves fails the vertical line test (e.g., a sideways parabola like $x=y^2$), it cannot be written as a single function $y=f(x)$. To integrate with respect to $x$, you would need to split the curve into a top half ($y=\sqrt{x}$) and a bottom half ($y=-\sqrt{x}$) and set up two separate integrals. Expressing the curve as $x=g(y)$ allows for a single, simpler integral with respect to $y$.

2. When the bounding curves change: Consider a region where the "top" and "bottom" bounding functions change partway through the interval. This would require splitting the area calculation into two or more integrals with respect to $x$. However, if the "right" and "left" bounding functions remain consistent throughout the region's vertical span, a single integral with respect to $y$ can find the entire area.

In some cases, a complex region may need to be divided into subregions regardless of the variable of integration. The total area is then found by summing the areas of these subregions.

Core Concepts & Rules

• The area between two curves, $x=f(y)$ and $x=g(y)$, is found by integrating the difference between the right-hand curve and the left-hand curve with respect to $y$.

• The definite integral is set up as ${\int }_c^d(\text{right function}-\text{left function})dy$. The bounds of integration, $c$ and $d$, must be $y$-values.

• To use this method, you must ensure that for the entire interval $[c,d]$, one curve is consistently on the right ($f(y)$) and the other is consistently on the left ($g(y)$).

• Integrating with respect to $y$ is a strategic choice to simplify calculations, particularly when the boundaries are more naturally described as $x$ in terms of $y$.

• If the right and left boundary curves switch roles within the desired area, the region must be split into subregions at the $y$-value(s) where they intersect, and the total area is the sum of the areas of the subregions.

Step-by-Step Example 1: Basic Application

Problem: Find the area of the region enclosed by the graphs of $x=y^2-3$ and $x=y+3$.

Step 1: Find the points of intersection.

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

$$y^2-3=y+3$$

$$y^2-y-6=0$$

$$(y-3)(y+2)=0$$

The points of intersection occur at $y=-2$ and $y=3$. These will be our lower and upper bounds of integration, $c=-2$ and $d=3$.

Step 2: Determine the right and left curves.

Pick a test value for $y$ within the interval $[-2,3]$, such as $y=0$.

• For $x=y^2-3$: $x=(0{)}^2-3=-3$.

• For $x=y+3$: $x=(0)+3=3$.

Since $3>-3$, the curve $x=y+3$ is the right curve and $x=y^2-3$ is the left curve on the interval $[-2,3]$.

Step 3: Set up the definite integral.

The area $A$ is the integral of (right curve - left curve) from $y=-2$ to $y=3$.

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

$$A={\int }_{-2}^3(-y^2+y+6)dy$$

Step 4: Evaluate the integral.

Find the antiderivative and evaluate it at the bounds.

$$A={\left[-\frac{y^3}{3}+\frac{y^2}{2}+6y\right]}_{-2}^3$$

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

$$A=\left(-9+\frac{9}{2}+18\right)-\left(\frac{8}{3}+2-12\right)$$

$$A=\left(9+\frac{9}{2}\right)-\left(\frac{8}{3}-10\right)$$

$$A=\frac{27}{2}-\left(-\frac{22}{3}\right)$$

$$A=\frac{27}{2}+\frac{22}{3}=\frac{81}{6}+\frac{44}{6}=\frac{125}{6}$$

The area of the region is $\frac{125}{6}$.

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 graph of $y=\sqrt{x}$, and the graph of $y=x-2$.

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

Let's sketch the region. The boundaries are $x=0$, $y=\sqrt{x}$, and $y=x-2$.

• If we integrate with respect to $x$ (vertical rectangles), the bottom boundary changes. From $x=0$ to $x=2$, the bottom is the x-axis ($y=0$). From $x=2$ to the intersection point, the bottom is the line $y=x-2$. This would require two separate integrals.

• Let's try integrating with respect to $y$ (horizontal rectangles). This requires expressing the boundaries as functions of $y$.

  • $y=\sqrt{x}⟹x=y^2$

  • $y=x-2⟹x=y+2$

  • The y-axis is $x=0$.

Step 2: Express boundaries as functions of $y$ and find intersection points.

We have the curves $x=y^2$ and $x=y+2$. Let's find their intersection.

$$y^2=y+2⟹y^2-y-2=0⟹(y-2)(y+1)=0$$

The intersections are at $y=2$ and $y=-1$. Since the region is in the first quadrant, we are interested in $y\ge 0$. The relevant intersection is at $y=2$.

The region is bounded below by the x-axis ($y=0$) and above by the intersection point $y=2$. So our bounds of integration are $c=0$ and $d=2$.

Step 3: Determine the right and left curves for the integral.

For any $y$ in the interval $[0,2]$, the right boundary of the region is the line $x=y+2$ and the left boundary is the curve $x=y^2$.

Self-check: At $y=1$, $x=1+2=3$ and $x=1^2=1$. The line is to the right of the parabola. This confirms our setup.

Step 4: Set up and evaluate the definite integral.

The area is the integral of (right curve - left curve) from $y=0$ to $y=2$.

$$A={\int }_0^2\left((y+2)-y^2\right)dy$$

$$A={\int }_0^2(-y^2+y+2)dy$$

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

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

$$A=\left(-\frac{8}{3}+\frac{4}{2}+4\right)-(0)$$

$$A=-\frac{8}{3}+2+4=-\frac{8}{3}+6=-\frac{8}{3}+\frac{18}{3}=\frac{10}{3}$$

The area of the region is $\frac{10}{3}$. This single integral was much more efficient than the two integrals required for an approach with respect to $x$.

Using Your Calculator

A graphing calculator can be used to visualize the region and evaluate the definite integral for the area.

Problem: Find the area of the region enclosed by $x=\cos (y)$ and $x=y^2-2$.

Step 1: Visualize the Region

Since calculators graph functions of $x$, you can swap the variables to visualize the region. Graph $Y_1=\cos (X)$ and $Y_2=X^2-2$. The region you see is a rotated version of the actual region. This helps you identify the "right" and "left" functions (which will appear as "top" and "bottom" on the graph) and find the intersection points.

Step 2: Find Intersection Points (Bounds)

• Use the CALC menu (2nd + TRACE) and select $5:intersect$.

• Move the cursor near one intersection point and press ENTER three times. The calculator will display the coordinates. For this problem, the intersections are at approximately $y\approx -1.455$ and $y\approx 1.455$.

• Store these values for accuracy. For example, after finding the first intersection, go to the home screen and press `X,T,\theta,n$ $STO\to$$A$. Store the second value as $B$.

Step 3: Evaluate the Integral

• On the interval $[-1.455,1.455]$, the graph of $\cos (y)$ is to the right of $y^2-2$.

• Use the numerical integration function, fnInt (found under MATH -> 9: fnInt).

• The syntax is fnInt(expression, variable, lower, upper).

• Enter the integral as:

  fnInt(cos(Y) - (Y^2 - 2), Y, A, B)

  (Note: Use the variable $Y$. The bounds $A$ and $B$ are the stored intersection values.)

• The calculator will return the area, which is approximately $5.633$.

AP Exam Quick Hit

Common Question Types

• Direct Calculation: You are given two equations, either in the form $x=f(y)$ or easily solvable for $x$, and asked to find the area of the region enclosed by them.

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

• Setup Only: You are given a region, often one that is much simpler to integrate with respect to $y$, and asked to choose the correct definite integral that represents its area from a list of options.

  • Example: "Which of the following integrals gives the area of the region bounded by $x=y^2$ and $x=y+2$?"

• Choosing the Best Method: You are given a region bounded by several curves (e.g., $y=\ln x$, $y=2$, $y=0$, $x=0$) and must first decide whether to integrate with respect to $x$ or $y$ before setting up and solving.

Common Mistakes

• Incorrect Integrand: Using $(top-bottom)$ instead of $(right-left)$. Remember, for a $dy$ integral, the integrand must represent the horizontal width of the rectangles.

• Incorrect Bounds: Using $x$-values for the limits of integration in a $dy$ integral. The bounds $c$ and $d$ must always be $y$-values that define the vertical extent of the region.

• Algebraic Errors: Making a mistake when solving an equation for $x$ in terms of $y$. For example, writing $x=y^2$ as $y=\sqrt{x}$ and forgetting the $y=-\sqrt{x}$ portion, or vice-versa.

• Sign Errors: Incorrectly distributing the negative sign when subtracting the left function from the right function, i.e., writing $(right)-(lef{t}_{t}erm1+lef{t}_{t}erm2)$ as $right - left_term1 + left_term2`.

• Ignoring the Result: Calculating a negative area and moving on. Area must be a positive quantity. A negative result is a sign that the left and right curves were likely reversed in the integral setup.