Volume with Washer Method: Revolving Around the $x$- or $y$-Axis - AP Calculus BC Study Guide

The Core Idea: Volume with Washer Method: Revolving Around the $x$- or $y$-Axis

The washer method is a technique for finding the volume of a solid of revolution that has a hole or cavity in the center. This situation arises when the region being revolved around an axis does not touch the axis along its entire inner boundary. The fundamental concept is to think of the solid as being composed of an infinite number of thin "washers" (disks with a hole in the center) stacked along the axis of revolution.

The volume of the entire solid is found by integrating the area of a representative washer cross-section. The area of a single washer is calculated by taking the area of the larger, outer disk and subtracting the area of the smaller, inner disk that forms the hole. This process extends the disk method to regions bounded by two distinct functions, allowing for the calculation of volumes for more complex solids.

Key Formulas

The volume of a solid of revolution using the washer method is determined by one of two formulas, depending on the axis of revolution.

1. Revolution Around the $x$-Axis

For a region bounded by $y=R(x)$ and $y=r(x)$ from $x=a$ to $x=b$, where $R(x)\ge r(x)$ for all $x$ in $[a,b]$, the volume of the solid generated by revolving the region about the $x$-axis is:

$$V=\pi {\int }_a^b\left([R(x){]}^2-[r(x){]}^2\right)dx$$

• $R(x)$ is the outer radius: the distance from the $x$-axis to the farther function.

• $r(x)$ is the inner radius: the distance from the $x$-axis to the closer function.

• The area of a single washer cross-section at a given $x$ is $A(x)=\pi [R(x){]}^2-\pi [r(x){]}^2$.

2. Revolution Around the $y$-Axis

For a region bounded by $x=R(y)$ and $x=r(y)$ from $y=c$ to $y=d$, where $R(y)\ge r(y)$ for all $y$ in $[c,d]$, the volume of the solid generated by revolving the region about the $y$-axis is:

$$V=\pi {\int }_c^d\left([R(y){]}^2-[r(y){]}^2\right)dy$$

• $R(y)$ is the outer radius: the distance from the $y$-axis to the farther function.

• $r(y)$ is the inner radius: the distance from the $y$-axis to the closer function.

• The area of a single washer cross-section at a given $y$ is $A(y)=\pi [R(y){]}^2-\pi [r(y){]}^2$.

Understanding the Radii

The most critical step in applying the washer method is correctly identifying the outer radius ($R$) and the inner radius ($r$). These radii are always measured perpendicularly from the axis of revolution to the boundary of the region.

• Outer Radius ($R$): This is the distance from the axis of revolution to the outer boundary of the region. For revolution about the $x$-axis, this is the "top" function; for revolution about the $y$-axis, this is the "right" function.

• Inner Radius ($r$): This is the distance from the axis of revolution to the inner boundary of the region. For revolution about the $x$-axis, this is the "bottom" function; for revolution about the $y$-axis, this is the "left" function.

A common mistake is to calculate the volume as $\pi \int (R-r{)}^{2}dx$. This is incorrect. The correct formula, $\pi \int (R^2-r^2)dx$, represents the integral of the difference of the areas of two circles, not the area of a single circle with radius $(R-r)$.

Core Concepts & Rules

• The washer method is used to find the volume of a solid of revolution with a central hole.

• The volume is calculated by integrating the area of washer-shaped cross-sections.

• The area of a single washer is the area of the outer disk minus the area of the inner disk: $A=\pi (\text{Outer Radius}{)}^2-\pi (\text{Inner Radius}{)}^2$.

• For revolution around the $x$-axis, the integral must be with respect to $x$. All functions must be in the form $y=f(x)$, and the limits of integration must be $x$-values.

• For revolution around the $y$-axis, the integral must be with respect to $y$. All functions must be in the form $x=g(y)$, and the limits of integration must be $y$-values.

• The outer radius, $R$, is always the function farther from the axis of revolution.

• The inner radius, $r$, is always the function closer to the axis of revolution.

Step-by-Step Example 1: Revolving Around the $x$-Axis

Problem: Let $R$ be the region enclosed by the graphs of $y=x+2$ and $y=x^2$. Find the volume of the solid generated when $R$ is revolved about the $x$-axis.

Step 1: Find the Bounds of Integration

Set the functions equal to each other to find their intersection points.

$x^2=x+2$

$x^2-x-2=0$

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

The intersection points are $x=-1$ and $x=2$. These will be our limits of integration, $a=-1$ and $b=2$.

Step 2: Identify the Outer and Inner Radii

The axis of revolution is the $x$-axis ($y=0$). We need to determine which function is greater on the interval $[-1,2]$. Let's test a point, say $x=0$.

• $y=0+2=2$

• $y=0^2=0$

Since $2>0$, the function $y=x+2$ is the "top" or outer function, and $y=x^2$ is the "bottom" or inner function.

• Outer Radius: $R(x)=x+2$

• Inner Radius: $r(x)=x^2$

Step 3: Set Up the Integral

Using the formula for revolution around the $x$-axis:

$$V=\pi {\int }_a^b\left([R(x){]}^2-[r(x){]}^2\right)dx$$

$$V=\pi {\int }_{-1}^2\left((x+2{)}^2-(x^2{)}^2\right)dx$$

Step 4: Evaluate the Integral

First, expand the integrand.

$(x+2{)}^2-(x^2{)}^2=(x^2+4x+4)-x^4=-x^4+x^2+4x+4$

Now, integrate:

$$V=\pi {\int }_{-1}^2(-x^4+x^2+4x+4)dx$$

$$V=\pi {\left[-\frac{1}{5}{x}^5+\frac{1}{3}{x}^3+2x^2+4x\right]}_{-1}^2$$

Evaluate at the limits:

$$V=\pi \left(\left(-\frac{32}{5}+\frac{8}{3}+8+8\right)-\left(\frac{1}{5}-\frac{1}{3}+2-4\right)\right)$$

$$V=\pi \left(\left(-\frac{96}{15}+\frac{40}{15}+16\right)-\left(\frac{3}{15}-\frac{5}{15}-2\right)\right)$$

$$V=\pi \left(\left(-\frac{56}{15}+\frac{240}{15}\right)-\left(-\frac{2}{15}-\frac{30}{15}\right)\right)$$

$$V=\pi \left(\frac{184}{15}-\left(-\frac{32}{15}\right)\right)=\pi \left(\frac{184+32}{15}\right)=\frac{216\pi }{15}=\frac{72\pi }{5}$$

Step-by-Step Example 2: Revolving Around the $y$-Axis

Problem: Let $R$ be the region in the first quadrant bounded by $y=x^2$, $x=0$, and $y=4$. Find the volume of the solid generated when $R$ is revolved about the $y$-axis.

Step 1: Express Functions in Terms of $y$

Since we are revolving around the $y$-axis, we need our functions in the form $x=g(y)$.

The function $y=x^2$ becomes $x=\sqrt{y}$ (we use the positive root because the region is in the first quadrant).

The line $x=0$ remains $x=0$.

Step 2: Identify the Bounds of Integration

The region is bounded by $y=4$ and starts at the origin where $y=0$. So, our limits of integration are $c=0$ and $d=4$.

Step 3: Identify the Outer and Inner Radii

The axis of revolution is the $y$-axis ($x=0$).

• The outer boundary is the curve farther from the $y$-axis, which is $x=\sqrt{y}$. So, $R(y)=\sqrt{y}$.

• The inner boundary is the line $x=0$. So, $r(y)=0$.

Note: When the inner radius is 0, the washer method simplifies to the disk method.

Step 4: Set Up the Integral

Using the formula for revolution around the $y$-axis:

$$V=\pi {\int }_c^d\left([R(y){]}^2-[r(y){]}^2\right)dy$$

$$V=\pi {\int }_0^4\left((\sqrt{y}{)}^2-(0{)}^2\right)dy$$

$$V=\pi {\int }_0^{4}ydy$$

Step 5: Evaluate the Integral

$$V=\pi {\left[\frac{1}{2}{y}^2\right]}_0^4$$

$$V=\pi \left(\frac{1}{2}(4{)}^2-\frac{1}{2}(0{)}^2\right)$$

$$V=\pi \left(\frac{16}{2}-0\right)=8\pi$$

Using Your Calculator

While setting up the integral is a required analytical skill, a graphing calculator is an essential tool for finding intersection points and evaluating complex definite integrals, especially on the AP Exam.

To find the volume from Example 1, $V=\pi {\int }_{-1}^2\left((x+2{)}^2-(x^4)\right)dx$:

1. Define Functions: In the Y= editor, enter:

   • $Y_1=x+2$ (This is your $R(x)$)

   • $Y_2=x^2$ (This is your $r(x)$)

2. Find Intersection Points (if unknown): Use the 2nd -> TRACE (CALC) menu, select $5:intersect$, and follow the prompts to find the bounds $x=-1$ and $x=2$.

3. Calculate the Integral: From the home screen, use the fnInt function (found under MATH -> 9: fnInt). The syntax is:

   fnInt(expression, variable, lower, upper)

   • Enter: π * fnInt( (Y₁)^2 - (Y₂)^2 , X, -1, 2)

   • Note: To access $Y_1$ and $Y_2$, press VARS -> Y-VARS -> 1: Function...

4. Get the Result: The calculator will return a decimal approximation, such as 45.2389.... You can verify this is close to the exact answer: \frac{72π}{5} ≈ 45.239.

AP Exam Quick Hit

Common Question Types

• Calculator-Active Volume: You will be given two functions, possibly complex ones like $f(x)=e^x\cos (x)$ and $g(x)=\ln (x+1)$, and asked to find the volume of the region between them revolved around the $x$-axis. The expectation is that you find the intersection points and evaluate the definite integral using your calculator.

• Integral Setup Only: A multiple-choice question will show a shaded region and ask you to choose the correct integral representing the volume when revolved about the $x$- or $y$-axis. This tests your ability to identify $R$ and $r$ and use the correct formula without any calculation. For example: "Which integral gives the volume of the region bounded by $y=f(x)$ and $y=g(x)$ revolved about the x-axis?"

• Revolving around the $y$-axis: A question that specifically requires you to solve for $x$ in terms of $y$ and integrate with respect to $y$. This tests your ability to reorient your thinking from vertical rectangles ($dx$) to horizontal rectangles ($dy$).

Common Mistakes

• Squaring $R-r$ instead of $R^2-r^2$: The most common error is setting up the integral as $\pi \int (R(x)-r(x){)}^{2}dx$. The correct formula is $\pi \int ([R(x){]}^2-[r(x){]}^2)dx$. Remember, it's the difference of the areas, not the area of the difference.

• Reversing Radii: Incorrectly identifying the outer and inner radii. Always visualize or test a point to confirm which function is farther from the axis of revolution ($R$) and which is closer ($r$). Subtracting in the wrong order will result in a negative volume.

• Forgetting to Rewrite for $y$-axis Revolution: When revolving around the $y$-axis, failing to rewrite equations from $y=f(x)$ to $x=g(y)$. You cannot integrate a function of $x$ with respect to $y$.

• Using Incorrect Limits: Using $x$-values as limits for a $dy$ integral, or vice-versa. The variable in the limits of integration must always match the differential ($dx$ or $dy$).

• Forgetting $\pi$: A simple but costly error is performing the entire integration correctly but omitting the $\pi$ constant in the final answer.