Volume with Washer Method: Revolving Around the $x$- or $y$-Axis - AP Calculus AB 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 in the center. This occurs when the region being revolved does not touch the axis of revolution along its entire length. 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 middle) stacked together.

The volume of the entire solid is found by integrating the area of a representative cross-sectional washer. The area of a single washer is calculated by finding the area of the outer circle and subtracting the area of the inner circle. This method extends the disk method to regions bounded between two functions, creating solids with a cavity.

Key Formulas

The formula for the volume of a solid of revolution using the washer method depends on the axis of revolution. In each case, the volume is the integral of the area of the cross-sectional washer, which is given by $A=\pi (\text{Outer Radius}{)}^2-\pi (\text{Inner Radius}{)}^2=\pi (R^2-r^2)$.

1. Revolution Around the $x$-axis

For a region bounded by $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$, where $f(x)\ge g(x)\ge 0$, the volume of the solid generated by revolving the region about the $x$-axis is:

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

• Outer Radius, $R(x)$: The distance from the axis of revolution ($x$-axis) to the outer curve, which is $f(x)$.

• Inner Radius, $r(x)$: The distance from the axis of revolution ($x$-axis) to the inner curve, which is $g(x)$.

2. Revolution Around the $y$-axis

For a region bounded by $x=f(y)$ and $x=g(y)$ from $y=c$ to $y=d$, where $f(y)\ge g(y)\ge 0$, the volume of the solid generated by revolving the region about the $y$-axis is:

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

• Outer Radius, $R(y)$: The distance from the axis of revolution ($y$-axis) to the outer curve (the one farther from the $y$-axis), which is $f(y)$.

• Inner Radius, $r(y)$: The distance from the axis of revolution ($y$-axis) to the inner curve (the one closer to the $y$-axis), which is $g(y)$.

Understanding the Radii

The most critical part of the washer method is correctly identifying the outer radius ($R$) and the inner radius ($r$). These radii represent the distances from the axis of revolution to the outer and inner boundaries of the region, respectively.

• For revolution around the $x$-axis: The radii are vertical distances, so they are functions of $x$. The outer radius, $R(x)$, is the "top" function, and the inner radius, $r(x)$, is the "bottom" function. You are integrating with respect to $x$.

• For revolution around the $y$-axis: The radii are horizontal distances, so they must be functions of $y$. The outer radius, $R(y)$, is the "right" function, and the inner radius, $r(y)$, is the "left" function. This often requires you to solve the original equations for $x$ in terms of $y$. You are integrating with respect to $y$.

The core formula $V=\pi \int ({\text{Outer Radius}}^2-{\text{Inner Radius}}^2)d(\text{variable})$ is universal. The challenge is to correctly define the radii and the variable of integration based on the axis of revolution.

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 a cross-sectional washer.

• The area of a single washer is $A=\pi (R^2-r^2)$, where $R$ is the outer radius and $r$ is the inner radius.

• When revolving around the $x$-axis, the integral is with respect to $x$. The radii are functions of $x$, representing vertical distances from the $x$-axis to the curves.

• When revolving around the $y$-axis, the integral is with respect to $y$. The radii are functions of $y$, representing horizontal distances from the $y$-axis to the curves.

• All functions defining the radii must be expressed in terms of the variable of integration. If integrating with respect to $y$ ($dy$), all functions must be in the form $x=f(y)$.

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

Problem: Find the volume of the solid generated by revolving the region bounded by the graphs of $y=\sqrt{x}$ and $y=x^2$ about the $x$-axis.

Step 1: Find the bounds of integration.

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

$$\sqrt{x}=x^2$$

$$x=x^4$$

$$x^4-x=0$$

$$x(x^3-1)=0$$

The intersection points are $x=0$ and $x=1$. These are our bounds of integration, $a=0$ and $b=1$.

Step 2: Identify the outer and inner radii.

For revolution around the $x$-axis, the radii are the function values. We need to determine which function is greater on the interval $[0,1]$. Let's test a point, say $x=0.25$.

• $\sqrt{0.25}=0.5$

• $(0.25{)}^2=0.0625$

Since $\sqrt{x}>x^2$ on $(0,1)$, $\sqrt{x}$ is the outer curve.

• Outer Radius: $R(x)=\sqrt{x}$

• Inner Radius: $r(x)=x^2$

Step 3: Set up the integral.

Use the formula $V=\pi {\int }_a^b([R(x){]}^2-[r(x){]}^2)dx$.

$$V=\pi {\int }_0^1\left((\sqrt{x}{)}^2-(x^2{)}^2\right)dx$$

$$V=\pi {\int }_0^1(x-x^4)dx$$

Step 4: Evaluate the integral.

Find the antiderivative and evaluate using the Fundamental Theorem of Calculus.

$$V=\pi {\left[\frac{1}{2}{x}^2-\frac{1}{5}{x}^5\right]}_0^1$$

$$V=\pi \left(\left(\frac{1}{2}(1{)}^2-\frac{1}{5}(1{)}^5\right)-\left(\frac{1}{2}(0{)}^2-\frac{1}{5}(0{)}^5\right)\right)$$

$$V=\pi \left(\frac{1}{2}-\frac{1}{5}\right)$$

$$V=\pi \left(\frac{5}{10}-\frac{2}{10}\right)=\frac{3\pi }{10}$$

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

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

Step 1: Find the bounds of integration.

The revolution is around the $y$-axis, so we need bounds for $y$. The problem states the region is bounded by $y=4$ and implicitly by the vertex of the parabola at $y=0$ (since $x=0$ is a boundary).

The bounds of integration are $c=0$ and $d=4$.

Step 2: Identify the outer and inner radii.

Since we are revolving around the $y$-axis, the radii are horizontal distances and must be functions of $y$.

The region is bounded on the left by the $y$-axis ($x=0$) and on the right by the parabola $y=x^2$.

• The distance from the $y$-axis to the left boundary is 0. This means we have a solid disk, not a washer. However, we can think of it as a washer where the inner radius is zero.

• Inner Radius: $r(y)=0$

• The distance from the $y$-axis to the right boundary is given by the function $y=x^2$. We must solve for $x$ in terms of $y$. Since we are in the first quadrant, $x=\sqrt{y}$.

• Outer Radius: $R(y)=\sqrt{y}$

Step 3: Set up the integral.

Use the formula $V=\pi {\int }_c^d([R(y){]}^2-[r(y){]}^2)dy$.

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

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

(Note: This is technically a disk method problem, which is a special case of the washer method where the inner radius is zero.)

Step 4: 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

A graphing calculator can be used to visualize the region, find intersection points, and evaluate the definite integral for the volume. This is especially useful when the functions are complex or the intersection points are not simple integers.

Example: Find the volume from Example 1, $V=\pi {\int }_0^1(x-x^4)dx$, using a calculator.

1. Graph the Radii Squared: It can be helpful to graph the expressions for the radii squared. Let $Y_1=(\sqrt{X}{)}^2=X$ and $Y_2=(X^2{)}^2=X^4$. Graph these to confirm that $Y_1\ge Y_2$ on the interval $[0,1]$.

2. Use the Definite Integral Function: Access the numerical integration function (often fnInt on TI-84 calculators, found under the MATH menu).

3. Enter the Expression: The syntax is typically fnInt(expression, variable, lower bound, upper bound).

   • Enter: π * fnInt(X - X^4, X, 0, 1)

   • Alternatively, if you have $Y_1=\sqrt{X}$ and $Y_2=X^2$, you can enter: π * fnInt((Y₁)^2 - (Y₂)^2, X, 0, 1). You can access $Y_1 and $Y_2$ through the VARS` menu.

4. Calculate: The calculator will return an approximation of the volume.

   • Result: 0.942477...

   • To check against the exact answer: 3\pi/10 \approx 0.942477...

AP Exam Quick Hit

Common Question Types

• Set Up, Do Not Evaluate: You are given functions that define a region and an axis of revolution. The question asks you to write, but not solve, an integral expression for the volume.

  • Example: "Let R be the region bounded by $y=\ln (x)$, $y=0$, and $x=e$. Write an integral expression for the volume of the solid generated when R is revolved about the $x$-axis." (Answer: $V=\pi {\int }_1^e(\ln (x){)}^{2}dx$)

• Calculator-Active Volume Calculation: You are given two or more functions (often transcendental) and asked to find the volume. You must use your calculator to find the intersection points (bounds) and then to compute the value of the definite integral.

  • Example: "Let R be the region enclosed by the graphs of $f(x)=4\cos (x)$ and $g(x)=x^2$. Find the volume of the solid generated when R is revolved about the $x$-axis."

Common Mistakes

• Forgetting to Square the Radii: Writing the integrand as $\pi (R-r)$ instead of $\pi (R^2-r^2)$. Volume is based on the area of a circle ($\pi r^2$), so the radii must be squared.

• Squaring the Difference Incorrectly: Writing the integrand as $\pi (R-r{)}^2$ instead of $\pi (R^2-r^2)$. Remember, $(a-b{)}^2\ne a^2-b^2$. You must square each radius individually before subtracting.

• Mixing Up Radii: Incorrectly identifying the outer and inner radii. This will result in the negative of the correct volume. Always be sure that $R\ge r$ over the interval.

• Using the Wrong Variable: Forgetting to rewrite equations in terms of $y$ (e.g., $x=f(y)$) when revolving around the $y$-axis. The variable in the functions must match the differential ($dx$ or $dy$).

• Omitting $\pi$: Forgetting to include the constant $\pi$ in the setup of the integral and in the final answer.