Volume with Washer Method: Revolving Around Other Axes - AP Calculus BC Study Guide

The Core Idea: Volume with Washer Method: Revolving Around Other Axes

The washer method is a technique for finding the volume of a three-dimensional solid that has a hole in the middle. This solid is generated by revolving a two-dimensional region, bounded by two different functions, around an axis. This topic extends the concept to axes of revolution that are not the x- or y-axis, such as a horizontal line $y=c$ or a vertical line $x=k$.

The fundamental principle is to slice the solid into infinitesimally thin "washers" (disks with a hole in the center) and sum their volumes using a definite integral. The volume of each individual washer is calculated by finding the area of its face—the area of the larger, outer circle minus the area of the smaller, inner hole—and multiplying by its thickness. The key to solving these problems is correctly identifying the outer radius ($R$) and the inner radius ($r$) of the washer, which are defined as the distances from the axis of revolution to the outer and inner curves of the original two-dimensional region.

Key Formulas

The volume of a solid of revolution generated by a region bounded by two functions can be found using one of two formulas, depending on the orientation of the axis of revolution. In both cases, $R$ represents the outer radius (distance from the axis of revolution to the farther function) and $r$ represents the inner radius (distance from the axis of revolution to the closer function).

Revolution Around a Horizontal Line $y=c$

For a region bounded above by $y=f(x)$ and below by $y=g(x)$ on the interval $[a,b]$, where $f(x)\ge g(x)$, the volume of the solid generated by revolving the region about the horizontal line $y=c$ is given by:

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

• The representative washer has a thickness of $\Delta x$.

• The outer radius is $R(x)$.

• The inner radius is $r(x)$.

Revolution Around a Vertical Line $x=k$

For a region bounded on the right by $x=f(y)$ and on the left by $x=g(y)$ on the interval $[c,d]$, where $f(y)\ge g(y)$, the volume of the solid generated by revolving the region about the vertical line $x=k$ is given by:

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

• The representative washer has a thickness of $\Delta y$.

• The outer radius is $R(y)$.

• The inner radius is $r(y)$.

Understanding Radii from an Axis of Revolution

The most critical step in applying the washer method for non-coordinate axes is correctly defining the outer radius $R$ and inner radius $r$. These radii are always measured as the perpendicular distance from the axis of revolution to the boundary curves of the region.

For a Horizontal Axis of Revolution $y=c$:

The radii are vertical distances, calculated as "Top curve minus Bottom curve."

• If the axis of revolution $y=c$ is below the region, the radii are:

  • Outer Radius $R(x)=(\text{farther curve})-(\text{axis of revolution})=f(x)-c$

  • Inner Radius $r(x)=(\text{closer curve})-(\text{axis of revolution})=g(x)-c$

• If the axis of revolution $y=c$ is above the region, the radii are:

  • Outer Radius $R(x)=(\text{axis of revolution})-(\text{farther curve})=c-g(x)$

  • Inner Radius $r(x)=(\text{axis of revolution})-(\text{closer curve})=c-f(x)$

For a Vertical Axis of Revolution $x=k$:

The radii are horizontal distances, calculated as "Right curve minus Left curve."

• If the axis of revolution $x=k$ is to the left of the region, the radii are:

  • Outer Radius $R(y)=(\text{farther curve})-(\text{axis of revolution})=f(y)-k$

  • Inner Radius $r(y)=(\text{closer curve})-(\text{axis of revolution})=g(y)-k$

• If the axis of revolution $x=k$ is to the right of the region, the radii are:

  • Outer Radius $R(y)=(\text{axis of revolution})-(\text{farther curve})=k-g(y)$

  • Inner Radius $r(y)=(\text{axis of revolution})-(\text{closer curve})=k-f(y)$

Core Concepts & Rules

• Method Foundation: The washer method calculates the volume of a solid of revolution with a hole by integrating the area of a representative washer, $A=\pi (R^2-r^2)$.

• Radii Definition: The outer radius $R$ and inner radius $r$ are always defined as the distances from the axis of revolution to the outer and inner boundary curves of the region, respectively.

• Horizontal Axis ($y=c$): When revolving around a horizontal line, the integral must be with respect to $x$. The radii, $R(x)$ and $r(x)$, are functions of $x$ and represent vertical distances. The volume formula is $V=\pi {\int }_a^b([R(x){]}^2-[r(x){]}^2)dx$.

• Vertical Axis ($x=k$): When revolving around a vertical line, the integral must be with respect to $y$. The functions must be expressed in the form $x=f(y)$, and the radii, $R(y)$ and $r(y)$, are functions of $y$ representing horizontal distances. The volume formula is $V=\pi {\int }_c^d([R(y){]}^2-[r(y){]}^2)dy$.

Step-by-Step Example 1: Revolving Around a Horizontal Line

Problem: Let $R$ be the region enclosed by the graphs of $f(x)=\sqrt{x}$ and $g(x)=x^2$. Find the volume of the solid generated when $R$ is revolved about the line $y=-1$.

Step 1: Sketch the region and axis of revolution.

The region is bounded by the square root function (top) and the parabola (bottom). The axis of revolution, $y=-1$, is a horizontal line below the region.

Step 2: Determine the variable of integration and find the bounds.

Since the axis of revolution is horizontal, we will integrate with respect to $x$. To find the bounds, set the functions equal to each other:

$\sqrt{x}=x^2⟹x=x^4⟹x^4-x=0⟹x(x^3-1)=0$

The points of intersection are $x=0$ and $x=1$. So, our interval is $[0,1]$.

Step 3: Define the outer and inner radii, $R(x)$ and $r(x)$.

The radii are the distances from the axis $y=-1$ to the curves.

• The outer curve is the one farther from $y=-1$, which is $f(x)=\sqrt{x}$.

  $R(x)=(\text{top curve})-(\text{axis})=\sqrt{x}-(-1)=\sqrt{x}+1$

• The inner curve is the one closer to $y=-1$, which is $g(x)=x^2$.

  $r(x)=(\text{bottom curve})-(\text{axis})=x^2-(-1)=x^2+1$

Step 4: Set up the definite integral.

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

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

Step 5: Evaluate the integral.

First, expand the squared terms:

$(\sqrt{x}+1{)}^2=x+2\sqrt{x}+1$

$(x^2+1{)}^2=x^4+2x^2+1$

Now, substitute back into the integral:

$V=\pi {\int }_0^1\left((x+2\sqrt{x}+1)-(x^4+2x^2+1)\right)dx$

$V=\pi {\int }_0^1(-x^4-2x^2+x+2x^{1/2})dx$

Find the antiderivative:

$V=\pi {\left[-\frac{1}{5}{x}^5-\frac{2}{3}{x}^3+\frac{1}{2}{x}^2+2\cdot \frac{2}{3}{x}^{3/2}\right]}_0^1$

$V=\pi {\left[-\frac{1}{5}{x}^5-\frac{2}{3}{x}^3+\frac{1}{2}{x}^2+\frac{4}{3}{x}^{3/2}\right]}_0^1$

Evaluate at the bounds:

$V=\pi \left(\left(-\frac{1}{5}-\frac{2}{3}+\frac{1}{2}+\frac{4}{3}\right)-(0)\right)$

$V=\pi \left(-\frac{1}{5}+\frac{2}{3}+\frac{1}{2}\right)=\pi \left(-\frac{6}{30}+\frac{20}{30}+\frac{15}{30}\right)=\frac{29\pi }{30}$

Step-by-Step Example 2: Revolving Around a Vertical Line

Problem: Let $R$ be the region enclosed by the graphs of $y=\sqrt{x}$ and $y=x^2$. Find the volume of the solid generated when $R$ is revolved about the line $x=2$.

Step 1: Sketch the region and axis of revolution.

The region is the same as in Example 1. The axis of revolution, $x=2$, is a vertical line to the right of the region.

Step 2: Determine the variable of integration and find the bounds.

Since the axis of revolution is vertical, we must integrate with respect to $y$. We need to rewrite our functions in terms of $y$:

$y=\sqrt{x}⟹x=y^2$

$y=x^2⟹x=\sqrt{y}$

The bounds for $y$ are from $y=0$ to $y=1$.

Step 3: Define the outer and inner radii, $R(y)$ and $r(y)$.

The radii are the horizontal distances from the axis $x=2$ to the curves.

• The outer curve is the one farther from $x=2$, which is the left-most curve, $x=y^2$.

  $R(y)=(\text{axis})-(\text{left curve})=2-y^2$

• The inner curve is the one closer to $x=2$, which is the right-most curve, $x=\sqrt{y}$.

  $r(y)=(\text{axis})-(\text{right curve})=2-\sqrt{y}$

Step 4: Set up the definite integral.

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

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

Step 5: Evaluate the integral.

First, expand the squared terms:

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

$(2-\sqrt{y}{)}^2=4-4\sqrt{y}+y$

Now, substitute back into the integral:

$V=\pi {\int }_0^1\left((4-4y^2+y^4)-(4-4\sqrt{y}+y)\right)dy$

$V=\pi {\int }_0^1(y^4-4y^2-y+4y^{1/2})dy$

Find the antiderivative:

$V=\pi {\left[\frac{1}{5}{y}^5-\frac{4}{3}{y}^3-\frac{1}{2}{y}^2+4\cdot \frac{2}{3}{y}^{3/2}\right]}_0^1$

$V=\pi {\left[\frac{1}{5}{y}^5-\frac{4}{3}{y}^3-\frac{1}{2}{y}^2+\frac{8}{3}{y}^{3/2}\right]}_0^1$

Evaluate at the bounds:

$V=\pi \left(\left(\frac{1}{5}-\frac{4}{3}-\frac{1}{2}+\frac{8}{3}\right)-(0)\right)$

$V=\pi \left(\frac{1}{5}+\frac{4}{3}-\frac{1}{2}\right)=\pi \left(\frac{6}{30}+\frac{40}{30}-\frac{15}{30}\right)=\frac{31\pi }{30}$

Using Your Calculator

The primary use of a calculator for this topic is to evaluate the definite integral after you have correctly set it up. You must still determine the bounds and the expressions for the radii analytically.

To solve Example 1 ($V=\pi {\int }_0^1((\sqrt{x}+1{)}^2-(x^2+1{)}^2)dx$) with a TI-84 style calculator:

1. Analytically determine the integrand:$(\sqrt{x}+1{)}^2-(x^2+1{)}^2$.

2. Access the numerical integration function: Press [MATH] and select 9: fnInt(.

3. Enter the arguments:

   • Lower bound: $0$

   • Upper bound: $1$

   • Expression: $(\surd (X)+1{)}^2-(X^2+1{)}^2$

   • Variable: $X$

4. The screen should look like: fnInt((√(X)+1)^2 - (X^2+1)^2, X, 0, 1)

5. Press [ENTER] to get the value of the integral (approximately $\mathrm{0.9666...}$).

6. Multiply by $\pi$: Do not forget to multiply the result by $\pi$. The final answer is $\pi \ast \mathrm{0.9666...}\approx 3.0368$.

AP Exam Quick Hit

Common Question Types

• Integral Setup (Multiple Choice): You will be given a description of a region and an axis of revolution and asked to choose the correct definite integral that represents the volume. These questions test your ability to correctly identify the variable of integration ($dx$ vs. $dy$) and formulate the expressions for $R$ and $r$.

  • Example: "The region bounded by $y=e^x$, $y=1$, and $x=2$ is revolved about the line $y=-2$. Which of the following integrals gives the volume of the solid?"

• Volume Calculation (Free Response, Calculator-Active): A part of an FRQ will define a region and ask you to find the volume of a solid generated by revolving it around a specified horizontal or vertical line. You are expected to write the correct integral setup on your paper and then use your calculator to find the decimal value.

  • Example: "Let R be the region bounded by $y=\ln (x)$ and $y=x-2$. Find the volume of the solid generated when R is revolved about the line $x=-1$."

Common Mistakes

• Incorrect Radii Calculation: The most frequent error is forgetting to measure the radii from the axis of revolution. For a region bounded by $f(x)$ and $g(x)$ revolved around $y=c$, students might incorrectly use $R(x)=f(x)$ instead of the correct $R(x)=|f(x)-c|$.

• Squaring $R-r$ instead of $R^2-r^2$: A major conceptual error is setting up the integral as $V=\pi \int (R-r{)}^{2}dx$. The correct formula is $V=\pi \int (R^2-r^2)dx$, which represents the area of the outer disk minus the area of the inner disk.

• Using $x$-variables in a $dy$ Integral: When revolving around a vertical axis, all functions and bounds must be in terms of $y$. A common mistake is to leave the functions as $f(x)$ and $g(x)$ inside a $dy$ integral.

• Reversing Outer and Inner Radii: Incorrectly identifying which function is farther from the axis of revolution. This leads to $r^2-R^2$ in the integrand, resulting in a negative volume. Volume must always be positive.

• Forgetting $\pi$: A simple but costly mistake is to correctly set up and evaluate the integral but forget to multiply the final result by $\pi$.