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

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

This topic expands on the concept of finding volumes of solids of revolution. While the disk and washer methods are initially learned with respect to the x- or y-axis, many solids are generated by revolving a region around other horizontal or vertical lines (e.g., $y=3$ or $x=-1$). The fundamental principle remains the same: volume is found by integrating the area of a cross-section. For the washer method, this cross-section is a "washer" with an outer and inner radius.

The critical adjustment for revolving around other axes is the definition of these radii. The outer radius, $R$, and the inner radius, $r$, are no longer simply the function values themselves. Instead, they must be defined as the distances from the axis of revolution to the outer and inner boundaries of the region being revolved. This requires careful consideration of the geometry, typically involving subtraction to find the correct lengths for $R$ and $r$ before setting up the integral.

Key Formulas

The volume of a solid generated by revolving a region about a horizontal or vertical line is found by integrating the area of a representative washer, $A=\pi (R^2-r^2)$.

1. Revolving around a Horizontal Line $y=k$:

   For a region bounded by $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$, where $f(x)\ge g(x)$, the volume is:

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

   • $R(x)$ is the outer radius: the distance from the line $y=k$ to the farther function.

   • $r(x)$ is the inner radius: the distance from the line $y=k$ to the closer function.

2. Revolving around a Vertical Line $x=h$:

   For a region bounded by $x=f(y)$ and $x=g(y)$ from $y=c$ to $y=d$, where $f(y)\ge g(y)$, the volume is:

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

   • $R(y)$ is the outer radius: the distance from the line $x=h$ to the farther function.

   • $r(y)$ is the inner radius: the distance from the line $x=h$ to the closer function.

Understanding the Radii

The most important skill for this topic is correctly defining the outer radius $R$ and inner radius $r$. These radii are always positive lengths representing the distance from the axis of revolution to the boundary of the region. A sketch is essential.

For a horizontal axis of revolution $y=k$ (integrating with respect to $x$):

• The radii are vertical distances, calculated using "Top - Bottom".

• The outer radius $R(x)$ is the distance from $y=k$ to the curve that is farther away.

• The inner radius $r(x)$ is the distance from $y=k$ to the curve that is closer.

Example: If revolving a region between $y=f(x)$ (top curve) and $y=g(x)$ (bottom curve) around the line $y=-2$ (which is below the region), the radii are:

• $R(x)=f(x)-(-2)=f(x)+2$ (distance from far curve to axis)

• $r(x)=g(x)-(-2)=g(x)+2$ (distance from near curve to axis)

For a vertical axis of revolution $x=h$ (integrating with respect to $y$):

• The radii are horizontal distances, calculated using "Right - Left".

• The outer radius $R(y)$ is the distance from $x=h$ to the curve that is farther away.

• The inner radius $r(y)$ is the distance from $x=h$ to the curve that is closer.

Example: If revolving a region between $x=f(y)$ (right curve) and $x=g(y)$ (left curve) around the line $x=5$ (which is to the right of the region), the radii are:

• $R(y)=5-g(y)$ (distance from axis to far curve)

• $r(y)=5-f(y)$ (distance from axis to near curve)

Core Concepts & Rules

• The washer method is used when the solid of revolution has a hole or cavity in the middle.

• The formula for the volume is derived by integrating the area of a washer: $\pi \int ({\text{Outer Radius}}^2-{\text{Inner Radius}}^2)d(\text{variable})$.

• A horizontal axis of revolution (e.g., $y=k$) implies an integral with respect to $x$. All functions and radii must be in terms of $x$.

• A vertical axis of revolution (e.g., $x=h$) implies an integral with respect to $y$. All functions and radii must be in terms of $y$.

• The outer radius $R$ is always the distance from the axis of revolution to the farther boundary of the region.

• The inner radius $r$ is always the distance from the axis of revolution to the closer boundary of the region.

• Radii are always found by subtracting the smaller coordinate from the larger coordinate to ensure a positive distance (e.g., Top - Bottom, Right - Left).

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

Problem: Let R be the region enclosed by the graphs of $y=x^2$ and $y=2x$. Find the volume of the solid generated when R is revolved about the line $y=-1$.

Step 1: Sketch the Region and Find Bounds

• Sketch the parabola $y=x^2$ and the line $y=2x$.

• Find the points of intersection by setting the functions equal: $x^2=2x⟹x^2-2x=0⟹x(x-2)=0$. The intersection points are at $x=0$ and $x=2$. These are our bounds of integration, $a=0$ and $b=2$.

• Draw the horizontal axis of revolution, $y=-1$. Notice it is below the region R.

Step 2: Determine the Radii

• The axis is horizontal, so we will integrate with respect to $x$. The radii are vertical distances.

• Outer Radius $R(x)$: The top curve of the region is $y=2x$, which is farther from the axis $y=-1$. The distance is $Top-Bottom$.

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

• Inner Radius $r(x)$: The bottom curve of the region is $y=x^2$, which is closer to the axis $y=-1$. The distance is $Top-Bottom$.

  $$r(x)=(x^2)-(-1)=x^2+1$$

Step 3: Set Up the Integral

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

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

Step 4: Evaluate the Integral

• Expand the squared terms:

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

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

• Substitute back into the integral:

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

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

• Find the antiderivative:

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

• Apply the Fundamental Theorem of Calculus:

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

  $$V=\pi \left(-\frac{32}{5}+\frac{16}{3}+8\right)=\pi \left(-\frac{96}{15}+\frac{80}{15}+\frac{120}{15}\right)=\frac{104\pi }{15}$$

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

Problem: Let R be the region in the first quadrant bounded by $y=x^3$, the y-axis, and the line $y=8$. Set up, but do not evaluate, an integral expression for the volume of the solid generated when R is revolved about the vertical line $x=3$.

Step 1: Sketch the Region and Find Bounds

• Sketch the curve $y=x^3$, the y-axis ($x=0$), and the horizontal line $y=8$.

• The region is bounded on the left by $x=0$ and on the right by $y=x^3$. It is bounded below by $y=0$ and above by $y=8$.

• Draw the vertical axis of revolution, $x=3$. Notice it is to the right of the region R.

Step 2: Determine the Radii

• The axis is vertical, so we must integrate with respect to $y$.

• Rewrite the function in terms of $y$: $y=x^3⟹x=\sqrt[3]{y}$.

• The bounds of integration are along the y-axis, from $c=0$ to $d=8$.

• The radii are horizontal distances, calculated using "Right - Left".

• Outer Radius $R(y)$: The axis of revolution is $x=3$. The left boundary of the region is $x=0$, which is farther from the axis. The distance is $Right-Left$.

  $$R(y)=3-0=3$$

• Inner Radius $r(y)$: The right boundary of the region is $x=\sqrt[3]{y}$, which is closer to the axis $x=3$. The distance is $Right-Left$.

  $$r(y)=3-\sqrt[3]{y}$$

Step 3: Set Up the Integral

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

  $$V=\pi {\int }_0^8((3{)}^2-(3-\sqrt[3]{y}{)}^2)dy$$

  $$V=\pi {\int }_0^8(9-(3-y^{1/3}{)}^2)dy$$

Using Your Calculator

A graphing calculator is extremely useful for finding volumes, especially when the resulting integral is difficult or impossible to evaluate by hand. The primary steps are setting up the integral correctly by hand and then using the calculator to evaluate it.

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

1. Define the Radii: It can be helpful to store the radii functions.

   • In the Y= editor, enter Y1 = 2X+1 (for R(x)).

   • Enter Y2 = X^2+1 (for r(x)).

2. Use the Numerical Integration Function:

   • From the home screen, press MATH and select 9: fnInt(.

   • The syntax is $fnInt(expression,variable,lowerbound,upperbound)$.

   • Enter the expression for the area of the washer, (Y1)^2 - (Y2)^2. You can access Y1 and Y2 by pressing VARS, going to the Y-VARS menu, and selecting 1: Function.

   • Your entry should look like this: fnInt((Y1)^2 - (Y2)^2, X, 0, 2)

   • Press ENTER. The result will be approximately $34.765$.

3. Multiply by $\pi$:

   • The calculator gives the value of the integral. Don't forget to multiply this result by $\pi$.

   • $\mathrm{34.765...}\ast \pi \approx 109.22$$.

   • To get the exact fraction, you can enter (104/15)*π to verify.

The calculator can also be used to find the bounds of integration ($a,b$) by graphing the functions and using the $CALC$ menu's $5$: intersect feature.

AP Exam Quick Hit

Common Question Types

• FRQ Setup: You will be given two or more functions, a region, and an axis of revolution. The question will ask you to "Write, but do not evaluate, an integral expression that gives the volume." This tests your ability to choose the correct variable of integration ($dx$ or $dy$) and correctly define $R$ and $r$.

• FRQ Calculator-Active Evaluation: You will be given functions (often complex, like trigonometric or logarithmic) and an axis of revolution. You will need to use your calculator to find the intersection points (bounds) and then use the numerical integration feature (fnInt) to compute the final volume.

• MCQ Integral Setup: A multiple-choice question will show a region and an axis of revolution and provide four or five integral options. You must select the one with the correct radii ($R$ and $r$), variable of integration, and bounds.

Common Mistakes

• Incorrect Radii Definition: The most frequent error is forgetting that the radii are distances to the axis of revolution. Students often just use the functions themselves, for example writing $R(x)=2x$ instead of $R(x)=2x-(-1)$. Always think "distance."

• Algebraic Error in Squaring: Calculating $\pi \int (R-r{)}^{2}dx$ instead of the correct $\pi \int (R^2-r^2)dx$. Remember to square each radius before you subtract.

• Mixing Up $R$ and $r$: Incorrectly identifying which function is farther from the axis. A quick, simple sketch of the region and the axis of revolution will always prevent this error.

• Using the Wrong Variable of Integration: Integrating with respect to $x$ when revolving around a vertical axis, or vice-versa. Remember: Horizontal Axis $⟹dx$, Vertical Axis $⟹dy$.

• Forgetting to Multiply by $\pi$: A simple but costly mistake, especially on free-response questions. The formula for the area of the washer is $A=\pi (R^2-r^2)$, so $\pi$ must be part of the volume integral.