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

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

This topic extends the fundamental concept of finding volumes of solids of revolution. Previously, volumes were calculated for regions revolved around the primary coordinate axes ($x=0$ or $y=0$). Here, we generalize this process to find the volume of a solid generated by revolving a region around any horizontal line ($y=k$) or any vertical line ($x=h$).

The core principle remains unchanged: we slice the solid into an infinite number of infinitesimally thin circular cross-sections (discs), find the area of a representative disc, and then integrate this area across the appropriate interval to sum the volumes of all the discs. The critical new step is correctly defining the radius ($R$) of each disc. The radius is no longer simply the function's value but is the distance from the new axis of revolution to the boundary of the region being revolved. This requires careful consideration of the geometry of the region relative to the axis of revolution.

Key Formulas

The volume of a solid of revolution is found by integrating the area of its circular cross-sections, $A={\text{piR}}^2$. The specific formula depends on the orientation of the axis of revolution.

Revolution Around a Horizontal Line $y=k$

For a region bounded by $y=f(x)$ and the line $y=k$ between $x=a$ and $x=b$, where the region is flush against the line $y=k$, the volume of the solid generated by revolving the region about $y=k$ is given by:

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

where the radius, $R(x)$, is the distance from the axis of revolution $y=k$ to the curve $y=f(x)$.

$$R(x)=|f(x)-k|$$

Revolution Around a Vertical Line $x=h$

For a region bounded by $x=g(y)$ and the line $x=h$ between $y=c$ and $y=d$, where the region is flush against the line $x=h$, the volume of the solid generated by revolving the region about $x=h$ is given by:

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

where the radius, $R(y)$, is the distance from the axis of revolution $x=h$ to the curve $x=g(y)$.

$$R(y)=|g(y)-h|$$

Understanding the Radius

The most critical skill in this topic is correctly defining the radius, $R$. The radius is always a measure of distance from the axis of revolution to the outer edge of the solid at a particular point.

• Orientation is Key: The orientation of the axis of revolution dictates the variable of integration.

  • A horizontal axis of revolution (e.g., $y=2$, $y=-5$) means the radius is a vertical distance. The setup will be in terms of $x$, and the integral will be $...dx$.

  • A vertical axis of revolution (e.g., $x=1$, $x=-3$) means the radius is a horizontal distance. The setup will be in terms of $y$, and the integral will be $...dy$.

• Calculating the Distance: The radius $R$ is the distance between the function that bounds the region and the axis of revolution.

  • For a horizontal axis $y=k$ and a function $y=f(x)$, the radius is the vertical distance $R(x)=|f(x)-k|$. You can think of this as $(Topy-value)-(Bottomy-value)$. If $f(x)$ is above the axis $y=k$, the radius is $f(x)-k$. If $f(x)$ is below the axis $y=k$, the radius is $k-f(x)$.

  • For a vertical axis $x=h$ and a function $x=g(y)$, the radius is the horizontal distance $R(y)=|g(y)-h|$. You can think of this as $(Rightx-value)-(Leftx-value)$. If $g(y)$ is to the right of the axis $x=h$, the radius is $g(y)-h$. If $g(y)$ is to the left of the axis $x=h$, the radius is $h-g(y)$.

The disc method specifically applies when the region being revolved is flush against the axis of revolution, creating a solid with no hole in the middle.

Core Concepts & Rules

• The volume of a solid of revolution is found by integrating the area of its disc-shaped cross-sections: $V=\int A(x)dx$ or $V=\int A(y)dy$.

• The area of each disc is $A={\text{piR}}^2$, where $R$ is the radius of the disc.

• When revolving around a horizontal line $y=k$, the integral must be with respect to $x$. The radius $R(x)$ is a function of $x$ representing the vertical distance from the line $y=k$ to the boundary curve.

• When revolving around a vertical line $x=h$, the integral must be with respect to $y$. The radius $R(y)$ is a function of $y$ representing the horizontal distance from the line $x=h$ to the boundary curve.

• The radius $R$ is always calculated as the distance from the axis of revolution to the outer boundary of the region. For a region bounded by a curve and an axis of revolution $y=k$ or $x=h$, the radius is $|curve-axis|$.

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

Problem: Let R be the region bounded by the graph of $y=x^2+1$, the line $y=1$, and the line $x=2$. Find the volume of the solid generated when R is revolved about the line $y=1$.

Step 1: Visualize the Region and Axis of Revolution

Sketch the parabola $y=x^2+1$, which opens upwards with its vertex at $(0,1)$. The region is bounded by this parabola, the horizontal line $y=1$, and the vertical line $x=2$. The axis of revolution is $y=1$, which is one of the boundaries of the region. This confirms the disc method is appropriate.

Step 2: Determine the Variable of Integration and Limits

The axis of revolution, $y=1$, is a horizontal line. Therefore, we must integrate with respect to $x$. The region is bounded by $x=0$ (where the parabola meets the line $y=1$) and $x=2$. So, our limits of integration are from $a=0$ to $b=2$.

Step 3: Define the Radius $R(x)$

The radius is the vertical distance from the axis of revolution ($y=1$) to the outer boundary of the region ($y=x^2+1$).

$$R(x)=(\text{Top Curve})-(\text{Bottom Curve/Axis})$$

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

$$R(x)=x^2$$

Step 4: Set Up the Volume Integral

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

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

$$V=\pi {\int }_0^2{x}^{4}dx$$

Step 5: Evaluate the Integral

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

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

$$V=\pi \left(\frac{32}{5}-0\right)$$

$$V=\frac{32\pi }{5}$$

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

Problem: Let R be the region in the first quadrant bounded by the graph of $x=\sqrt{y}+2$, the line $x=2$, and the line $y=4$. Find the volume of the solid generated when R is revolved about the line $x=2$.

Step 1: Visualize the Region and Axis of Revolution

Sketch the curve $x=\sqrt{y}+2$, which is a parabola opening to the right with its vertex at $(2,0)$. The region is bounded by this curve, the vertical line $x=2$, and the horizontal line $y=4$. The axis of revolution is $x=2$, which is one of the boundaries of the region. This confirms the disc method is appropriate.

Step 2: Determine the Variable of Integration and Limits

The axis of revolution, $x=2$, is a vertical line. Therefore, we must integrate with respect to $y$. The region is bounded by $y=0$ and $y=4$. So, our limits of integration are from $c=0$ to $d=4$.

Step 3: Define the Radius $R(y)$

The radius is the horizontal distance from the axis of revolution ($x=2$) to the outer boundary of the region ($x=\sqrt{y}+2$).

$$R(y)=(\text{Right Curve})-(\text{Left Curve/Axis})$$

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

$$R(y)=\sqrt{y}$$

Step 4: Set Up the Volume Integral

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

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

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

Step 5: Evaluate the Integral

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

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

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

$$V=8\pi$$

Using Your Calculator

The primary skill assessed is setting up the correct definite integral. A calculator is then used for efficient and accurate evaluation, especially when the resulting integral is complex or on a calculator-active portion of the exam.

To find the volume from Example 1, $V=\int [0,2]\pi (x^2{)}^{2}dx$, on a TI-84 style calculator:

1. Analytically determine the integrand: First, you must perform the analysis to find that the integrand is $\pi (x^2{)}^2$ or ${\text{pix}}^4$. The calculator cannot determine the radius for you.

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

3. Enter the arguments: The syntax is $fnInt(expression,variable,lowerbound,upperbound)$.

   • $expression$: $\pi \ast X^4$

   • $variable$: $X$

   • $lowerbound$: $0$

   • $upperbound$: $2$

4. Your screen should look like this:fnInt(π*X^4, X, 0, 2)

5. Press [ENTER]: The calculator will return the approximate decimal value, $\mathrm{20.106...}$. You can verify this is $32\pi /5$.

AP Exam Quick Hit

Common Question Types

• Setup, No Solve: A multiple-choice question provides a description of a region and an axis of revolution and asks you to select the definite integral that represents the volume.

  • Example: "The region R is bounded by $y=\cos (x)+2$, $y=2$, $x=0$, and $x=\pi /2$. Which integral gives the volume of the solid formed by revolving R about the line $y=2$?" The correct setup would be $V=\int [0,\pi /2]\pi (\cos (x){)}^{2}dx$.

• FRQ Part (Calculator Active): A free-response question defines a region with functions that may be difficult to integrate by hand. You are expected to set up the correct integral and use your calculator to find the final decimal answer.

  • Example: "Let R be the region bounded by $y=e^x-1$ and $y=\ln (x+1)$. Find the volume of the solid generated by revolving the portion of R in the first quadrant that is bounded below by $y=0$ and above by $y=e^x-1$ about the line $y=-1$." (Note: This would be a washer problem, a common extension. A pure disc problem would have the region bounded by $y=e^x-1$ and $y=0$ revolved around $y=0$).

Common Mistakes

• Incorrect Radius Definition: Forgetting to subtract the axis of revolution. For a region bounded by $y=f(x)$ and $y=k$ revolved around $y=k$, using $R(x)=f(x)$ instead of the correct $R(x)=f(x)-k$.

• Forgetting to Square the Radius: A very common algebraic error is to set up the integral as $V=\int \pi R(x)dx$ instead of the correct $V=\int \pi [R(x){]}^{2}dx$. The formula is based on the area of a circle, ${\text{pir}}^2$.

• Incorrect Variable of Integration: Using $dx$ for a vertical axis of revolution or $dy$ for a horizontal one. Remember: Horizontal Axis ($y=k$) → $dx$; Vertical Axis ($x=h$) → $dy$.

• Errors in Squaring Binomials: When the radius is a binomial, such as $R(x)=x+1$, students may incorrectly square it as $x^2+1$ instead of the correct $(x+1{)}^2=x^2+2x+1$.

• Confusing Disc and Washer Methods: Applying the disc method formula (${\text{piR}}^2$) to a solid that has a hole in it. The disc method is only for solids where the region of revolution is flush against the axis of revolution. If there is a gap, the washer method must be used.