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

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

The fundamental concept behind the disc method is to calculate the volume of a three-dimensional solid by treating it as a collection of an infinite number of infinitesimally thin circular discs. This method is used when we take a two-dimensional region in the plane and revolve it around an axis, creating a solid of revolution.

Imagine a single, thin rectangle within the 2D region, perpendicular to the axis of revolution. When this rectangle is revolved around the axis, it sweeps out a thin cylinder, or "disc." The volume of this disc can be found using the formula for the volume of a cylinder, $V=\pi r^{2}h$. In this context, the radius $r$ is the height of the rectangle (the function's value), and the height $h$ is the infinitesimally small width of the rectangle ($dx$ or $dy$). By integrating the volume of these discs over the entire interval, we are essentially summing them up to find the total volume of the solid.

Key Formulas

The specific formula used for the disc method depends on whether the region is revolved around a horizontal or vertical axis.

1. Revolution Around the $x$-axis:

   For a region bounded by the graph of a function $y=f(x)$, the $x$-axis, and the vertical lines $x=a$ and $x=b$, the volume of the solid generated by revolving the region about the $x$-axis is:

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

2. Revolution Around the $y$-axis:

   For a region bounded by the graph of a function $x=g(y)$, the $y$-axis, and the horizontal lines $y=c$ and $y=d$, the volume of the solid generated by revolving the region about the $y$-axis is:

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

Understanding the Radius

The most critical part of setting up a volume integral with the disc method is correctly identifying the radius of revolution. The term $[f(x){]}^2$ or $[g(y){]}^2$ in the formulas represents the square of the radius, $R^2$, which comes from the area formula of a circle, $A=\pi R^2$.

• For revolution around the $x$-axis: The axis of revolution is horizontal. We use vertical representative rectangles of width $dx$. The radius, $R$, of each disc is the distance from the $x$-axis to the curve. This distance is simply the function value, $y=f(x)$. Therefore, $R(x)=f(x)$.

• For revolution around the $y$-axis: The axis of revolution is vertical. We use horizontal representative rectangles of width $dy$. The radius, $R$, of each disc is the distance from the $y$-axis to the curve. This distance is the $x$-value, which must be expressed as a function of $y$, so $x=g(y)$. Therefore, $R(y)=g(y)$.

It is essential to ensure that the function, the limits of integration, and the differential ($dx$ or $dy$) are all in terms of the same variable. If revolving around the $y$-axis, you must solve the original equation for $x$ in terms of $y$.

Core Concepts & Rules

• The disc method is used to find the volume of a solid of revolution where the cross-sections perpendicular to the axis of revolution are solid circles (discs).

• The method is derived by summing the volumes of an infinite number of thin cylinders, where the volume of a single cylinder is $V_{cylinder}=\pi (\text{radius}{)}^2(\text{height})$.

• When revolving a region around the $x$-axis, the integral must be set up with respect to $x$.

  • The radius is a function of $x$: $R(x)=f(x)$.

  • The volume is $V=\pi {\int }_a^b[R(x){]}^{2}dx$.

  • The bounds of integration, $a$ and $b$, must be $x$-values.

• When revolving a region around the $y$-axis, the integral must be set up with respect to $y$.

  • The radius is a function of $y$: $R(y)=g(y)$. This may require rewriting an equation from $y=f(x)$ form to $x=g(y)$ form.

  • The volume is $V=\pi {\int }_c^d[R(y){]}^{2}dy$.

  • The bounds of integration, $c$ and $d$, must be $y$-values.

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

Problem: Find the volume of the solid formed by revolving the region bounded by the graph of $y=x^2+1$, the $x$-axis, $x=0$, and $x=2$ about the $x$-axis.

Step 1: Identify the Axis of Revolution and Integration Variable

The region is revolved around the $x$-axis. Therefore, we will integrate with respect to $x$, and our integral will use $dx$.

Step 2: Determine the Radius Function

The radius, $R(x)$, is the distance from the axis of revolution (the $x$-axis) to the outer edge of the region. This distance is given by the function itself.

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

Step 3: Set Up the Definite Integral

Using the formula $V=\pi {\int }_a^b[R(x){]}^{2}dx$, we substitute our radius function and the given bounds of integration, from $x=0$ to $x=2$.

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

Step 4: Evaluate the Integral

First, expand the integrand.

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

Now, substitute the expanded form back into the integral.

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

Find the antiderivative.

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

Apply the Fundamental Theorem of Calculus.

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

$$V=\pi \left(\frac{32}{5}+\frac{16}{3}+2-0\right)$$

Find a common denominator (15) to simplify.

$$V=\pi \left(\frac{96}{15}+\frac{80}{15}+\frac{30}{15}\right)=\pi \left(\frac{206}{15}\right)$$

The volume is $\frac{206\pi }{15}$.

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

Problem: Find the volume of the solid formed by revolving the region in the first quadrant bounded by the graph of $y=x^3$ and the line $y=8$ about the $y$-axis.

Step 1: Identify the Axis of Revolution and Integration Variable

The region is revolved around the $y$-axis. Therefore, we must integrate with respect to $y$, and our integral will use $dy$.

Step 2: Determine the Radius Function

The radius, $R(y)$, is the distance from the $y$-axis to the curve. This distance is the $x$-coordinate. We must express `x$ as a function of $y$.

Given $y=x^3$, we solve for $x$:

$$x=\sqrt[3]{y}=y^{1/3}$$

So, our radius function is:

$$R(y)=y^{1/3}$$

Step 3: Determine the Bounds of Integration

The region is bounded by $y=8$ and the curve $y=x^3$. The lower bound is where the curve intersects the $y$-axis, which is at $y=0$. So, the bounds are from $y=0$ to $y=8$.

Step 4: Set Up the Definite Integral

Using the formula $V=\pi {\int }_c^d[R(y){]}^{2}dy$, we substitute our radius function and bounds.

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

Step 5: Evaluate the Integral

Simplify the integrand.

$$(y^{1/3}{)}^2=y^{2/3}$$

Now, integrate.

$$V=\pi {\int }_0^8{y}^{2/3}dy$$

Find the antiderivative using the power rule.

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

Apply the Fundamental Theorem of Calculus.

$$V=\pi \left(\frac{3}{5}(8{)}^{5/3}-\frac{3}{5}(0{)}^{5/3}\right)$$

$$V=\pi \left(\frac{3}{5}(\sqrt[3]{8}{)}^5-0\right)=\pi \left(\frac{3}{5}(2{)}^5\right)=\pi \left(\frac{3}{5}(32)\right)$$

The volume is $\frac{96\pi }{5}$.

Using Your Calculator

A graphing calculator is an excellent tool for evaluating the definite integrals that arise from volume problems, especially on the calculator-active section of the AP exam. However, you must be able to correctly set up the integral by hand first.

To calculate the volume from Example 1, $V=\pi {\int }_0^2(x^2+1{)}^{2}dx$:

1. Press the math` key and select `fnInt(` (usually option 9) or press `ALPHA` `WINDOW` and select `fnInt(`. 2. The screen will show a template for a definite integral. 3. Enter the lower bound: $0.

2. Enter the upper bound: $2$.

3. Enter the integrand. Be careful with parentheses: $(X^2+1{)}^2$.

4. Enter the variable of integration: $X$.

5. Press ENTER. The calculator will return a decimal value (e.g., $\mathrm{43.1445...}$).

6. Crucially, do not forget to multiply this result by $\pi$. The fnInt command only computes the integral part. You can either multiply the result by \pi` or include it in the initial expression: `π * fnInt((X^2+1)^2, X, 0, 2)`. 9. To verify your exact answer, you can calculate $206\pi/15 on your calculator and see if the decimal approximations match.

AP Exam Quick Hit

Common Question Types

• Direct Calculation: You will be given a function and a bounded region and asked to find the volume of the solid generated by revolving it around the $x$- or $y$-axis. This is common in both multiple-choice and free-response sections.

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

• Setup Only: You will be asked to identify the correct integral that represents the volume, without needing to calculate it. This tests your understanding of the setup (radius, bounds, variable) directly.

  • Example: "The region bounded by $y=e^x$, the $y$-axis, and the line $y=3$ is revolved about the $y$-axis. Which of the following integrals gives the volume of the solid?" (The correct answer would involve an integral with $dy$ and $R(y)=\ln (y)$).

• Part of a Larger FRQ: A volume calculation is often one part of a multi-part free-response question that also asks for the area of the region, perimeter, or volumes of solids with other cross-sections.

Common Mistakes

• Forgetting to Square the Radius: A very frequent error is to integrate $\pi \cdot R(x)$ instead of $\pi \cdot [R(x){]}^2$. Remember that the formula comes from the area of a circle, $\pi r^2$.

• Omitting $\pi$: Students often correctly set up and evaluate the integral part but forget to multiply the final answer by $\pi$.

• Using the Wrong Variable or Bounds: When revolving around the $y$-axis, it is a common mistake to use an integral with $dx$ or to use $x$-values for the bounds of a $dy$ integral. Always ensure your function, differential ($dx/dy$), and bounds are all in terms of the same variable.

• Incorrectly Rewriting the Function: For $y$-axis revolutions, you must solve for $x$ in terms of $y$. Algebraic errors in this step (e.g., incorrectly taking a square root) will lead to the wrong setup.

• Algebraic Errors in Expansion: When squaring a binomial radius, such as $(x^2+1{)}^2$, students may incorrectly write $x^4+1$ instead of the correct expansion $x^4+2x^2+1$.