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

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

The fundamental concept of this topic is to determine the volume of a three-dimensional solid that is generated by revolving a two-dimensional region around an axis. The method involves conceptualizing this solid as being composed of an infinite number of infinitesimally thin circular discs stacked along the axis of revolution.

The volume of a single representative disc is the area of its circular face, $\pi r^2$, multiplied by its thickness ($dx$ or $dy$). By summing the volumes of all these discs across the region using a definite integral, we can find the exact volume of the entire solid. This process translates the geometric problem of finding a solid's volume into the analytical problem of evaluating a definite integral. The key is to correctly identify the radius of a representative disc as a function of its position along the axis of revolution.

Key Formulas

The volume of a solid of revolution using the disc method is calculated with one of two primary formulas, depending on the axis of revolution.

Revolution Around the $x$-Axis

For a region bounded by the graph of a function $y=f(x)$, the $x$-axis ($y=0$), and the vertical lines $x=a$ and $x=b$, the volume of the solid generated by revolving this region about the $x$-axis is given by:

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

In this setup:

• The integration is performed with respect to $x$.

• The limits of integration, $a$ and $b$, are $x$-values.

• The radius of a representative disc at any given $x$ is $r=f(x)$. The formula integrates the area of these discs, $A(x)=\pi [f(x){]}^2$, over the interval $[a,b]$.

Revolution Around the $y$-Axis

For a region bounded by the graph of a function $x=g(y)$, the $y$-axis ($x=0$), and the horizontal lines $y=c$ and $y=d$, the volume of the solid generated by revolving this region about the $y$-axis is given by:

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

In this setup:

• The integration is performed with respect to $y$.

• The limits of integration, $c$ and $d$, are $y$-values.

• The radius of a representative disc at any given $y$ is $r=g(y)$. The formula integrates the area of these discs, $A(y)=\pi [g(y){]}^2$, over the interval $[c,d]$.

Understanding the Representative Disc

The entire disc method is built upon the concept of a "representative disc." This is a single, infinitesimally thin slice of the solid of revolution, taken perpendicular to the axis of revolution.

• Axis of Revolution and Disc Orientation: If you revolve a region around the $x$-axis (a horizontal axis), the representative discs must be oriented vertically. Their thickness is a small change in $x$, which we denote as $dx$. This is why the integral is with respect to $x$.

• If you revolve a region around the $y$-axis (a vertical axis), the representative discs must be oriented horizontally. Their thickness is a small change in $y$, which we denote as $dy$. This is why the integral is with respect to $y$.

• The Radius ($r$): The radius of each disc is the distance from the center of the disc (which lies on the axis of revolution) to the outer edge of the disc (which lies on the boundary curve).

  • When revolving around the $x$-axis, the radius $r$ is the vertical distance from the $x$-axis to the curve. This distance is simply the function's value, $r=f(x)$.

  • When revolving around the $y$-axis, the radius $r$ is the horizontal distance from the $y$-axis to the curve. This distance is given by the function's value, $r=g(y)$.

The volume of one such representative disc is $dV=(\text{Area of circular face})\times (\text{thickness})$.

• For revolution about the $x$-axis: $dV=\pi [f(x){]}^{2}dx$.

• For revolution about the $y$-axis: $dV=\pi [g(y){]}^{2}dy$.

The definite integral simply sums these infinitesimal volumes ($dV$) over the entire interval to find the total volume ($V$).

Core Concepts & Rules

• Summation of Discs: The volume of a solid of revolution is calculated by integrating the area of a representative circular cross-section, $A=\pi r^2$.

• Axis Determines Integration Variable: The axis of revolution dictates the variable of integration. Revolution around the horizontal $x$-axis requires integration with respect to $x$. Revolution around the vertical $y$-axis requires integration with respect to $y$.

• Radius is Key: The radius, $r$, of a representative disc is the distance from the axis of revolution to the outer boundary of the region.

• Formula for $x$-Axis Revolution: To find the volume of a solid formed by revolving a region bounded by $y=f(x)$ and the $x$-axis from $x=a$ to $x=b$ about the $x$-axis, use the formula $V=\pi {\int }_a^b[f(x){]}^{2}dx$.

• Formula for $y$-Axis Revolution: To find the volume of a solid formed by revolving a region bounded by $x=g(y)$ and the $y$-axis from $y=c$ to $y=d$ about the $y$-axis, use the formula $V=\pi {\int }_c^d[g(y){]}^{2}dy$.

• Function and Limits Must Match: When integrating with respect to $x$, the radius function and the limits of integration must be in terms of $x$. When integrating with respect to $y$, the radius function and the limits of integration must be in terms of $y$.

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 graph of $y=2\sqrt{x}$, the $x$-axis, and the line $x=9$ about the $x$-axis.

Step 1: Identify the Axis and Integration Variable

The region is revolved about the $x$-axis. Therefore, the representative discs are vertical, their thickness is $dx$, and we must integrate with respect to $x$.

Step 2: Determine the Limits of Integration

The region is bounded by the line $x=9$ and implicitly starts where $2\sqrt{x}$ intersects the $x$-axis ($y=0$), which is at $x=0$. The limits of integration are from $a=0$ to $b=9$.

Step 3: Define the Radius Function

The radius, $r$, of a representative disc at a given $x$ is the distance from the axis of revolution ($x$-axis) to the outer curve.

$$r=f(x)=2\sqrt{x}$$

Step 4: Set Up the Definite Integral

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

$$V=\pi {\int }_0^9(2\sqrt{x}{)}^{2}dx$$

Step 5: Evaluate the Integral

First, simplify the integrand:

$$(2\sqrt{x}{)}^2=4x$$

Now, set up the simplified integral:

$$V=\pi {\int }_0^{9}4xdx$$

Find the antiderivative:

$$V=4\pi {\int }_0^{9}xdx=4\pi {\left[\frac{1}{2}{x}^2\right]}_0^9$$

Apply the Fundamental Theorem of Calculus:

$$V=4\pi \left(\frac{1}{2}(9{)}^2-\frac{1}{2}(0{)}^2\right)$$

$$V=4\pi \left(\frac{81}{2}-0\right)$$

$$V=\frac{324\pi }{2}=162\pi$$

The volume of the solid is $162\pi$ cubic units.

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 the graph of $y=x^3$, the $y$-axis, and the line $y=8$ about the $y$-axis.

Step 1: Identify the Axis and Integration Variable

The region is revolved about the $y$-axis. Therefore, the representative discs are horizontal, their thickness is $dy$, and we must integrate with respect to $y$.

Step 2: Determine the Limits of Integration

The region is bounded by the line $y=8$ and the $y$-axis. The lower bound is where the region starts on the $y$-axis, which is $y=0$. The limits of integration are from $c=0$ to $d=8$.

Step 3: Define the Radius Function

The radius, $r$, of a representative disc at a given $y$ is the horizontal distance from the axis of revolution ($y$-axis) to the outer curve. The function must be expressed as $x$ in terms of $y$.

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

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

So, the radius function is:

$$r=g(y)=y^{1/3}$$

Step 4: Set Up the Definite Integral

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

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

Step 5: Evaluate the Integral

First, simplify the integrand:

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

Now, set up the simplified integral:

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

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:

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

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

$$V=\pi \left(\frac{3}{5}(2{)}^5\right)=\pi \left(\frac{3}{5}\cdot 32\right)$$

$$V=\frac{96\pi }{5}$$

The volume of the solid is $\frac{96\pi }{5}$ cubic units.

Using Your Calculator

A graphing calculator can be used to evaluate the definite integral once it has been correctly set up. It is a tool for computation, not for conceptual understanding. You must be able to determine the correct formula, radius, and limits of integration by hand.

To evaluate the integral from Example 1:$V=\pi {\int }_0^9(2\sqrt{x}{)}^{2}dx$

1. On the home screen, access the numerical integration function. For TI-84 style calculators, press MATH and select 9: fnInt(.

2. Enter the lower limit, upper limit, integrand, and variable of integration.

   • $fnInt((2\surd (X){)}^2,X,0,9)$

3. Press ENTER. The calculator will return $162$.

4. This is the value of the integral. The final volume is this value multiplied by $\pi$.

   • Answer: $162\pi$.

To evaluate the integral from Example 2:$V=\pi {\int }_0^8(y^{1/3}{)}^{2}dy$

1. Access the numerical integration function: MATH -> 9: fnInt(.

2. Even though the variable is $y$, use $X$ on the calculator. The procedure is identical.

   • fnInt((X^(1/3))^2, X, 0, 8)

3. Press ENTER. The calculator will return $19.2$.

4. Multiply this result by $\pi$. To get a fractional answer, you can convert the decimal: $19.2\ast \pi$ -> $(96/5)\pi$.

   • Answer: $\frac{96\pi }{5}$.

AP Exam Quick Hit

Common Question Types

• Direct Calculation from a Function (No-Calculator): You will be given a simple function, such as a polynomial or root function, and an interval. You must set up the correct integral for the volume of revolution and evaluate it by hand using the Fundamental Theorem of Calculus.

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

• Integral Setup Only (Calculator or No-Calculator): You will be given a more complex function or region and asked only to write the integral expression for the volume, not to evaluate it. This tests your ability to identify the correct radius, limits, and formula without getting bogged down in computation.

  • Example: "Let R be the region bounded by $g(x)=\cos (x)$ and the $x$-axis from $x=0$ to $x=\pi /2$. Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is revolved about the $x$-axis."

• Revolution Around the $y$-Axis: You will be given a function, often in the form $y=f(x)$, and asked to find the volume of revolution about the $y$-axis. This requires you to solve the function for $x$ in terms of $y$ ($x=g(y)$) and use the appropriate limits and formula.

  • Example: "Find the volume of the solid generated by revolving the region bounded by $y=\ln x$, the $y$-axis, $y=0$, and $y=1$ about the $y$-axis."

Common Mistakes

• Forgetting to Square the Radius: The most frequent error is integrating $\pi r$ instead of $\pi r^2$. The formula is based on the area of a circle, $A=\pi r^2$. Always write $[f(x){]}^2$ or $[g(y){]}^2$ in your setup.

• Omitting the Constant $\pi$: Students correctly set up and evaluate the integral but forget to include the factor of $\pi$ in the final answer. It's best practice to write $\pi$ outside the integral from the very first step.

• Using Incorrect Limits of Integration: When revolving around the $y$-axis, the limits of integration must be $y$-values. Students often mistakenly use the $x$-intercepts of the region's boundary for a $dy$ integral.

• Mixing Up Variables: Using $dx$ for a revolution around the $y$-axis or using a function of $x$ in a $dy$ integral. The variable in the integrand, the differential ($dx$ or $dy$), and the limits of integration must all be consistent.

• Incorrectly Identifying the Radius: When the axis of revolution is the $x$- or $y$-axis, the radius is simply the function value ($f(x)$ or $g(y)$). However, students can become confused, especially when also learning the washer method. For the disc method, the region must be flush against the axis of revolution.