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

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

This topic expands the disc method to calculate the volume of a solid of revolution when the axis of revolution is a horizontal or vertical line other than the x-axis or y-axis. The fundamental principle remains the same: we are summing the volumes of an infinite number of infinitesimally thin circular discs. The key challenge is correctly defining the radius of each disc.

The radius, $R$, is no longer simply the function value itself. Instead, it is defined as the distance from the new axis of revolution to the outer boundary of the region being revolved. This requires careful consideration of vertical distances for horizontal axes of revolution ($y=k$) and horizontal distances for vertical axes of revolution ($x=h$). The core task is to express this distance as a function, square it to find the area of a disc, and then integrate that area to find the total volume.

Key Formulas

The volume of a solid of revolution is found by integrating the area of a representative disc, $A=\pi R^2$, across the appropriate interval. The specific formula depends on the orientation of the axis of revolution.

1. Revolution Around a Horizontal Line $y=k$:

   For a region bounded by $y=f(x)$, the x-axis, and the lines $x=a$ and $x=b$, revolved around the horizontal line $y=k$, the volume is:

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

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

2. Revolution Around a Vertical Line $x=h$:

   For a region bounded by $x=g(y)$, the y-axis, and the lines $y=c$ and $y=d$, revolved around the vertical line $x=h$, the volume is:

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

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

Understanding the Radius

The most critical concept in this topic is correctly determining the radius, $R$. The Essential Knowledge states that the radius $R$ is the distance from the axis of revolution to the boundary of the region.

To calculate this distance:

• For a horizontal axis of revolution $y=k$: The radius $R(x)$ is a vertical distance. You find it by taking the y-coordinate of the "top" curve and subtracting the y-coordinate of the "bottom" curve. The axis of revolution $y=k$ acts as one of these "curves."

  • If the region is above the axis of revolution, the radius is $R(x)=(\text{function})-(\text{axis})=f(x)-k$.

  • If the region is below the axis of revolution, the radius is $R(x)=(\text{axis})-(\text{function})=k-f(x)$.

  • In general, $R(x)=|f(x)-k|$. Since the radius is squared in the volume formula, the order of subtraction does not affect the final answer, but establishing the correct distance is conceptually important.

• For a vertical axis of revolution $x=h$: The radius $R(y)$ is a horizontal distance. You find it by taking the x-coordinate of the "right" curve and subtracting the x-coordinate of the "left" curve. The axis of revolution $x=h$ acts as one of these "curves."

  • If the region is to the right of the axis of revolution, the radius is $R(y)=(\text{function})-(\text{axis})=g(y)-h$.

  • If the region is to the left of the axis of revolution, the radius is $R(y)=(\text{axis})-(\text{function})=h-g(y)$.

  • In general, $R(y)=|g(y)-h|$.

Core Concepts & Rules

• The volume of a solid of revolution is calculated by integrating the area of a representative cross-section, which for the disc method is a circle with area $A=\pi R^2$.

• When revolving around a horizontal line ($y=k$), the representative discs are vertical (perpendicular to the x-axis), so the integral must be with respect to $x$. The volume formula is $V=\pi {\int }_a^b(R(x){)}^{2}dx$.

• When revolving around a vertical line ($x=h$), the representative discs are horizontal (perpendicular to the y-axis), so the integral must be with respect to $y$. The volume formula is $V=\pi {\int }_c^d(R(y){)}^{2}dy$.

• The radius $R$ is always the distance from the axis of revolution to the boundary of the region. It is crucial to set up this distance correctly before integrating.

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

Problem: Let R be the region bounded by the graph of $y=x^2$, the x-axis, and the line $x=2$. 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

First, visualize the region. It's the area under the parabola $y=x^2$ from $x=0$ to $x=2$. The axis of revolution is the horizontal line $y=-1$, which is below the region.

Step 2: Determine the Variable of Integration

The axis of revolution is horizontal ($y=-1$), so we will use vertical discs and integrate with respect to $x$. The bounds of integration are given as $x=0$ to $x=2$.

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

The radius is the distance from the axis of revolution ($y=-1$) to the boundary curve ($y=x^2$).

Since this is a vertical distance, we use "top y-value minus bottom y-value".

Top: $y=x^2$

Bottom: $y=-1$

So, the radius is $R(x)=x^2-(-1)=x^2+1$.

Step 4: Set Up the Integral

Use the volume formula for revolution around a horizontal axis:

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

Substitute the bounds and the radius function:

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

Step 5: Evaluate the Integral

First, expand the integrand:

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

Now, integrate:

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

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

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

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

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

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

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

Problem: Let R be the region in the first quadrant bounded by the graph of $x=\sqrt{y}$, the y-axis, and the line $y=4$. Find the volume of the solid generated when R is revolved about the line $x=3$.

Step 1: Sketch the Region and Axis of Revolution

The region is bounded on the left by the y-axis ($x=0$), on the right by the curve $x=\sqrt{y}$, and on top by $y=4$. The axis of revolution is the vertical line $x=3$, which is to the right of the region.

Step 2: Determine the Variable of Integration

The axis of revolution is vertical ($x=3$), so we will use horizontal discs and integrate with respect to $y$. The bounds of integration are from $y=0$ to $y=4$.

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

The radius is the distance from the axis of revolution ($x=3$) to the boundary curve ($x=\sqrt{y}$).

Since this is a horizontal distance, we use "right x-value minus left x-value".

Right: $x=3$

Left: $x=\sqrt{y}$

So, the radius is $R(y)=3-\sqrt{y}$.

Step 4: Set Up the Integral

Use the volume formula for revolution around a vertical axis:

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

Substitute the bounds and the radius function:

$$V=\pi {\int }_0^4(3-\sqrt{y}{)}^{2}dy$$

Step 5: Evaluate the Integral

First, expand the integrand:

$$(3-\sqrt{y}{)}^2=(3-y^{1/2})(3-y^{1/2})=9-6y^{1/2}+y$$

Now, integrate:

$$V=\pi {\int }_0^4(9-6y^{1/2}+y)dy$$

$$V=\pi {\left[9y-6\cdot \frac{y^{3/2}}{3/2}+\frac{1}{2}{y}^2\right]}_0^4$$

$$V=\pi {\left[9y-4y^{3/2}+\frac{1}{2}{y}^2\right]}_0^4$$

$$V=\pi \left(\left(9(4)-4(4{)}^{3/2}+\frac{1}{2}(4{)}^2\right)-(0)\right)$$

$$V=\pi \left(36-4(8)+\frac{1}{2}(16)\right)$$

$$V=\pi (36-32+8)=12\pi$$

The volume is $12\pi$.

Using Your Calculator

The primary skill for this topic is setting up the correct definite integral. A calculator is then used to evaluate the integral, especially when the resulting antiderivative is complex or when a numerical answer is requested on a calculator-active portion of the exam.

**To find the volume from Example 2, V = \pi \int_{0}^{4} (3 - \sqrt{y})^2 \,dy`, using a TI-84 style calculator:** 1. **Access the numerical integration function.** Press `MATH` and select `9: fnInt(`. 2. **Enter the arguments.** The calculator expects the format $fnInt(expression, variable, lower, upper).

*   **Expression**: $(3-\surd (X){)}^2$ (Note: Use $X$ as the variable regardless of whether the original problem used $x$, $y$, or another variable).

*   **Variable**: $X$

*   **Lower bound**: $0$

*   **Upper bound**: $4$

3. Your screen should look like this (in MathPrint mode):

   $${\int }_0^4(3-\sqrt{X}{)}^{2}dX$$

4. Press ENTER. The calculator will return a value, likely $12$.

5. Multiply by $\pi$. Remember that the $\pi$ is outside the integral. The final answer is $12\pi$. You can get a decimal approximation by multiplying the result by $\pi$.

AP Exam Quick Hit

Common Question Types

• Set Up, Do Not Evaluate: You are given functions defining a region and an axis of revolution and asked to write an integral expression for the volume of the resulting solid. Example: "Write, but do not evaluate, an integral expression that gives the volume of the solid generated when the region bounded by $y=e^x$, $y=1$, and $x=2$ is rotated about the line $y=4$."

• Calculator-Active Volume Calculation: You are given a region, often with intersection points that must be found with a calculator, and asked to find the volume. Example: "Find the volume of the solid generated when the region enclosed by $f(x)=\cos (x)$ and $g(x)=x^2-1$ is revolved about the line $y=-2$."

Common Mistakes

• Incorrect Radius Definition: The most frequent error. Students often forget to subtract the axis of revolution, using $R(x)=f(x)$ instead of $R(x)=f(x)-k$ or $R(x)=k-f(x)$. Always define the radius as the distance from the axis to the curve.

• Forgetting to Square the Radius: Writing $V=\pi \int R(x)dx$ instead of the correct $V=\pi \int (R(x){)}^{2}dx$. Remember that the formula comes from the area of a circle, $\pi r^2$.

• Mixing Up Variables: Using $dx$ for a revolution around a vertical axis or $dy$ for a revolution around a horizontal axis. Remember: horizontal axis ($y=k$) means $dx$; vertical axis ($x=h$) means $dy$.

• Incorrectly Squaring Binomials: A common algebraic mistake is to expand $(f(x)-k{)}^2$ as $(f(x){)}^2-k^2$. The correct expansion is $(f(x){)}^2-2k\cdot f(x)+k^2$.

• Ignoring $\pi$: Forgetting to include the factor of $\pi$ in the final answer for the volume.