8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

The disc method finds the volume of a solid made by revolving a region around the x- or y-axis. You slice the solid into thin discs perpendicular to the axis, write each disc's volume as $\pi(\text{radius})^2$ times its thickness, then integrate. For AP Calculus, match the slice direction to the axis of rotation before choosing $dx$ or $dy$.

Why This Matters for the AP Calculus Exam

Volumes of revolution show up in Unit 8, Applications of Integration, which carries a noticeable share of the AP Calculus exam (more weight on AB than BC). This topic asks you to calculate volumes of solids of revolution using definite integrals.

On the exam you will need to recognize a region, decide whether to integrate with respect to $x$ or $y$, set up a correct integral with proper notation, and evaluate it. Free-response questions in this unit reward a clearly written integral expression even before you compute, so practice writing the full setup with $\pi$, the squared radius, the bounds, and the correct differential. Getting comfortable here also sets you up for the disc method around other axes (8.10) and the washer method (8.11 and 8.12).

Key Takeaways

• Revolving around the x-axis: $V = \pi\int_a^b (f(x))^2\,dx$, where the radius is $f(x)$ and the thickness is $dx$.
• Revolving around the y-axis: $V = \pi\int_c^d (f(y))^2\,dy$, where the radius is $f(y)$ and the thickness is $dy$.
• The discs are sliced perpendicular to the axis of rotation, so the axis decides whether you integrate in $x$ or $y$.
• These formulas work only when one curve forms the radius and the solid has no hole. A hole means you need the washer method.
• For y-axis rotation, rewrite the curve as $x = f(y)$ before setting up the integral.
• Always sketch the region and shade it before writing the integral so you pick the right bounds.

Volumes of Solids of Revolution

When you find the volume of a solid of revolution, you are measuring how much space a 3D shape occupies. You take a curve, rotate it around an axis, and that rotation sweeps out a solid. The disc method calculates the volume of that solid by adding up many thin slices.

The Disc Method: X-Axis

The disc method slices the solid into infinitely thin discs perpendicular to the axis of rotation. Adding the volumes of these discs with a definite integral gives the total volume.

To find the volume of a solid rotated around the x-axis, you sum the volumes of many thin cross-sections. Each cross-section has a width of $dx$ (approaching 0) and a radius of $f(x)$. Each one is a very flat cylinder, with volume $w \times \pi r^2$, where $w$ is the width and $r$ is the radius. Plugging in the width and radius gives:

$$\pi (f(x))^2 dx$$

To add all of these volumes together, you use an integral:

$$\int_{a}^{b}\pi (f(x))^2dx$$

Here $a$ and $b$ are the boundaries for $f(x)$, given as $x = a$ and $x = b$.

The Disc Method: Y-Axis

Sometimes you rotate a region around the y-axis instead. The process is the same, except you replace $f(x)$ and $dx$ with $f(y)$ and $dy$:

$$\int_{c}^{d}\pi (f(y))^2dy$$

The boundaries $c$ and $d$ are given as $y = c$ and $y = d$.

Solving Using the Disc Method

To solve a rotated-solid problem around the x- or y-axis with just one equation, use these steps:

1. Determine the Axis: Identify whether the solid revolves around the x-axis or y-axis. This decides how you set up the integral.
2. Slice the Solid: Picture slicing the solid into thin discs perpendicular to that axis. Each disc is a tiny volume element.
3. Set Up the Integral: Use a definite integral to sum the volumes of all the discs along the interval, using the formula above.
4. Evaluate the Integral: Once it is set up, evaluate to find the total volume.

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Practicing with the Disc Method

Here is how to apply these steps to practice problems.

Disc Method: Practice Question 1

Calculate the volume of the solid obtained by revolving the region bounded by the curves $y=x^2$, $x=1$, and the y-axis about the x-axis.

Step 1: Determine the Axis

The question revolves the region about the x-axis, so you will integrate with respect to $x$.

Step 2: Slice the Solid

Picture slicing the region into thin discs perpendicular to the x-axis. Graph all of your equations and shade the area to be revolved.

Step 3: Set Up the Integral

Since you are revolving around the x-axis, integrate with respect to $x$ and use the general formula, where $f(x)$ defines the region and $a$ and $b$ are the interval of integration.

$$V= \int_{a}^{b}\pi (f(x))^2dx=\pi\int_{a}^{b}(f(x)^2)\, dx$$

Step 4: Evaluate the Integral

Integrate $y=x^2$ from $x=0$ to $x=1$. The lower bound is where $y=x^2$ meets the y-axis at $x=0$, and the upper bound is at $x=1$.

$$V=\pi\int_{0}^{1}(x^2)^2\, dx$$ $$V=\pi\int_{0}^{1}x^4\, dx$$ $$V=\pi[\frac15x^5]\vert_0^1$$ $$V=\pi(\frac15(1)^5-\frac15(0)^5)$$ $$V=\pi(\frac15)=\boxed{\frac{\pi}{5}}$$

Now try another one.

Disc Method: Practice Question 2

Find the volume of the solid found by rotating the region bounded by $y = 1$, $y=8$, and $y = x^3$ around the y-axis.

Step 1: Determine the Axis

You are rotating around the y-axis, so your integral should be in terms of $f(y)$ instead of $f(x)$.

Step 2: Slice the Solid

Draw out the region. You are rotating the area between the curve and the y-axis, bounded above and below by the horizontal lines $y = 8$ and $y = 1$.

Step 3: Set Up the Integral

Be careful here. For rotating around the y-axis, your function needs to be in terms of $f(y)$, meaning $x = f(y)$. The given equation is:

$$y = x^3$$

To rewrite this in terms of $y$, take the cube root of both sides:

$$\sqrt[3] {y}=x$$

This is the correct function to use in the general integral formula.

Step 4: Evaluate the Integral

The general format is:

$$V= \int_{a}^{b}\pi (f(y))^2dy=\pi\int_{a}^{b}(f(y)^2)\, dy$$

The bounds are $y = 1$ and $y=8$. Plugging in:

$$V= \int_{1}^{8}\pi (\sqrt[3]{y})^2dy=\pi\int_{1}^{8}(\sqrt[3]{y}^2)\, dy=\pi\int_{1}^{8}({y}^{2/3}) dy$$

This simplifies to:

$$V=\pi[\frac35y^{5/3}]\vert_1^8$$ $$V=\pi(\frac35(8)^{5/3}-\frac35(1)^{5/3})$$ $$V=\pi(\frac35(32)-\frac35(1))$$ $$V=\pi(\frac35(31))$$ $$V=\boxed{\frac{93}5 \pi}$$

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How to Use This on the AP Calculus Exam

Free Response

Write the full integral setup before you simplify. A clear expression like $V=\pi\int_{a}^{b}(f(x))^2\,dx$ with correct bounds and differential shows your reasoning and is important for clear exam work. If the region is revolved around the y-axis, show that you rewrote the curve as $x = f(y)$.

Problem Solving

• Sketch and shade the region first. The picture tells you the radius and the bounds.
• Match the differential to the axis: $dx$ for x-axis rotation, $dy$ for y-axis rotation.
• Square the entire radius, not just part of it. For $\sqrt[3]{y}$, squaring gives $y^{2/3}$.
• After integrating, plug in both bounds and subtract carefully.

Common Trap

If the region is revolved around the y-axis but your function is written as $y = f(x)$, you cannot just swap letters. Solve for $x$ first so your radius is a function of $y$.

Common Misconceptions

• Forgetting to square the radius. The disc area is $\pi r^2$, so the integrand must include $(f(x))^2$ or $(f(y))^2$, not just $f(x)$ or $f(y)$.
• Using the wrong differential. Rotating around the x-axis uses $dx$ and bounds in $x$. Rotating around the y-axis uses $dy$ and bounds in $y$. Mixing them gives the wrong setup.
• Using disc when there is a hole. The disc method assumes the solid is solid all the way through. If the region does not touch the axis of rotation, there is a gap and you need the washer method.
• Confusing volume with area. This topic finds volume, not area between curves. The $\pi$ and the squared radius are what turn the slice into a 3D disc.
• Not converting for y-axis rotation. When you rotate around the y-axis, rewrite the curve as $x = f(y)$ before plugging into the formula.

Related AP Calculus Guides

• Unit 8 Overview: Applications of Integration
• 8.1 Finding the Average Value of a Function on an Interval
• 8.7 Volumes with Cross Sections: Squares and Rectangles
• 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals
• 8.4 Finding the Area Between Curves Expressed as Functions of x
• 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts