Disc method around the x- or y-axis — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Derivation of the disc method volume formula, calculating volumes of solids of revolution around the x-axis and y-axis, setting up correct definite integrals, and identifying conditions where the disc method is appropriate.

You should already know: How to evaluate definite integrals; how to rearrange functions for a different variable; core properties of Riemann sums.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Disc method around the x- or y-axis?

This is an integral calculus technique used to find the volume of a solid of revolution: a 3D solid formed when a 2D region bounded by curves is rotated around a horizontal or vertical axis (most commonly the x-axis or y-axis). On the AP Calculus AB CED, this topic falls within Unit 8: Applications of Integration, which accounts for 10–15% of the total AP exam score. The disc method appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone volume question or as a component of a multi-part FRQ paired with area between curves or other integration applications. The method gets its name from the thin, circular disc-shaped cross-sections we use to approximate the solid’s total volume, analogous to how thin rectangles approximate area under a curve. It is also sometimes called the disk method; both spellings are accepted on the AP exam. The core intuition is simple: the volume of each thin disc is πr² (the area of the circular face) multiplied by the thickness of the disc. We then integrate (sum) these individual disc volumes across the entire solid to get the total volume.

2. Disc Method Around the X-Axis

When rotating a region bounded above by $y=f(x)$, below by the x-axis, between $x=a$ and $x=b$ around the x-axis, we slice the solid perpendicular to the axis of rotation. The x-axis is horizontal, so perpendicular slices are vertical with thickness $\Delta x$. For any $x$ between $a$ and $b$, the radius of the disc at that position is exactly the perpendicular distance from the x-axis to the curve, which is $f(x)$. The area of the disc face is $\pi r^2=\pi {\left[f(x)\right]}^2$, so the volume of the infinitesimal disc is $dV=\pi {\left[f(x)\right]}^{2}dx$. Integrating from the left bound $x=a$ to the right bound $x=b$ gives the total volume: $$V=\pi {\int }_a^b{\left[f(x)\right]}^{2}dx$$ This formula comes directly from Riemann sums: if we split $[a,b]$ into $n$ slices of width $\Delta x$, total volume is approximated by ${\sum }_{i=1}^n\pi {\left[f(x_i)\right]}^2\Delta x$. As $n\to \infty$, the sum becomes the definite integral above. This method only works when the region is adjacent to the axis of rotation (no gap between the region and axis). A disc is just a special case of a washer with an inner radius of 0.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2\sqrt{x}$, $y=0$, and $x=4$ is rotated around the x-axis.

1. Confirm conditions: The region is bounded by the curve, x-axis, and $x=4$, and is adjacent to the axis of rotation (x-axis). Bounds of integration are $x=0$ (where the curve meets $y=0$) to $x=4$.
2. The radius at any $x$ is $r=f(x)=2\sqrt{x}$.
3. Set up the volume integral using the x-axis disc formula: $$V=\pi {\int }_0^4{\left(2\sqrt{x}\right)}^{2}dx$$
4. Simplify the integrand: ${\left(2\sqrt{x}\right)}^2=4x$, so $V=4\pi {\int }_0^{4}xdx$.
5. Evaluate the definite integral: $V=4\pi {\left[\frac{x^2}{2}\right]}_0^4=4\pi \left(\frac{16}{2}-0\right)=32\pi$.

Exam tip: Always square the entire radius before integrating, not just the variable term. A common mistake is writing $(2\sqrt{x}{)}^2=2x$ instead of $4x$, which loses the square of the constant coefficient.

3. Disc Method Around the Y-Axis

When rotating around the y-axis (a vertical axis), we slice perpendicular to the axis of rotation, so slices are horizontal with thickness $dy$. This means we must express the radius as a function of $y$, not $x$, and integrate with respect to $y$. For a region bounded on the right by $x=g(y)$, on the left by the y-axis, between $y=c$ and $y=d$, the radius of each disc is the horizontal perpendicular distance from the y-axis to the curve, so $r=g(y)$. Following the same logic as the x-axis case, the total volume is: $$V=\pi {\int }_c^d{\left[g(y)\right]}^{2}dy$$ The only key difference from x-axis rotation is the variable of integration: you must rearrange the original function to solve for $x$ in terms of $y$ before setting up the integral. Skipping this step and incorrectly integrating with respect to $x$ is the most common error on this type of problem.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2\sqrt{x}$, $y=0$, and $x=4$ is rotated around the y-axis.

1. Confirm conditions: The region is between the y-axis ($x=0$) and $x=4$, bounded by $y=0$ and $y=4$ (the value of $y$ at $x=4$). It is adjacent to the axis of rotation (y-axis), so disc method applies. Bounds for $y$ are $0$ to $4$.
2. Rearrange $y=2\sqrt{x}$ to solve for $x$ in terms of $y$: $\frac{y}{2}=\sqrt{x}⟹x=\frac{y^2}{4}$, so $g(y)=\frac{y^2}{4}$.
3. Set up the volume integral: $$V=\pi {\int }_0^4{\left(\frac{y^2}{4}\right)}^{2}dy$$
4. Simplify and evaluate: ${\left(\frac{y^2}{4}\right)}^2=\frac{y^4}{16}$, so $V=\frac{\pi }{16}{\int }_0^4{y}^{4}dy=\frac{\pi }{16}{\left[\frac{y^5}{5}\right]}_0^4=\frac{\pi }{16}\left(\frac{1024}{5}\right)=\frac{64\pi }{5}$.

Exam tip: When rotating around the y-axis, double-check that you are integrating with respect to $y$, not $x$. Even if all your arithmetic is correct, integrating $dx$ for a y-axis rotation will always give the wrong answer.

4. Disc Method Around Shifted Horizontal/Vertical Axes

The disc method works for any horizontal or vertical axis of rotation, not just the x-axis ($y=0$) or y-axis ($x=0$). The core principle never changes: the radius $r$ is always the perpendicular distance between the curve and the axis of rotation. For a horizontal axis of rotation $y=k$, we still integrate with respect to $x$ (perpendicular slices are vertical, thickness $dx$). The radius is $r=|f(x)-k|$, so $r^2=(f(x)-k{)}^2$ (the absolute value can be dropped because squaring eliminates the sign). For a vertical axis of rotation $x=h$, we integrate with respect to $y$, and the radius is $r=|g(y)-h|$, so $r^2=(g(y)-h{)}^2$. This is just a generalization of the base cases we learned earlier: when $k=0$ (x-axis), $r=f(x)-0=f(x)$, which matches our original formula, and when $h=0$ (y-axis), $r=g(y)-0=g(y)$, which also matches. Always calculate the distance first, do not just use the function value out of habit.

Worked Example

Find the volume of the solid formed when the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=2$ is rotated around the horizontal axis $y=-1$.

1. Identify the axis: $y=-1$ is horizontal, so we integrate with respect to $x$, with bounds from $x=0$ to $x=2$.
2. Calculate the radius: The curve $y=x^2$ is above the axis $y=-1$, so the perpendicular distance is $r=x^2-(-1)=x^2+1$.
3. Set up the volume integral: $$V=\pi {\int }_0^2(x^2+1{)}^{2}dx$$
4. Expand the integrand: $(x^2+1{)}^2=x^4+2x^2+1$.
5. Evaluate the definite integral: $V=\pi {\left[\frac{x^5}{5}+\frac{2x^3}{3}+x\right]}_0^2=\pi \left(\frac{32}{5}+\frac{16}{3}+2\right)=\frac{206\pi }{15}$.

Exam tip: When the axis is shifted below the x-axis (e.g., $y=-k$) or to the right of the y-axis (e.g., $x=h$), the distance increases, so don’t forget to add the shift when calculating radius.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When rotating around the y-axis, you leave the function in terms of x and integrate with respect to x. Why: Students get used to integrating dx for x-axis rotation and forget to switch variables for y-axis rotation. Correct move: Always match the variable of integration to the axis: horizontal axis (constant y) → integrate dx; vertical axis (constant x) → integrate dy, so rewrite x as a function of y first.
• Wrong move: You forget to square the entire radius, e.g., $(2f(x){)}^2$ becomes $2f(x{)}^2$ instead of $4f(x{)}^2$. Why: Students rush squaring and only square the variable term, forgetting to square constant coefficients. Correct move: Always write the entire radius inside parentheses before squaring, then expand step-by-step to avoid missing constant terms.
• Wrong move: When rotating around a shifted axis, you use the original function value as radius instead of the distance between the curve and the axis. Why: Students memorize $r=f(x)$ for x-axis rotation and don’t adapt the formula for shifted axes. Correct move: For any axis, calculate radius as the absolute value of (curve position - axis position), then square that difference, regardless of where the axis is.
• Wrong move: You use the disc method when there is a gap between the region and the axis of rotation, leading to an undercalculated volume. Why: Students confuse disc method (for regions adjacent to the axis) with the washer method (for regions with a gap). Correct move: Before setting up the integral, check if there is any area between the region and the axis of rotation. If there is, use the washer method instead.
• Wrong move: You use the wrong bounds of integration, e.g., using x bounds as y bounds when rotating around the y-axis. Why: Students copy bounds from the problem statement without checking which variable they correspond to. Correct move: After setting up your integral in terms of the correct variable, convert bounds to that variable to get the correct upper and lower limits.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The region $R$ is bounded by $y=\ln x$, $y=0$, and $x=e$. What is the volume of the solid formed when $R$ is rotated around the x-axis? A) $\pi (e-1)$ B) $\pi (e-2)$ C) $\pi (e+1)$ D) $\pi \left(\frac{e^2}{2}-1\right)$

Worked Solution: The region $R$ spans from $x=1$ (where $\ln x=0$) to $x=e$, and is adjacent to the x-axis, so the disc method applies. The volume formula gives $V=\pi {\int }_1^e(\ln x{)}^{2}dx$. Use integration by parts twice to evaluate the integral: $\int (\ln x{)}^{2}dx=x(\ln x{)}^2-2x\ln x+2x+C$. Evaluating from $1$ to $e$ gives $(e-2e+2e)-(0-0+2)=e-2$. Multiply by $\pi$ to get $V=\pi (e-2)$. The correct answer is B.

Question 2 (Free Response)

Let $R$ be the region bounded by $y=\sqrt{4-x^2}$, the x-axis, and the y-axis. (a) Set up, but do not evaluate, a definite integral for the volume of the solid formed when $R$ is rotated around the x-axis using the disc method. (b) Set up, but do not evaluate, a definite integral for the volume of the solid formed when $R$ is rotated around the y-axis using the disc method. (c) Calculate the volume from part (a) and use geometric reasoning to confirm your result.

Worked Solution: (a) Rotating around the x-axis (horizontal), so we integrate with respect to $x$, with bounds from $x=0$ to $x=2$. Radius at each $x$ is $\sqrt{4-x^2}$, so the integral is: $$V=\pi {\int }_0^2{\left(\sqrt{4-x^2}\right)}^{2}dx=\pi {\int }_0^2(4-x^2)dx$$ (b) Rotating around the y-axis (vertical), so we rewrite $x=\sqrt{4-y^2}$, integrate with respect to $y$ from $y=0$ to $y=2$. The integral is: $$V=\pi {\int }_0^2{\left(\sqrt{4-y^2}\right)}^{2}dy=\pi {\int }_0^2(4-y^2)dy$$ (c) Evaluate the integral from (a): $$V=\pi {\left[4x-\frac{x^3}{3}\right]}_0^2=\pi \left(8-\frac{8}{3}\right)=\frac{16\pi }{3}$$ Geometrically, rotating the first quarter of a circle of radius 2 around the x-axis produces a half-sphere of radius 2. The volume of a full sphere is $\frac{4}{3}\pi r^3=\frac{32\pi }{3}$, so half the volume is $\frac{16\pi }{3}$, which matches our integral result.

Question 3 (Application / Real-World Style)

A civil engineer is designing a tapered parabolic support column for a pedestrian bridge. The column has a height of 10 meters, and its radius at a height $x$ meters above the ground is given by $r(x)=\frac{(10-x{)}^2}{100}$ meters, for $0\le x\le 10$. What is the total volume of concrete needed to cast the column, rounded to two decimal places?

Worked Solution: The column is formed by rotating the radius function around the central x-axis (the center line of the column). The region is adjacent to the axis of rotation, so the disc method applies. Set up the volume integral: $$V=\pi {\int }_0^{10}{\left(\frac{(10-x{)}^2}{100}\right)}^{2}dx=\frac{\pi }{10000}{\int }_0^{10}(10-x{)}^{4}dx$$ Use substitution $u=10-x$, $du=-dx$, to evaluate the integral: $$V=\frac{\pi }{10000}{\int }_0^{10}{u}^{4}du=\frac{\pi }{10000}{\left[\frac{u^5}{5}\right]}_0^{10}=\frac{\pi }{10000}\left(\frac{100000}{5}\right)=2\pi \approx 6.28$$ Interpretation: Approximately 6.28 cubic meters of concrete are needed to cast the tapered support column.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the disc method is a critical prerequisite for the next topics in Unit 8: volumes of revolution, starting with the washer method for regions with gaps between the region and the axis of rotation, and then the cylindrical shells method. Without understanding how radius and cross-sectional area work in the disc method, you will struggle to generalize these ideas to more complex volume problems, which regularly appear as 6–9 point FRQ questions on the AP exam. Beyond volumes of revolution, the core idea of slicing a 3D shape into infinitesimal cross-sections and integrating their areas to get total volume is a foundational concept that extends to volumes of solids with known cross-sections, another common AP exam topic.

• Washer method around the x- or y-axis
• Volumes of solids with known cross-sections
• Cylindrical shells method for volumes of revolution