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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Derivation and application of the disc method for calculating volumes of revolution around the x-axis and y-axis, including formula derivation, handling explicit functions of x and y, and identifying bounds for enclosed regions of revolution.

You should already know: How to evaluate definite integrals analytically and numerically. How to find intersection points of two curves. How to rewrite functions as x in terms of y for problems involving the y-axis.

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 BC 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?

The disc method (alternately spelled disk method, both are acceptable on the AP exam) is an integration technique to calculate the volume of a 3D solid formed by rotating a 2D region around a coordinate axis (the x-axis or y-axis). The method gets its name from the core approach: we slice the solid perpendicular to the axis of rotation, so each slice is a thin circular disc. We then sum the volumes of all these thin discs using integration to get the total volume of the solid. Unlike the washer method, which handles solids with a central hole, the disc method applies to solid solids of revolution with no internal cavity, where the entire cross-section perpendicular to the axis of rotation is solid. Per the AP Calculus Course and Exam Description (CED), this topic is part of Unit 8: Applications of Integration, which accounts for 10–15% of the total AP exam score. Disc method questions appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a core part of multi-part FRQ questions.

2. Disc Method for Rotation Around the x-Axis

When rotating a region around the x-axis (a horizontal axis), we slice perpendicular to the x-axis, so each slice has a thickness of $dx$. The radius of any disc at position $x$ is the vertical distance from the x-axis (the axis of rotation, $y=0$) up to the boundary curve $y=f(x)$. For a region bounded by $y=f(x)$, the x-axis, $x=a$, and $x=b$, the area of a single disc at position $x$ is $\pi r^2=\pi [f(x){]}^2$, and the volume of the thin disc is area times thickness, or $\pi [f(x){]}^{2}dx$. Integrating over all discs from $x=a$ to $x=b$ gives the total volume formula: $$V=\pi {\int }_a^b{\left[f(x)\right]}^{2}dx$$ This is the simplest case of the disc method, and it is the foundation for all other volume of revolution problems. No reworking of the original function is required here, since we integrate directly with respect to $x$, which matches the input variable of the function.

Worked Example

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

1. Identify bounds of integration: The region starts at $x=0$, where $y=\sqrt{x}$ meets the x-axis, and ends at the given boundary $x=4$, so $a=0$, $b=4$.
2. Identify the radius: At any $x$, the distance from the x-axis to the curve is $r(x)=f(x)=\sqrt{x}$.
3. Substitute into the disc method formula: $V=\pi {\int }_0^4{\left(\sqrt{x}\right)}^{2}dx=\pi {\int }_0^{4}xdx$.
4. Evaluate the definite integral: $\pi {\left[\frac{1}{2}{x}^2\right]}_0^4=\pi \left(\frac{1}{2}(16)-0\right)=8\pi$.

The total volume of the solid is $8\pi$ cubic units.

Exam tip: On AP FRQ, always write out the full integral setup before evaluating. You will earn most points for the correct setup, even if you make a small arithmetic error in the final calculation.

3. Disc Method for Rotation Around the y-Axis

When rotating a region around the y-axis (a vertical axis), the disc method requires slicing perpendicular to the y-axis, so each slice has thickness $dy$, and we integrate with respect to $y$. This means we must rewrite the boundary function as $x=g(y)$, a function of $y$, because the radius of the disc at position $y$ is the horizontal distance from the y-axis (the axis of rotation, $x=0$) out to the curve. For a region bounded by $x=g(y)$, the y-axis, $y=c$, and $y=d$, the area of a disc at position $y$ is $\pi [g(y){]}^2$, so the total volume formula is: $$V=\pi {\int }_c^d{\left[g(y)\right]}^{2}dy$$ A common point of confusion here: the shell method for rotation around the y-axis integrates with respect to $x$, but the disc method always integrates with respect to the same variable as the axis of rotation. This means disc method around y-axis always requires rewriting the function in terms of $y$.

Worked Example

Find the volume of the solid formed when the region bounded by $y=x^3$, the y-axis, and $y=8$ is rotated around the y-axis.

1. Rewrite the function as $x$ in terms of $y$: $y=x^3⟹x=y^{1/3}=g(y)$.
2. Identify bounds of integration: The region runs from $y=0$, where the curve meets the y-axis, to the given boundary $y=8$, so $c=0$, $d=8$.
3. Radius at any $y$ is $r(y)=g(y)=y^{1/3}$, the distance from the y-axis to the curve.
4. Substitute into the formula: $V=\pi {\int }_0^8{\left(y^{1/3}\right)}^{2}dy=\pi {\int }_0^8{y}^{2/3}dy$.
5. Evaluate the integral: $\pi {\left[\frac{y^{5/3}}{5/3}\right]}_0^8=\pi \cdot \frac{3}{5}(8^{5/3})=\frac{3\pi }{5}(32)=\frac{96\pi }{5}$.

The total volume of the solid is $\frac{96\pi }{5}$ (or $19.2\pi$) cubic units.

Exam tip: If your function cannot be easily rewritten as $x$ in terms of $y$, the disc method is not the most efficient approach for rotation around the y-axis, and you should use the shell method instead. AP problems will never force you to use disc when it is impractical.

4. Finding Bounds from Intersecting Curves

Most AP exam problems do not give you explicit bounds for integration. Instead, you are given a region bounded by two or more curves, and you must first find their intersection points to get the limits of integration. For the disc method, the region must be bounded on one side by the axis of rotation (x-axis or y-axis) and on the other side by the curve, so the intersection points of the curve with the axis or with other boundary curves give you your bounds. Always confirm the region is between the axis of rotation and the curve: if there is a gap between the region and the axis, you have a hole, and you need the washer method instead of disc.

Worked Example

Find the volume of the solid formed when the region bounded by $y=4-x^2$ and the x-axis is rotated around the x-axis.

1. Find bounds by solving for where the curve intersects the x-axis ($y=0$): $0=4-x^2⟹x=-2$ and $x=2$, so $a=-2$, $b=2$.
2. Confirm the region is between the x-axis (axis of rotation) and the curve: For all $x$ between $-2$ and $2$, $y=4-x^2$ is non-negative, so the entire region sits between $y=0$ (axis) and the curve, so disc method applies.
3. Radius at any $x$ is $r(x)=4-x^2$, so the volume integral is $V=\pi {\int }_{-2}^2(4-x^2{)}^{2}dx$.
4. Simplify and evaluate: $(4-x^2{)}^2=16-8x^2+x^4$, so $V=\pi {\left[16x-\frac{8}{3}{x}^3+\frac{1}{5}{x}^5\right]}_{-2}^2=\frac{512\pi }{15}$.

The total volume of the solid is $\frac{512\pi }{15}$ cubic units.

Exam tip: If the region is symmetric across the y-axis (like this example), you can integrate from $0$ to $2$ and double the result to cut your calculation time in half.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Rotating around the y-axis with the disc method, integrate with respect to x using the original $y=f(x)$ function, leaving the integral as $V=\pi {\int }_a^b[f(x){]}^{2}dx$. Why: Confuses disc method around y-axis with shell method around y-axis, which does integrate with respect to x. Correct move: Always rewrite your function as x in terms of y, set up the integral with respect to y, and confirm all terms are in y before integrating.
• Wrong move: Forgets to square the radius when setting up the integral, writing $V=\pi {\int }_a^{b}f(x)dx$ instead of $V=\pi {\int }_a^b[f(x){]}^{2}dx$. Why: Confuses the area of a circle ($\pi r^2$) with the circumference ($2\pi r$) from the shell method. Correct move: Always add the step "Area of disc = πr²" to your FRQ setup to remind yourself to square the radius.
• Wrong move: Uses the disc method when the region being rotated does not touch the axis of rotation, leading to an undercalculated volume. Why: Forgets that disc method is for solid cross-sections with no hole; a gap between the region and axis creates a hole that needs the washer method. Correct move: Before setting up the integral, check if any point along the axis of rotation between your bounds is inside the region; if there is a gap between the region and the axis, use washer method instead of disc.
• Wrong move: Uses x-bounds for a disc method integral around the y-axis. Why: Mixes up which variable matches which axis of rotation for disc method. Correct move: Follow the rule: for disc method, integrate with respect to the same variable as the axis of rotation: x-axis → integrate dx, y-axis → integrate dy.
• Wrong move: Squares a binomial incorrectly, writing $(x+2{)}^2=x^2+4$ instead of $x^2+4x+4$. Why: Forgets the middle term when expanding squared binomials, a common algebraic error. Correct move: Always expand squared binomials step by step, or use your calculator to evaluate the integral of the unexpanded squared function to avoid this error.
• Wrong move: Squares the formula's π when the radius already contains a π, writing $V={\pi }^2\int f(x{)}^{2}dx$ when $r(x)=\pi f(x)$. Why: Confuses the constant π from the area formula with constants inside the radius. Correct move: Separate the formula's π from any internal constants: if $r(x)=\pi f(x)$, then $V=\pi (\pi f(x){)}^{2}dx={\pi }^3\int f(x{)}^{2}dx$.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

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

Worked Solution: We use the disc method for rotation around the x-axis, since the region is bounded by the axis of rotation. Substitute into the formula: $V=\pi {\int }_0^2(e^{x/2}{)}^{2}dx$. Simplify the integrand: $(e^{x/2}{)}^2=e^x$, so the integral becomes $\pi {\int }_0^2{e}^{x}dx$. Evaluate: $\pi [e^x{]}_0^2=\pi (e^2-e^0)=\pi (e^2-1)$. The distractors correspond to common errors: B comes from failing to square the radius, C comes from incorrectly doubling the exponent, D comes from forgetting to subtract the lower bound. The correct answer is A.

Question 2 (Free Response)

Let $R$ be the region bounded by $y=2x$, $y=0$, and $x=3$. (a) Find the volume of the solid generated when $R$ is rotated around the x-axis. (b) Find the volume of the solid generated when $R$ is rotated around the y-axis using the disc method. (c) Show that your result from (b) can be written as $18\pi$, and explain why the volume is smaller when rotating around the y-axis than the x-axis for this region.

Worked Solution: (a) Bounds for x are $a=0$, $b=3$, radius $r(x)=2x$. $$V_x=\pi {\int }_0^3(2x{)}^{2}dx=4\pi {\int }_0^3{x}^{2}dx=4\pi {\left[\frac{x^3}{3}\right]}_0^3=4\pi (9)=36\pi$$ Volume is $36\pi$ cubic units.

(b) Rewrite $y=2x$ as $x=\frac{y}{2}=g(y)$. Bounds for y are $c=0$, $d=6$ (when $x=3$, $y=6$). $$V_y=\pi {\int }_0^6{\left(\frac{y}{2}\right)}^{2}dy=\frac{\pi }{4}{\int }_0^6{y}^{2}dy=\frac{\pi }{4}{\left[\frac{y^3}{3}\right]}_0^6=\frac{\pi }{12}(216)=18\pi$$ Volume via disc method is $18\pi$ cubic units.

(c) Simplifying the result from (b) gives $\frac{216\pi }{12}=18\pi$, which matches the required form. The volume is smaller when rotating around the y-axis because the average radius of the discs is smaller: the region extends further vertically than it extends horizontally from the y-axis, so the area of each cross-sectional disc is smaller on average, leading to a smaller total volume.

Question 3 (Application / Real-World Style)

A civil engineer is designing a solid concrete support pillar that tapers linearly from the base to the top. The pillar is formed by rotating the region bounded by $y=2\left(1-\frac{x^2}{100}\right)$ (x in meters, y in meters), the x-axis, and $x=10$ around the x-axis, where $x=0$ is the base of the pillar and $x=10$ is the top. Concrete has a uniform density of 2400 kg per cubic meter. Find the total mass of the pillar, rounded to the nearest whole number.

Worked Solution: Use the disc method around the x-axis to find the volume, with bounds $x=0$ to $x=10$, radius $r(x)=2\left(1-\frac{x^2}{100}\right)$: $$V=\pi {\int }_0^{10}{\left(2-\frac{x^2}{50}\right)}^{2}dx=\pi {\int }_0^{10}\left(4-\frac{2x^2}{25}+\frac{x^4}{2500}\right)dx$$ $$V=\pi {\left[4x-\frac{2x^3}{75}+\frac{x^5}{12500}\right]}_0^{10}=\pi \left(40-\frac{80}{3}+8\right)=\frac{64\pi }{3}\approx 67.0206{\text{ m}}^3$$ Total mass is density times volume: $M=2400\ast \frac{64\pi }{3}=51200\pi \approx 160850$ kg. In context, the total mass of the finished concrete pillar is approximately 160,850 kilograms.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the disc method is an essential prerequisite for the washer method, which extends the disc method to solids of revolution with a central hole, and the cylindrical shells method, another common approach to calculating volumes of revolution around the x- and y-axis. Without correctly setting up disc method integrals and understanding the core slicing intuition, you will struggle to distinguish between disc, washer, and shell methods, a commonly tested skill on both AP Calculus MCQ and FRQ sections. This topic is part of the larger unit on applications of integration, which also includes finding arc length, surface area of revolution, and work done by a variable force, all of which rely on the same slicing-and-summing-via-integration framework you learned here.

Next topics to study: Washer method around the x- or y-axis Cylindrical shells method for volumes of revolution Arc length of a planar curve Surface area of revolution