Disc method around other axes — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: This chapter covers the general disc method for finding volumes of solids of revolution around horizontal non-x-axes and vertical non-y-axes, deriving the general radius formula, and solving setup and evaluation problems for the AP Calculus exam.

You should already know: The basic disc method for revolution around the x-axis and y-axis, how to calculate distance between two points on the coordinate plane, how to set up and evaluate definite integrals.

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 other axes?

The disc method is an integration technique to calculate the volume of a solid of revolution, formed when a 2-dimensional region is rotated around a fixed straight axis. When the axis of rotation does not coincide with the x-axis ($y=0$) or y-axis ($x=0$) — a common scenario on the AP exam, accounting for nearly half of all volume of revolution problems — we must adjust the basic radius formula to account for the shifted axis, which is the core of this topic. Per the AP Calculus CED, this topic is part of Unit 8: Applications of Integration, and contributes approximately 2-3% of the total exam score, appearing in both multiple-choice (MCQ) and free-response (FRQ) sections. Synonyms sometimes used include "disk method around shifted axes" or "volume of revolution around arbitrary axes," but College Board consistently refers to it as the disc method. Unlike the washer method (used for hollow solids), the disc method applies when the rotated region is adjacent to the axis of rotation, leaving no hollow space in the resulting solid. This topic extends the basic disc method you already know to any horizontal or vertical axis of rotation, the most common form tested on the AP exam.

2. Revolution around a horizontal shifted axis

A horizontal axis of rotation always has the form $y=k$, where $k$ is a non-zero constant. To apply the disc method, we take slices perpendicular to the axis of rotation: for a horizontal axis, this means vertical slices parallel to the y-axis, so we integrate with respect to $x$. Each vertical slice becomes a circular disc when rotated around the axis, with area $\pi r^2$, where $r$ is the radius of the disc. The only adjustment from the basic disc method (where $k=0$, the x-axis) is calculating $r$ correctly: $r$ is the distance between the edge of the region (the function) and the axis of rotation. Since distance is always positive, and squaring any real number gives a non-negative result, we do not need to keep absolute value after squaring. The general formula for volume when rotating the region bounded by $a\le x\le b$ and $k\le y\le f(x)$ around $y=k$ is: $$V=\pi {\int }_a^b{\left[f(x)-k\right]}^{2}dx$$ Intuition: shifting the axis up or down only changes the length of the radius, not the overall structure of the integral. The integration variable and bounds stay the same as the basic case, only the radius changes.

Worked Example

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

1. Identify the axis orientation and integration variable: The axis is horizontal $y=-1$, so we integrate with respect to $x$, the standard variable for this function.
2. Find bounds of integration: The region runs from $x=0$ (where $y=x^2$ meets $y=0$) to $x=2$, given.
3. Calculate the radius: The distance between the function $y=x^2$ and the axis $y=-1$ is $r=x^2-(-1)=x^2+1$, which is positive over $[0,2]$.
4. Set up and evaluate the integral: $$V=\pi {\int }_0^2(x^2+1{)}^{2}dx=\pi {\int }_0^2(x^4+2x^2+1)dx$$ $$=\pi {\left[\frac{x^5}{5}+\frac{2x^3}{3}+x\right]}_0^2=\pi \left(\frac{32}{5}+\frac{16}{3}+2\right)=\frac{206\pi }{15}$$

Exam tip: If you ever need to check your radius setup, confirm that when $k=0$ (rotation around the x-axis), the formula simplifies to the basic disc method you already know. This 10-second check catches most setup errors.

3. Revolution around a vertical shifted axis

A vertical axis of rotation has the form $x=h$, where $h$ is a non-zero constant. For the disc method, we still need slices perpendicular to the axis of rotation, so for a vertical axis we use horizontal slices parallel to the x-axis, meaning we integrate with respect to $y$. This requires rewriting any function given as $y=f(x)$ into the form $x=g(y)$, and finding bounds for $y$ instead of $x$. The radius is still the distance between the edge of the region (the function $x=g(y)$) and the axis $x=h$, so the general formula when rotating the region bounded by $c\le y\le d$ and $h\le x\le g(y)$ around $x=h$ is: $$V=\pi {\int }_c^d{\left[g(y)-h\right]}^{2}dy$$ The core logic is identical to the horizontal shifted axis case, only the integration variable changes to match the perpendicular slice requirement. The most common student error here is defaulting to integrating with respect to $x$ out of habit, so always confirm the integration variable before setting up the integral.

Worked Example

Find the volume of the solid formed when the region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ is rotated around the line $x=5$.

1. Identify axis orientation: The axis is vertical $x=5$, so we integrate with respect to $y$.
2. Rewrite the function and find bounds: $y=\sqrt{x}⟹x=y^2$. Bounds for $y$ are $y=0$ (at $x=0$) to $y=2$ (at $x=4$).
3. Calculate the radius: The entire region is to the left of $x=5$, so the distance between $x=y^2$ and $x=5$ is $r=5-y^2$, which is positive over $[0,2]$.
4. Set up and evaluate the integral: $$V=\pi {\int }_0^2(5-y^2{)}^{2}dy=\pi {\int }_0^2(25-10y^2+y^4)dy$$ $$=\pi {\left[25y-\frac{10y^3}{3}+\frac{y^5}{5}\right]}_0^2=\pi \left(50-\frac{80}{3}+\frac{32}{5}\right)=\frac{446\pi }{15}$$

Exam tip: When rewriting $y=f(x)$ as $x=g(y)$, double-check your inverse function algebra. Small errors here lead to completely wrong integrals, even if your radius setup is correct.

4. Radius adjustment for regions on the opposite side of the axis

A common exam variation is having the entire rotated region located below a horizontal axis or to the left of a vertical axis, rather than the more common above/right position. The core rule for radius does not change: $r$ is always the positive distance between the edge of the region and the axis. If the region is entirely below the horizontal axis $y=k$, then $f(x)<k$ for all $x$ in the interval, so $r=k-f(x)$ (to get a positive radius). Similarly, if the region is entirely to the left of the vertical axis $x=h$, then $g(y)<h$ for all $y$, so $r=h-g(y)$. While squaring $(f(x)-k)$ gives the same result as squaring $(k-f(x))$, using a positive radius makes it easier to catch setup errors when checking your work.

Worked Example

Find the volume of the solid formed when the region bounded by $y=2x+1$, $x=0$, $x=3$, and $y=0$ is rotated around the horizontal line $y=10$.

1. Axis orientation: horizontal $y=10$, so integrate with respect to $x$ from $x=0$ to $x=3$.
2. Confirm region position: The maximum value of $y=2x+1$ on $[0,3]$ is $7<10$, so the entire region is below the axis.
3. Calculate radius: $r=10-(2x+1)=9-2x$, positive over $[0,3]$.
4. Evaluate the integral: $$V=\pi {\int }_0^3(9-2x{)}^{2}dx=\pi {\int }_0^3(81-36x+4x^2)dx$$ $$=\pi {\left[81x-18x^2+\frac{4x^3}{3}\right]}_0^3=\pi (243-162+36)=117\pi$$

Exam tip: Always sketch a rough 10-second graph of the region and the axis to confirm which side of the axis the region sits on. This eliminates 90% of radius sign errors.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Integrating with respect to $x$ when rotating around a vertical axis $x=h$ with the disc method. Why: Students default to $x$ integration because most functions are written as $y=f(x)$, and forget that disc method requires slices perpendicular to the axis of rotation. Correct move: Confirm orientation: horizontal axis ($y=k$) → integrate with respect to $x$; vertical axis ($x=h$) → integrate with respect to $y$, and rewrite the function in terms of $y$ first.
• Wrong move: Using $r=f(x)$ instead of $r=f(x)-k$ when rotating around $y=k\ne 0$. For example, writing $r=x^2$ for rotation around $y=-1$. Why: Students memorize the basic formula for the x-axis and forget to shift the radius when the axis moves. Correct move: Every time, write $r=|y_{\text{function}}-y_{\text{axis}}|$ before squaring, regardless of where the axis is located.
• Wrong move: Using the disc method when there is a gap between the region and the axis of rotation. Why: Students associate any solid of revolution around a non-standard axis with the disc method, regardless of whether the solid is hollow. Correct move: Confirm the region touches the axis along its entire length. If there is a gap, use the washer method instead.
• Wrong move: Panicking when you get a negative radius before squaring, or changing integral bounds to fix the sign. Why: Students expect the function to be larger than the axis constant, so they get confused when the opposite is true. Correct move: Reverse the order of subtraction to get a positive radius, then square and integrate as normal.
• Wrong move: Keeping absolute value around the radius when integrating, leading to incorrect splitting of the integral. Why: Students remember radius is positive, so they keep absolute value out of caution. Correct move: Drop the absolute value after squaring, since $(|r|{)}^2=r^2$ for any real $r$.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The region $R$ is bounded by $y=\ln x$, $y=0$, and $x=e$. What is the correct integral expression for the volume of the solid formed when $R$ is rotated around the line $y=2$? A) $\pi {\int }_1^e(2-\ln x)dx$ B) $\pi {\int }_0^1(2-e^y{)}^{2}dy$ C) $\pi {\int }_1^e(2-\ln x{)}^{2}dx$ D) $\pi {\int }_0^1(e^y-2{)}^{2}dy$

Worked Solution: The axis of rotation $y=2$ is horizontal, so we use vertical slices perpendicular to the axis and integrate with respect to $x$. This eliminates options B and D, which integrate with respect to $y$. The area of each disc is $\pi r^2$, so the radius must be squared, eliminating option A which omits the square. The region runs from $x=1$ (where $\ln x=0$) to $x=e$, and the radius is the positive distance from $y=\ln x$ to $y=2$, which is $2-\ln x$. The correct answer is C.

Question 2 (Free Response)

Let $R$ be the region bounded by $y=x^3$, $y=0$, and $x=2$. (a) Write, but do not evaluate, an integral expression for the volume of the solid formed when $R$ rotated around the horizontal line $y=-2$. (b) Write, but do not evaluate, an integral expression for the volume of the solid formed when $R$ is rotated around the vertical line $x=-3$. (c) Find the exact value of the volume described in part (a).

Worked Solution: (a) The axis is horizontal, so integrate with respect to $x$ from $0$ to $2$. The radius is $x^3-(-2)=x^3+2$, so: $$V=\pi {\int }_0^2(x^3+2{)}^{2}dx$$ (b) The axis is vertical, so rewrite $y=x^3$ as $x=y^{1/3}$, integrate with respect to $y$ from $0$ to $8$. The radius is $y^{1/3}-(-3)=y^{1/3}+3$, so: $$V=\pi {\int }_0^8{\left(y^{1/3}+3\right)}^{2}dy$$ (c) Expand the integrand: $(x^3+2{)}^2=x^6+4x^3+4$. Integrate term-by-term: $${\int }_0^2(x^6+4x^3+4)dx={\left[\frac{x^7}{7}+x^4+4x\right]}_0^2=\frac{128}{7}+16+8=\frac{296}{7}$$ Multiply by $\pi$: $V=\frac{296\pi }{7}$

Question 3 (Application / Real-World Style)

A civil engineer is designing a tapered cylindrical cooling tower, whose interior wall can be modeled by rotating the region bounded by $x=\frac{1}{20}{y}^2+10$, $y=0$, and $y=100$ around the $y$-axis (the central vertical axis of the tower). All measurements are in meters. What is the volume of the interior of the cooling tower, to the nearest cubic meter?

Worked Solution: The axis of rotation is the vertical line $x=0$, so we integrate with respect to $y$ from $0$ to $100$. The radius is the distance from $x=\frac{y^2}{20}+10$ to $x=0$, so $r=\frac{y^2}{20}+10$. The volume formula gives: $$V=\pi {\int }_0^{100}{\left(\frac{y^2}{20}+10\right)}^{2}dy=\pi {\int }_0^{100}\left(\frac{y^4}{400}+y^2+100\right)dy$$ $$=\pi {\left[\frac{y^5}{2000}+\frac{y^3}{3}+100y\right]}_0^{100}=\pi \left(50+\frac{1000000}{3}+10000\right)\approx 343383.33\pi \approx 1078720$$ The interior of the cooling tower has an approximate volume of 1,078,720 cubic meters, enough to hold the required cooling water for a 1 GW power plant.

7. Quick Reference Cheatsheet

8. What's Next

This topic extends the basic disc method to all horizontal and vertical axes of rotation, and is a direct prerequisite for the washer method around non-standard axes, which you will learn next. The washer method uses the same core radius calculation you learned here, subtracting the area of the inner disc from the outer disc to account for hollow solids, so mastering the radius setup for shifted axes is critical to avoiding errors there. This topic also feeds into the larger Unit 8: Applications of Integration, where volume calculations are a major component of both MCQ and FRQ sections, and often appear in multi-part questions that connect to other integration techniques like integration by parts or u-substitution. Without mastering the radius setup for shifted axes, you will not be able to correctly set up most volume problems tested on the AP exam.

• Washer method around other axes
• Cylindrical shells method
• Volumes of solids with known cross sections
• Integration by parts