Disc method around other axes — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This chapter covers deriving and applying the disc method to find volumes of solids of revolution around horizontal axes other than the x-axis and vertical axes other than the y-axis, including correct radius calculation and definite integral setup for bounded planar regions.

You should already know: Basic disc method for revolution around the x-axis and y-axis. How to find intersection points of two functions. Definite integral evaluation via the Fundamental Theorem of Calculus.

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 other axes?

The disc method is an integration technique to find the volume of a solid formed by rotating a planar region around a fixed axis, where each cross-section perpendicular to the axis of rotation is a solid disc. When first learning the disc method, you typically practiced rotating around the coordinate axes ($y=0$ or $x=0$), but on the AP Calculus AB exam, you will frequently be asked to rotate around any horizontal or vertical line that is not a coordinate axis.

This topic is part of Unit 8: Applications of Integration, which accounts for 10-15% of the total AP Calculus AB exam score. Problems involving disc method around other axes appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often as part of a multi-part volume question worth 3-4 points on FRQ. The core idea of the disc method remains the same regardless of the axis: volume is the integral of the area of cross-sectional discs, so $V=\int \pi r^{2}dx$ (if integrating with respect to $x$) or $V=\int \pi r^{2}dy$ (if integrating with respect to $y$). The only key difference from the basic method is how we calculate the radius $r$, which is the distance between the bounding function and the axis of rotation.

2. Revolution Around a Horizontal Axis Other Than $y=0$

When rotating around a horizontal line of the form $y=k$ where $k\ne 0$, we always integrate with respect to $x$. This is because the axis of rotation is horizontal, so cross-sections perpendicular to the axis are vertical slices, which run parallel to the y-axis and are integrated along the x-axis.

The radius is the vertical distance between the curve that bounds the region and the axis of rotation $y=k$. Since radius is a distance, it must always be positive, so we can write $r=|f(x)-k|$. When we square the radius for the area formula, the absolute value can be dropped because squaring removes any negative sign, so $r^2=(f(x)-k{)}^2$ regardless of whether the curve is above or below the axis.

For a region bounded between $x=a$ and $x=b$ rotated around $y=k$, the volume formula is: $$V=\pi {\int }_a^b{\left[f(x)-k\right]}^{2}dx$$ Intuition: Each vertical slice at position $x$ becomes a disc with thickness $dx$ and radius equal to the distance from the slice to the axis, so area is $\pi r^2$, and integrating over all slices gives total volume.

Worked Example

Problem: Find the volume of the solid formed by rotating the region bounded by $y=x+2$, $x=0$, $x=2$, and $y=-1$, around the line $y=-1$.

Solution:

1. Confirm the axis of rotation is the horizontal line $y=k=-1$, so we integrate with respect to $x$, with given bounds $x=0$ to $x=2$.
2. Calculate the radius: The region is bounded above by $y=x+2$ and below by the axis $y=-1$, so distance gives $r=(x+2)-(-1)=x+3$.
3. Write the volume integral: $V=\pi {\int }_0^2(x+3{)}^{2}dx$.
4. Expand and evaluate the integral: $(x+3{)}^2=x^2+6x+9$, so ${\int }_0^2{x}^2+6x+9dx={\left[\frac{x^3}{3}+3x^2+9x\right]}_0^2=\frac{8}{3}+12+18=\frac{98}{3}$.
5. Multiply by $\pi$ to get the final volume: $V=\frac{98}{3}\pi$.

Exam tip: Always check that your radius is positive before squaring. Even though squaring will accidentally give the correct result if you swap terms, AP exam graders will deduct points for an incorrect setup on FRQ, so confirm the order of subtraction first.

3. Revolution Around a Vertical Axis Other Than $x=0$

When rotating around a vertical line of the form $x=c$ where $c\ne 0$, the axis of rotation is vertical, so cross-sections perpendicular to the axis are horizontal slices, meaning we integrate with respect to $y$.

The radius here is the horizontal distance between the curve $x=g(y)$ (expressed as a function of $y$) and the axis $x=c$. Again, radius is a positive distance, so $r=|g(y)-c|$, and $r^2=(g(y)-c{)}^2$ regardless of which side of the axis the curve sits on. If your original function is given as $y=f(x)$, you simply rearrange it to solve for $x$ as a function of $y$ to get $g(y)$.

For a region bounded between $y=d$ and $y=e$ rotated around $x=c$, the volume formula is: $$V=\pi {\int }_d^e{\left[g(y)-c\right]}^{2}dy$$ This follows the exact same logic as rotation around the y-axis ($x=0$), just adjusted for the new position of the axis.

Worked Example

Problem: Find the volume of the solid formed by rotating the region bounded by $y=x^2$, $y=0$, $y=4$, in the first quadrant, around the line $x=3$.

Solution:

1. Axis of rotation is the vertical line $x=c=3$, so we integrate with respect to $y$, with given bounds $y=0$ to $y=4$.
2. Rewrite the function to get $x$ as a function of $y$: $y=x^2⟹x=\sqrt{y}$ for the first quadrant.
3. Calculate the radius: The curve $x=\sqrt{y}$ is always left of the axis $x=3$ for $0\le y\le 4$, so distance gives $r=3-\sqrt{y}$.
4. Set up and expand the integral: $V=\pi {\int }_0^4(3-\sqrt{y}{)}^{2}dy=\pi {\int }_0^4(9-6y^{1/2}+y)dy$.
5. Evaluate the integral: ${\int }_0^{4}9-6y^{1/2}+ydy={\left[9y-4y^{3/2}+\frac{y^2}{2}\right]}_0^4=36-32+8=12$.
6. Multiply by $\pi$: $V=12\pi$.

Exam tip: If rotating around a vertical axis, you must integrate with respect to $y$. A common mistake is trying to keep the integral in terms of $x$, which leads to an incorrect setup. Remember: you slice perpendicular to the axis of rotation, so perpendicular to vertical is horizontal, which means $dy$.

4. Disc Method for Regions Bounded by Two Intersecting Curves

Many AP exam problems ask you to rotate a region bounded by two intersecting curves (not a curve and a given coordinate bound) around a non-coordinate axis. When one side of the region is the axis of rotation, the disc method still applies (there is no hole in the middle, so inner radius is zero, unlike the washer method).

The key extra step here is finding the bounds of integration by solving for the points of intersection of the two curves. Once you have bounds, you identify which curve forms the outer edge of the region, then calculate the radius as the distance between this outer curve and the axis of rotation, just as you did for bounded regions with given bounds.

Worked Example

Problem: Find the volume of the solid formed by rotating the region bounded by $y=x^2$ and $y=2x$ around the line $y=4$.

Solution:

1. Find intersection points of the two curves to get x bounds: set $x^2=2x⟹x(x-2)=0$, so intersections at $x=0$ and $x=2$, which are our bounds.
2. Confirm the position of the region: for $0<x<2$, $2x>x^2$, so the region between the curves is bounded above by $y=2x$ and below by $y=x^2$, and the entire region lies below the axis $y=4$.
3. Calculate the radius: distance between the top edge of the region ($y=2x$) and the axis $y=4$ is $r=4-2x$.
4. Set up and evaluate the integral: $V=\pi {\int }_0^2(4-2x{)}^{2}dx=\pi {\int }_0^2(16-16x+4x^2)dx=\pi {\left[16x-8x^2+\frac{4x^3}{3}\right]}_0^2=\pi \left(32-32+\frac{32}{3}\right)=\frac{32}{3}\pi$.

Exam tip: Always sketch a quick 10-second graph of the region and axis before calculating the radius. A sketch will prevent you from accidentally using the wrong curve to calculate the radius, a very common mistake on exam questions.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When rotating around $y=k$, uses $r=f(x)$ instead of $r=f(x)-k$. Why: Students memorize the basic disc formula for rotation around $y=0$, so they forget to adjust the radius for a non-zero axis. Correct move: Always subtract the axis value from the function value to get distance; this works even for $y=0$, since $f(x)-0=f(x)$.
• Wrong move: When rotating around a vertical non-coordinate axis, integrates with respect to $x$. Why: Students get used to always integrating with respect to $x$, and forget that slicing is always perpendicular to the axis of rotation. Correct move: For any horizontal axis, integrate with respect to $x$; for any vertical axis, integrate with respect to $y$.
• Wrong move: When calculating radius, uses the distance between the two bounding curves instead of the distance between the outer curve and the axis of rotation. Why: Students confuse disc method for two-curve regions with washer method, and misidentify what the radius measures. Correct move: For disc method, radius is always the distance from the edge of the disc (the outer bounding curve) to the center of the disc (the axis of rotation).
• Wrong move: Forgets to rewrite $y=f(x)$ as $x=g(y)$ when rotating around a vertical axis, leaving the integral in terms of $x$. Why: Students avoid inverse function rearrangement and incorrectly keep the original variable. Correct move: Always solve for $x$ as a function of $y$ before setting up the integral for rotation around a vertical axis.
• Wrong move: Leaves radius as a negative value after subtraction, leading to confusion about the sign of the final volume. Why: Students don't think of radius as a positive distance, and just subtract in the order they read the problem. Correct move: Check that $r>0$ before squaring; if it's negative, swap the order of subtraction.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

What is the volume of the solid formed when the region bounded by $y=\ln x$, $y=0$, $y=1$, in the first quadrant, is rotated around the line $x=-2$? A) $\pi {\int }_0^1(\ln x+2{)}^{2}dx$ B) $\pi {\int }_0^1(e^y+2{)}^{2}dy$ C) $\pi {\int }_0^e(e^y+2{)}^{2}dy$ D) $\pi {\int }_0^1(\ln y+2{)}^{2}dy$

Worked Solution: First, the axis of rotation $x=-2$ is vertical, so we must integrate with respect to $y$, eliminating option A. The region is bounded between $y=0$ and $y=1$, so the integral limits are 0 to 1, eliminating option C. Next, rewrite $y=\ln x$ as $x=e^y$, so the function in terms of $y$ is $x=e^y$, eliminating option D, which incorrectly writes the function as $\ln y$. The radius is the distance between $x=e^y$ and $x=-2$, so $r=e^y-(-2)=e^y+2$, giving the volume integral shown in option B. The correct answer is B.

Question 2 (Free Response)

Let R be the region in the first quadrant bounded by $y=4-x^2$, $y=0$, and $x=0$. (a) Set up, but do not evaluate, a definite integral for the volume of the solid formed by rotating R around the line $y=5$. (b) Set up, but do not evaluate, a definite integral for the volume of the solid formed by rotating R around the line $x=2$. (c) The integrand from part (a), after expanding the squared radius, can be written as $A{x}^4+B{x}^2+C$. Find the values of $A$, $B$, and $C$.

Worked Solution: (a) First, find the x-bound by solving for the x-intercept: $0=4-x^2⟹x=2$ (first quadrant). Axis is horizontal $y=5$, so integrate with respect to $x$ from 0 to 2. Radius is $r=5-(4-x^2)=1+x^2$. The integral is: $$V=\pi {\int }_0^2(1+x^2{)}^{2}dx$$

(b) Axis is vertical $x=2$, so integrate with respect to $y$. Rewrite $y=4-x^2$ as $x=\sqrt{4-y}$, bounds from $y=0$ to $y=4$. Radius is $r=2-\sqrt{4-y}$. The integral is: $$V=\pi {\int }_0^4(2-\sqrt{4-y}{)}^{2}dy$$

(c) Expand $(1+x^2{)}^2=1+2x^2+x^4=x^4+2x^2+1$. Matching to $A{x}^4+B{x}^2+C$, we get $A=1$, $B=2$, $C=1$.

Question 3 (Application / Real-World Style)

A civil engineer is designing a curved support pillar for a pedestrian bridge. The cross-section of the pillar (in the x-y plane) is the region bounded by $x=0.3+0.05y$, $x=0.1$, $y=0$, and $y=10$, where all lengths are measured in meters. The pillar is formed by rotating this region around the vertical line $x=0.1$, which is the offset central axis of the casting formwork. Calculate the total volume of concrete required for the pillar, in cubic meters.

Worked Solution: Axis of rotation is vertical $x=0.1$, so integrate with respect to $y$ from 0 to 10. Radius is the distance between the right boundary $x=0.3+0.05y$ and the axis: $r=(0.3+0.05y)-0.1=0.2+0.05y$. The volume integral is: $$V=\pi {\int }_0^{10}(0.2+0.05y{)}^{2}dy=\pi {\int }_0^{10}(0.04+0.02y+0.0025y^2)dy$$ Evaluate the integral: ${\left[0.04y+0.01y^2+\frac{0.0025}{3}{y}^3\right]}_0^{10}=0.4+1+\frac{2.5}{3}\approx 2.233$. Multiply by $\pi$ to get $V\approx 7.01$ cubic meters.

Interpretation: The engineer needs approximately 7 cubic meters of concrete to cast the curved support pillar.

7. Quick Reference Cheatsheet

8. What's Next

Mastering disc method around non-coordinate axes is a critical prerequisite for the washer method around other axes, which extends the disc method to solids with a hole in the middle, and for the shell method, another common volume of revolution technique. This topic builds directly on the basic disc method you learned earlier, and feeds into the larger Unit 8 learning objective of finding volumes of solids of revolution, which is a heavily tested topic on the AP Calculus AB exam.

Without correctly calculating the radius for non-coordinate axes, you will not be able to set up the correct integrals for more complex volume problems, which often make up a full FRQ question on the exam. After this topic, you will move on to problems involving washer method and volumes with general cross-sections, which rely entirely on the radius calculation and integral setup skills you practiced here.

• Washer method around other axes
• Volumes of solids with known cross-sections
• Shell method for volumes of revolution
• Fundamental Theorem of Calculus