8.10 Volume with Disc Method: Revolving Around Other Axes

The disc method still works when you revolve a region around a line other than the x- or y-axis, like $y=2$ or $x=3$. The radius becomes the distance from the function to that shifted axis, such as $R=|f(x)-k|$ for a horizontal line $y=k$ or $R=|g(y)-h|$ for a vertical line $x=h$. For AP Calculus, define the radius as a distance before squaring it in the volume integral.

Why This Matters for the AP Calculus Exam

Volumes of solids of revolution are a regular part of AP Calculus, and shifting the axis of rotation is a common twist on the standard disc problem. On free-response questions, you are expected to write a correct integral with proper notation before you compute, so being able to set up an expression like $V = \pi \int_a^b [f(x) - k]^2 \, dx$ matters for clear exam work. These problems test whether you can read a region from a graph, pick the right variable to integrate with respect to, and adjust the radius for an axis that is not at the origin.

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Key Takeaways

• The radius is the perpendicular distance from the curve to the axis of rotation, not just the function value.
• For a horizontal axis $y=k$, integrate with respect to $x$: $V = \pi \int_a^b [f(x) - k]^2 \, dx$.
• For a vertical axis $x=h$, integrate with respect to $y$: $V = \pi \int_c^d [g(y) - h]^2 \, dy$.
• Use absolute value when needed so the radius stays positive, and remember the square keeps it positive anyway.
• Find your limits of integration from given values or from where curves intersect each other or the axis.
• Always draw the region and the axis first so you can see the radius clearly.

Revolving Around Other Axes

"Revolving around other axes" means a region spins around a horizontal line like $y=5$ or a vertical line like $x=3$ instead of the x- or y-axis. AP Calculus sticks to horizontal and vertical lines, so you will not see rotation around slanted lines.

If you want a refresher on the standard disc method first, check the 8.9 guide.

The Core Idea

Volume comes from stacking up many thin circular cross sections (discs). Each disc has area $\pi R^2$, so the volume is the integral of that area over the region:

The only thing that changes when the axis moves is how you find $R$.

Step 1: Pick Your Variable

• Rotate around a horizontal line (like $y=7$ or the x-axis): integrate with respect to $x$.
• Rotate around a vertical line (like $x=4$ or the y-axis): integrate with respect to $y$.

Quick memory hook: horizontal rotation goes with $x$, vertical rotation goes with $y$.

Step 2: Find the Radius

The radius is the distance between the curve and the axis of rotation, not the height of the function by itself. That is the part that changes for a shifted axis.

• Axis $y = k$ (horizontal): $R(x) = |f(x) - k|$
• Axis $x = h$ (vertical): $R(y) = |g(y) - h|$

For example, if you revolve $y = f(x)$ around the line $y = 2$, the radius is $f(x) - 2$. If the curve sits below the axis, subtracting still works as long as you square the result, which keeps the area positive.

Step 3: Set Up the Integral

Combine the radius with $\pi R^2$:

Use the first form for a horizontal axis $y = b$ and the second for a vertical axis $x = a$.

Finding the Limits

The bounds $c$ and $d$ are either given directly in the problem or come from intersection points. To find them algebraically, set $f(x) = g(x)$, or set $f(x) = 0$, or set the function equal to whatever line the problem names. Graphing the region makes these limits easy to see.

How to Use This on the AP Calculus Exam

Free Response

Write the integral with correct notation before evaluating. A clean setup like

shows your reasoning and is important for clear exam work. Make sure the square is outside the parentheses and that you used the distance to the shifted axis, not just $f(x)$.

Problem Solving

1. Sketch the region and the axis of rotation.
2. Decide horizontal (use $x$) or vertical (use $y$).
3. Write the radius as the distance from the curve to the axis.
4. Find the limits from given values or intersections.
5. Integrate and, if needed, evaluate with a calculator.

Common Trap

Subtracting a negative constant flips to addition. If the axis is $y = -3$ and the curve is $f(x)$, then $R = f(x) - (-3) = f(x) + 3$. Forgetting this sign change is one of the most common setup errors.

Practice Problem

Rotate the region bounded by $y=10e^{2x}$ between $x=-2$ and $x=-1$ about the line $y=-3$.

Try it, then check the solution below.

Solution

The general form is $\int_{c}^{d}\pi (f(x)-b)^2 \, dx$ or $\int_{c}^{d}\pi (f(y)-a)^2 \, dy$. Since we revolve around a horizontal line, use the first form with $x$.

The bounds are given directly: $-2$ and $-1$. Here $f(x) = 10e^{2x}$ and the axis is $y = -3$, so $b = -3$. Because we subtract a negative, the radius becomes $10e^{2x} + 3$.

For a refresher on evaluating integrals, see the integration and accumulation guide.

Pull out the constant using $\int a f(x) \, dx = a \int f(x) \, dx$:

Expand the square:

Integrate term by term:

If you want to check, take the derivative of the antiderivative and confirm it matches the integrand.

Plug in $-1$ and $-2$ and subtract:

which simplifies to

This is about $40.7$. Don't forget the factor of $\pi$ at the end.

Common Misconceptions

• Using $f(x)$ as the radius when the axis is shifted. The radius is always the distance from the curve to the axis, so subtract the axis line. For $y = k$, use $f(x) - k$, not just $f(x)$.
• Mixing up the integration variable. Horizontal axes pair with $dx$ and vertical axes pair with $dy$. Choosing the wrong one usually means you cannot even set up the integral correctly.
• Sign mistakes with negative axes. Subtracting a negative number adds. Revolving around $y = -3$ gives a radius of $f(x) + 3$, not $f(x) - 3$.
• Forgetting to square the whole radius. The area of a disc is $\pi R^2$, so the entire expression $(f(x) - k)$ gets squared, not just $f(x)$.
• Confusing the disc method with the washer method. If the axis passes through the region or the region has a gap from the axis, you get a hole and need washers instead of solid discs.
• Dropping the $\pi$. Every disc area carries a factor of $\pi$, so keep it through the whole calculation.

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