8.10 Volume with Disc Method: Revolving Around Other Axes

8.10 Volume with Disc Method: Revolving Around Other Axes

A region does not have to spin around a coordinate axis. A bowl, a bead or a machine part may be modeled by revolving a region around a line such as $y=1$ or $x=-1$. The disc method from Topic 8.9 still applies, as long as each slice is perpendicular to the line and its radius is measured from that line.

The central question is: What is the area of a cross section perpendicular to the axis of rotation? If the cross section is a filled circle, its area is

$$ A=\pi r^2, $$

where $r$ is the distance from the axis of rotation to the boundary of the region. Adding these cross-sectional areas across an interval gives

$$ V=\int A,d(\text{position}). $$

Topic 8.10: revolving around other axes

Topic 8.10 extends the disc-method idea to horizontal or vertical axes that are not the coordinate axes. The radius is always a distance to the axis, not automatically the value of the function.

For a horizontal axis $y=k$,

$$ r=\lvert f(x)-k\rvert. $$

For a vertical axis $x=h$, use horizontal slices and measure

$$ r=\lvert g(y)-h\rvert. $$

The suggested skill 2.D — Connecting Representations is visible here: the graph identifies the radius, the algebra expresses that distance, and the integral accumulates the resulting areas.

Worked example: rotation around $y=1$

Rotate the region between $y=1$ and $y=1+\sqrt{x}$, for $0\le x\le4$, around the line $y=1$. Each vertical slice touches the axis of rotation, so it produces a disc with radius

$$ r=(1+\sqrt{x})-1=\sqrt{x}. $$

Hence,

$$ \begin{aligned} V&=\int_0^4 \pi(\sqrt{x})^2,dx\ &=\int_0^4 \pi x,dx\ &=\pi\left[\frac{x^2}{2}\right]_0^4\ &=\boxed{8\pi\text{ cubic units}}. \end{aligned} $$

A vertical axis can be handled similarly by rewriting the region in terms of $y$. For example, rotate the region

$$ 0\le y\le2,\qquad -1\le x\le y-1 $$

around $x=-1$. A horizontal slice has radius

$$ r=(y-1)-(-1)=y, $$

so

$$ V=\int_0^2 \pi y^2,dy=\frac{8\pi}{3}. $$

Misconception check: “The limits always come from $x$”

Misconception: Every volume integral should use $dx$.
Correction: The variable is determined by the direction of the slices. Rotation around a horizontal axis uses vertical slices and usually $dx$; rotation around a vertical axis uses horizontal slices and usually $dy$.

Before integrating, verify three choices:

1. Are the slices perpendicular to the axis?
2. Is the radius a distance to that axis?
3. Do the limits describe the full range of the slicing variable?

The justification skill 3.D — Justification requires more than writing an integral: the radius, limits, and cross-sectional area must match the geometry.

Retrieval check: A region under $y=f(x)$ from $x=a$ to $x=b$ is rotated around $y=k$. If every vertical slice reaches the axis, what integral represents the volume?

The answer is

$$ \boxed{V=\int_a^b \pi\bigl(f(x)-k\bigr)^2,dx} $$

when $f(x)\ge k$ on the interval. The function supplies the radius only after its distance from the axis has been identified.