Revolution About A Shifted Vertical Axis

Introduction

A curve written as $x=g(y)$ describes horizontal position as a function of vertical position. If this curve is revolved around a vertical line $x=c$, each point on the curve travels in a circle centered on that axis.

The circle’s radius is the horizontal distance from the point to the axis, not its vertical coordinate. The surface-area integral must therefore combine this perpendicular distance with the length of the generating curve, and its bounds must describe precisely the portion being revolved.

Key idea: For a smooth profile $x=g(y)$ revolved about $x=c$, the radius is $\lvert g(y)-c\rvert$, and the surface-area element is $2\pi\lvert g(y)-c\rvert\,ds$.

Radius and Arc Length

Fix a value of $y$. The point on the profile is $(g(y),y)$, while the axis is the vertical line $x=c$. The shortest distance from this point to the axis is horizontal, so its length is $\lvert g(y)-c\rvert$. The absolute value is essential: distance is nonnegative whether the profile is to the left or to the right of the axis.

As $y$ changes by a small amount $dy$, the profile moves both horizontally and vertically. Since $dx=g'(y)\,dy$, the corresponding arc-length element is

ds=\sqrt{(dx)^2+(dy)^2} =\sqrt{1+\bigl(g'(y)\bigr)^2}\,dy.

A small piece of the curve sweeps out a narrow band when revolved. Its circumference is approximately $2\pi r$, where $r$ is its distance from the axis, and its width along the surface is approximately $ds$. Thus the band’s area is approximately $2\pi r\,ds$. Adding the areas of increasingly narrow bands gives the surface-area integral.

Surface-Area Formula: If $g$ is continuously differentiable on $[a,b]$, and the selected curve segment sweeps out the surface once, then its area when revolved about $x=c$ is $$ S=2\pi\int_a^b \lvert g(y)-c\rvert\sqrt{1+\bigl(g'(y)\bigr)^2}\,dy. $$ Here $a$ and $b$ are the y-coordinates of the endpoints of the selected generating segment.

Why the Formula Works

The surface can be approximated by bands formed from short pieces of the profile. If a short piece has arc length $\Delta s$, and its endpoint radii are $r_1$ and $r_2$, the band is approximated by a frustum. The lateral area of a frustum is $\pi(r_1+r_2)\Delta s$. When the piece is short, both endpoint radii are close to the radius $r$ along that piece, so this area is close to $2\pi r\,\Delta s$.

Divide the chosen curve segment into smaller and smaller pieces. The sum of the band areas approaches the integral of $2\pi r$ with respect to arc length. Substituting $r=\lvert g(y)-c\rvert$ and $ds=\sqrt{1+(g'(y))^2}\,dy$ gives the stated formula. This is the same band principle used in the earlier material on frustum bands, with the radius now measured to a shifted vertical axis.

Choosing the Bounds

The variable of integration is $y$, so the limits must be the vertical coordinates of the endpoints of the curve segment being revolved. They are not chosen from the value of the axis $c$, and they do not automatically equal the curve’s x-coordinates.

If a problem specifies the segment between $y=a$ and $y=b$, use those limits directly. If instead it specifies points by their x-coordinates, determine the corresponding y-values on the stated profile. When the profile has more than one y-value for a given x-coordinate, use the values that bound the requested branch and segment.

Identify the profile and axis.
Write the profile as $x=g(y)$, and record the axis $x=c$.

Set the segment’s y-bounds.
Use the endpoint y-coordinates of the portion specified in the problem.

Find the perpendicular radius.
Use $r(y)=\lvert g(y)-c\rvert$, not $g(y)$ alone unless the axis is $x=0$ and the profile is on the right.

Find the arc-length factor.
Differentiate $g$ and form $\sqrt{1+(g'(y))^2}$.

Integrate and check.
Substitute the radius and arc-length factor into the formula, evaluate over the selected bounds, and verify the calculation.

Worked Examples

Worked Example: A Linear Profile to the Right of the Axis

Revolve the segment $x=2y+1$, $0\le y\le2$, about the vertical line $x=-1$.

The profile is $g(y)=2y+1$, and the axis coordinate is $c=-1$. The radius is

r(y)=\lvert (2y+1)-(-1)\rvert =\lvert 2y+2\rvert =2y+2 \qquad (0\le y\le2).

The derivative is $g'(y)=2$, so the arc-length factor is

\sqrt{1+\bigl(g'(y)\bigr)^2} =\sqrt{1+2^2} =\sqrt{5}.

Therefore,

\begin{aligned} S &=2\pi\int_0^2(2y+2)\sqrt{5}\,dy\\ &=2\pi\sqrt{5}\int_0^2(2y+2)\,dy\\ &=2\pi\sqrt{5}\,[y^2+2y]_0^2\\ &=2\pi\sqrt{5}\bigl((4+4)-(0+0)\bigr)\\ &=16\pi\sqrt{5}. \end{aligned}

Check: the antiderivative $y^2+2y$ differentiates to $2y+2$, the radius used in the integrand. Also, the radius runs from $2$ at $y=0$ to $6$ at $y=2$, consistent with the profile lying to the right of $x=-1$ throughout the selected segment.

Worked Example: Bounds from a Specified x-Interval

Consider $x=y+1$, revolved about $x=5$. Find the area generated by the part of the profile whose x-coordinates lie between $2$ and $4$.

First convert the specified x-coordinates into y-coordinates using $x=y+1$. At $x=2$, $y=2-1=1$; at $x=4$, $y=4-1=3$. Thus the selected segment has $1\le y\le3$. The profile is left of the axis on this interval, so its radius is

r(y)=\lvert (y+1)-5\rvert =\lvert y-4\rvert =4-y \qquad (1\le y\le3).

Since $g'(y)=1$, the arc-length factor is $\sqrt{1+1^2}=\sqrt{2}$. The area is

\begin{aligned} S &=2\pi\int_1^3(4-y)\sqrt{2}\,dy\\ &=2\pi\sqrt{2}\,[4y-y^2/2]_1^3\\ &=2\pi\sqrt{2}\bigl((12-9/2)-(4-1/2)\bigr)\\ &=2\pi\sqrt{2}(15/2-7/2)\\ &=8\pi\sqrt{2}. \end{aligned}

Check: differentiating $4y-y^2/2$ gives $4-y$, the radius factor in the integral. The selected y-bounds also give $x(1)=2$ and $x(3)=4$, as required.

Worked Example: A Nonlinear Profile

Revolve the segment $x=\sqrt{y}$, $1\le y\le4$, about the y-axis, which is $x=0$.

Here $g(y)=\sqrt{y}$, so the radius is $r(y)=\sqrt{y}$. The derivative and arc-length factor are

g'(y)=\frac{1}{2\sqrt{y}}, \qquad \sqrt{1+\bigl(g'(y)\bigr)^2} =\sqrt{1+\frac{1}{4y}} =\frac{\sqrt{4y+1}}{2\sqrt{y}}.

Multiplying the radius by the arc-length factor simplifies the integrand:

\sqrt{y}\cdot\frac{\sqrt{4y+1}}{2\sqrt{y}} =\frac{\sqrt{4y+1}}{2}.

Consequently,

\begin{aligned} S &=2\pi\int_1^4\frac{\sqrt{4y+1}}{2}\,dy\\ &=\pi\int_1^4\sqrt{4y+1}\,dy\\ &=\frac{\pi}{6}\left[(4y+1)^{3/2}\right]_1^4\\ &=\frac{\pi}{6}\left(17^{3/2}-5^{3/2}\right)\\ &=\frac{\pi}{6}\left(17\sqrt{17}-5\sqrt{5}\right). \end{aligned}

Check: differentiating $\frac{\pi}{6}(4y+1)^{3/2}$ gives $\pi\sqrt{4y+1}$, the integrand after simplification. The endpoint radii are $1$ and $2$, matching $x=\sqrt{y}$ at $y=1$ and $y=4$.

A Common Setup Error

A frequent mistake is to use the profile’s x-coordinate as the radius without measuring from the actual axis. For example, in the first worked example, using $2y+1$ as the radius would treat the axis as $x=0$, even though the given axis is $x=-1$. That incorrect setup would use $2y+1$ rather than $2y+2$, producing the wrong area.

\text{Correct radius: }\lvert (2y+1)-(-1)\rvert=2y+2.

The correction is to subtract the axis coordinate from the profile coordinate, then take the absolute value. A second common error is to use x-coordinates as the limits of a $dy$ integral. In the second example, the x-interval $[2,4]$ led to y-limits $[1,3]$, not $[2,4]$.

What to Keep in View

The setup has two independent geometric choices: the axis determines the radius, and the selected curve segment determines the bounds. Keeping those choices separate helps prevent errors. For a vertical axis, the distance is horizontal; for a profile written as $x=g(y)$, the natural integration variable is $y$.

Takeaway: For a selected smooth segment $a\le y\le b$, revolving $x=g(y)$ about $x=c$ gives $$ S=2\pi\int_a^b \lvert g(y)-c\rvert\sqrt{1+\bigl(g'(y)\bigr)^2}\,dy, $$ provided the swept surface is counted once. Check the axis distance and the segment endpoints before evaluating.

Check Your Understanding

Use the radius-to-axis distance and the y-coordinates of the selected segment in each setup.

For $x=3y+2$, $0\le y\le1$, revolved about $x=-2$, what are the radius function and arc-length factor?
The profile $x=y+4$ is revolved about $x=1$. If the selected x-coordinates run from $3$ to $6$, what are the y-bounds?
For $x=2-y$, $1\le y\le2$, revolved about $x=5$, write the complete surface-area integral, including the absolute-value radius.
Why is $g(y)-c$ not always the radius, while $\lvert g(y)-c\rvert$ is?
For $x=\sqrt{y}$, $1\le y\le4$, about $x=0$, identify the radius and the arc-length factor before multiplying them.