Surface-Area Problems With Shifted Axes

Why a Shifted Axis Changes the Setup

A surface-of-revolution integral depends on the distance from the profile to the axis of rotation. When the axis is a coordinate axis, that distance may be given directly by a coordinate. For a shifted axis such as \(y=c\) or \(x=c\), the radius is the distance from the profile to that line instead.

This distinction matters especially when the axis crosses the profile’s range. A radius is never negative, so the correct expression is an absolute value whenever the profile can lie on either side of the axis. If the axis intersects the profile, the integral may need to be split at the intersection.

Key rule: The radius of a surface element is the perpendicular distance from the profile point to the axis of rotation. For a horizontal axis \(y=c\), this distance is \(|f(x)-c|\). For a vertical axis \(x=c\), it is \(|g(y)-c|\) when the profile is written as \(x=g(y)\).

Surface-Area Formulas for Shifted Axes

Suppose a smooth profile is given by \(y=f(x)\) on \(a\le x\le b\), and is revolved about the horizontal line \(y=c\). Its arc-length element is \(ds=\sqrt{1+[f'(x)]^2}\,dx\). A small piece of the profile sweeps out a band whose circumference is \(2\pi|f(x)-c|\). Therefore its surface-area element is circumference times arc length:

S=2\pi\int_a^b |f(x)-c|\sqrt{1+[f'(x)]^2}\,dx.

For a profile given by \(x=g(y)\) on \(p\le y\le q\), revolved about the vertical line \(x=c\), the arc-length element is \(ds=\sqrt{1+[g'(y)]^2}\,dy\), and the radius is \(|g(y)-c|\). Thus

S=2\pi\int_p^q |g(y)-c|\sqrt{1+[g'(y)]^2}\,dy.

These formulas assume a smooth, regular profile on the interval. If the axis does not intersect the profile, the distance expression has one sign throughout the interval, so the absolute value can be replaced by the corresponding positive expression. If the profile meets or crosses the axis, keep the absolute value or split the integral at the crossing points.

A vertical axis can also be used with a profile written as \(y=f(x)\): the radius is then \(|x-c|\), and the same arc-length element in \(x\) is used. Writing the profile as \(x=g(y)\) is often convenient, but it is not required if the original parametrization already makes the integral straightforward.

Why the Radius Must Be an Absolute Distance

The formula follows from the area element of the surface traced by a rotating curve. For the horizontal-axis case, view the axis \(y=c\) as the \(x\)-direction in three-dimensional space. Rotating the point \((x,f(x))\) gives the parametrization

\mathbf{X}(x,\theta) = \bigl(x,\ c+(f(x)-c)\cos\theta,\ (f(x)-c)\sin\theta\bigr), \qquad 0\le\theta\le2\pi.

The parameter \(x\) moves along the profile, while \(\theta\) moves around the axis. The area element of a parametrized surface is the magnitude of the cross product of its two tangent vectors, integrated over the parameter region.

Differentiate with respect to \(x\).
\(\mathbf{X}_x=(1,f'(x)\cos\theta,f'(x)\sin\theta)\).

Differentiate with respect to \(\theta\).
\(\mathbf{X}_\theta=(0,-(f(x)-c)\sin\theta,(f(x)-c)\cos\theta)\).

Compute the area factor.
The cross product has magnitude \(|f(x)-c|\sqrt{1+[f'(x)]^2}\). This is independent of \(\theta\).

Integrate around the circle.
Integrating over \(0\le\theta\le2\pi\) contributes a factor of \(2\pi\), giving the horizontal-axis formula.

The absolute value appears because the cross-product magnitude is a length and cannot be negative. If the profile touches the axis, the parametrization degenerates at that point, but the area integral remains valid by splitting the interval there and evaluating each side. The same argument with coordinates exchanged gives the formula for a vertical axis.

Worked Example: A Profile Revolved About a Horizontal Axis Outside Its Range

Worked Example: A Line Segment About \(y=-1\)

Revolve \(y=2x+1\), for \(0\le x\le2\), about \(y=-1\). The profile has \(1\le y\le5\), so the axis lies below the entire profile. The radius is

$$ r(x)=(2x+1)-(-1)=2x+2. $$

Differentiate the profile and compute its arc-length factor:

f'(x)=2, \qquad \sqrt{1+[f'(x)]^2}=\sqrt{1+4}=\sqrt5.

Substitute into the surface-area formula and evaluate:

\begin{aligned} S &=2\pi\int_0^2(2x+2)\sqrt5\,dx\\ &=2\pi\sqrt5\,[x^2+2x]_0^2\\ &=2\pi\sqrt5\bigl((4+4)-0\bigr)\\ &=16\pi\sqrt5. \end{aligned}

As a geometric check, the endpoint radii are \(2\) and \(6\), and the slant length of the line segment is \(\sqrt{2^2+4^2}=2\sqrt5\). The lateral area of the resulting frustum is \(\pi(2+6)(2\sqrt5)=16\pi\sqrt5\), agreeing with the integral.

Worked Example: A Profile Crossing a Horizontal Axis

Worked Example: A Line Segment About \(y=1\)

Revolve \(y=x+1\), for \(-1\le x\le1\), about \(y=1\). The profile meets the axis at \(x=0\). Its radius is the distance

$$ r(x)=|(x+1)-1|=|x|. $$

Since \(f'(x)=1\), the arc-length factor is \(\sqrt{1+1^2}=\sqrt2\). Split the integral at the crossing, where the expression inside the absolute value changes sign:

\begin{aligned} S &=2\pi\sqrt2\int_{-1}^1|x|\,dx\\ &=2\pi\sqrt2\left(\int_{-1}^0(-x)\,dx+\int_0^1x\,dx\right)\\ &=2\pi\sqrt2\left(\left[-\frac{x^2}{2}\right]_{-1}^0+ \left[\frac{x^2}{2}\right]_0^1\right)\\ &=2\pi\sqrt2\left(\frac12+\frac12\right)\\ &=2\pi\sqrt2. \end{aligned}

The two parts of the profile form a cone on either side of the point where the profile meets the axis. Each cone has base radius \(1\) and slant length \(\sqrt2\), so each has lateral area \(\pi\sqrt2\). Their total, \(2\pi\sqrt2\), matches the integral. The surfaces meet at the vertex but do not overlap in area.

Worked Example: A Vertical Axis Outside the Profile’s Range

Worked Example: A Line Segment About \(x=-1\)

Write the profile as \(x=3-y\), for \(0\le y\le2\), and revolve it about \(x=-1\). Since \(1\le x\le3\), the axis lies to the left of the entire profile. The radius is

$$ r(y)=(3-y)-(-1)=4-y. $$

Here \(g'(y)=-1\), so the arc-length factor is \(\sqrt{1+(-1)^2}=\sqrt2\). Therefore

\begin{aligned} S &=2\pi\int_0^2(4-y)\sqrt2\,dy\\ &=2\pi\sqrt2\,[4y-y^2/2]_0^2\\ &=2\pi\sqrt2\bigl((8-2)-0\bigr)\\ &=12\pi\sqrt2. \end{aligned}

The endpoint radii are \(4\) and \(2\), and the slant length is \(\sqrt{2^2+2^2}=2\sqrt2\). The frustum check gives \(\pi(4+2)(2\sqrt2)=12\pi\sqrt2\), confirming the result.

Worked Example: A Profile Crossing a Vertical Axis

Worked Example: A Line Segment About \(x=2\)

Revolve \(x=y+2\), for \(-1\le y\le1\), about \(x=2\). The profile meets the axis at \(y=0\), and its radius is

$$ r(y)=|(y+2)-2|=|y|. $$

Since \(g'(y)=1\), the arc-length factor is \(\sqrt2\). Splitting at \(y=0\) gives

\begin{aligned} S &=2\pi\sqrt2\int_{-1}^1|y|\,dy\\ &=2\pi\sqrt2\left(\int_{-1}^0(-y)\,dy+\int_0^1y\,dy\right)\\ &=2\pi\sqrt2\left(\frac12+\frac12\right)\\ &=2\pi\sqrt2. \end{aligned}

The result describes two conical pieces meeting at the point on the axis. Each has radius \(1\) and slant length \(\sqrt2\), giving total area \(2\pi\sqrt2\), as calculated.

A Common Error: Keeping a Signed Radius

For the horizontal crossing example, using \(x\) instead of \(|x|\) would give

2\pi\sqrt2\int_{-1}^1x\,dx = 2\pi\sqrt2\left[\frac{x^2}{2}\right]_{-1}^1 = 0.

This cannot be a surface area. The error is treating the signed displacement from the axis as a radius. On the left side of the crossing, \(x<0\), but the distance from the profile to the axis is \(-x>0\). The corrected integral uses \(|x|\), or equivalently splits the interval and uses \(-x\) on \([-1,0]\) and \(x\) on \([0,1]\).

Takeaway: First find the perpendicular distance to the shifted axis. Then multiply by \(2\pi\) and the profile’s arc-length element. If the profile crosses the axis, use a nonnegative radius on every part of the interval.

Interpreting the Surface and Checking the Setup

The integral computes the area swept out as the chosen profile segment rotates. If different parts of the profile generate the same patch of surface, adding their contributions counts that patch more than once. In that situation, distinguish the area traced with multiplicity by the integral from the area of the geometric union. Axis crossings are a natural place to check for such overlap; the actual geometry depends on the profile and the segment being revolved.

Before evaluating a shifted-axis surface-area integral, verify the following:

The axis is horizontal or vertical, and the radius is measured perpendicular to it.
The radius is nonnegative: use an absolute value if the profile can lie on both sides of the axis.
The arc-length factor matches the variable of integration.
If the profile meets the axis, split the integral at each crossing when needed.
The computed area agrees with a geometric check when the surface is a familiar shape such as a cone or frustum.

Check Your Understanding

For each question, identify the radius before attempting to calculate the surface area.

For \(y=x^2+3\) on \(0\le x\le1\), revolved about \(y=1\), what radius expression should appear in the integral?
For \(x=y-2\) on \(-1\le y\le1\), revolved about \(x=-2\), where does the profile meet the axis, and where must the integral be split?
For \(y=3x+2\) on \(0\le x\le2\), revolved about \(y=-1\), is the radius signed or absolute on the whole interval? Explain.
What arc-length factor is used when a profile is written as \(x=g(y)\)?
Why can integrating a signed displacement across an axis crossing produce an incorrect result?