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Surface Area of Revolution · Tutorial 433 of 998

Rotating Coordinates For Oblique-Axis Surface Area

Rotate the coordinate system so an oblique axis becomes horizontal, then derive and apply the surface-area formula using perpendicular distance and profile arc length.

Intermediate 10 min read

What You'll Learn

  • Rotate Cartesian coordinates to align a chosen oblique axis with the horizontal coordinate.
  • Express the signed perpendicular distance from a profile point to the axis in the rotated system.
  • Derive the surface-area element from narrow frustum bands and profile arc length.
  • Set up surface-area integrals for profiles described in Cartesian or rotated coordinates.
  • Recognize why an axis-crossing profile must be split and why coincident swept patches cannot be counted twice.

Why Rotate the Coordinates?

A curve can be revolved about any straight line, not only a horizontal or vertical axis. The geometry is simplest when the axis of rotation is horizontal: the radius of a band is then the curve’s perpendicular distance from that horizontal axis, and the length of the band comes from the curve’s arc length.

A rotation of coordinates makes this simplification available for an oblique axis. Because rotation preserves distances and lengths, it changes the coordinates used to describe the curve but does not change the surface being measured.

Key idea: In coordinates aligned with the axis, let \(v\) be signed perpendicular distance from the axis and let \(ds\) be an element of profile arc length. A narrow band has area approximately \(2\pi |v|\,ds\). Thus, for a profile segment that generates a surface without overlapping another part of the profile, its area is \(2\pi\int |v|\,ds\).

Coordinates Along and Perpendicular to the Axis

Suppose the axis has direction angle \(\theta\) measured from the positive \(x\)-axis. Its unit direction vector is \((\cos\theta,\sin\theta)\), and a perpendicular unit vector is \((-\sin\theta,\cos\theta)\). For a point \((x,y)\), define rotated coordinates \(u\) and \(v\) by projecting onto these two directions:

$$ u=x\cos\theta+y\sin\theta,\qquad v=-x\sin\theta+y\cos\theta. $$

Here \(u\) measures position along the axis, and \(v\) is signed perpendicular distance from the axis when the axis passes through the origin. The axis is \(v=0\). If the axis does not pass through the origin, translate the origin to a point on the axis first; equivalently, include the corresponding constant in the perpendicular coordinate.

The inverse rotation is

$$ x=u\cos\theta-v\sin\theta,\qquad y=u\sin\theta+v\cos\theta. $$

Since this is a rotation, it preserves length. In particular, for a differentiable profile parametrized by \(t\),

$$ ds=\sqrt{\left(\frac{du}{dt}\right)^2+ \left(\frac{dv}{dt}\right)^2}\,dt. $$

If the original profile is a graph \(y=f(x)\), the rotated coordinates along it are

$$ u(x)=x\cos\theta+f(x)\sin\theta,\qquad v(x)=-x\sin\theta+f(x)\cos\theta. $$

Although \(u(x)\) and \(v(x)\) can have different derivatives from the original coordinates, their combined arc-length element is unchanged:

$$ ds=\sqrt{1+\bigl(f'(x)\bigr)^2}\,dx. $$

Why the Surface-Area Element Has This Form

First consider a short straight segment of the profile lying entirely on one side of the axis. Suppose its endpoint distances from the axis are \(r_1\) and \(r_2\), and its length is \(\ell\). When the segment has different \(u\)-coordinates at its endpoints, its revolution forms the sloping side of a frustum. The lateral area of that frustum is

$$ \pi(r_1+r_2)\ell. $$

For a short band, the radius changes only a little. If \(r\) is a representative radius and \(\ell\) the short profile length, then \(r_1+r_2\) is approximately \(2r\), so the band area is approximately \(2\pi r\ell\). Shrinking the bands gives the integral of circumference times profile arc length.

The same local area element follows directly from parametrizing the surface. On a portion with \(v\geq 0\), rotate a profile \((u(t),v(t))\) through an angle \(\phi\):

$$ \mathbf{X}(t,\phi)= \bigl(u(t),v(t)\cos\phi,v(t)\sin\phi\bigr), \qquad 0\leq\phi\leq 2\pi. $$

The two tangent directions are \(\mathbf{X}_t\) and \(\mathbf{X}_\phi\). Their cross product has magnitude \(v(t)\sqrt{u'(t)^2+v'(t)^2}\). Therefore, the area element is \(v\,ds\,d\phi\). Integrating around a full revolution gives \(2\pi v\,ds\). If the profile is on the other side of the axis, its radius is \(|v|\), so the magnitude is \(2\pi|v|\,ds\).

Theorem: For a differentiable profile segment that stays on one side of the axis and whose revolution does not overlap another generated portion, its surface area is $$ S=2\pi\int |v|\,ds. $$ For a graph \(y=f(x)\) revolved about an oblique axis, this becomes $$ S=2\pi\int_a^b |v(x)|\sqrt{1+\bigl(f'(x)\bigr)^2}\,dx, $$ provided the chosen graph segment generates the surface being measured without duplicate coverage.

Worked Example: A Line Revolved About an Oblique Axis

Worked Example: A Line Revolved About an Oblique Axis

Revolve the segment \(y=x\), \(0\leq x\leq 1\), about the line \(3x+4y=0\). A direction vector for the axis is \((4,-3)\), so take \(\cos\theta=4/5\) and \(\sin\theta=-3/5\). The rotated coordinates are

$$ u=\frac45x-\frac35y,\qquad v=\frac35x+\frac45y. $$

On the profile \(y=x\), this gives \(u=x/5\) and \(v=7x/5\). The perpendicular distance is \(r=|v|=7x/5\) on this interval. The profile arc-length element is \(ds=\sqrt{1+1^2}\,dx=\sqrt2\,dx\). Therefore,

$$ S=2\pi\int_0^1 \frac{7x}{5}\sqrt2\,dx =\frac{14\pi\sqrt2}{5}\left[\frac{x^2}{2}\right]_0^1 =\frac{7\pi\sqrt2}{5}. $$

As a check, the endpoint at \(x=0\) is on the axis, and the endpoint at \(x=1\) has distance \(7/5\) from it. The segment length is \(\sqrt2\). Its revolution is a cone with lateral area \(\pi(0+7/5)\sqrt2=7\pi\sqrt2/5\), agreeing with the integral.

Worked Example: A Horizontal Profile and an Oblique Axis

Worked Example: A Horizontal Profile and an Oblique Axis

Revolve the segment \(y=0\), \(0\leq x\leq2\), about \(y=x\). The axis direction has angle \(\theta=\pi/4\). Thus

$$ u=\frac{x+y}{\sqrt2}=\frac{x}{\sqrt2}, \qquad v=\frac{-x+y}{\sqrt2}=-\frac{x}{\sqrt2}. $$

The radius is \(|v|=x/\sqrt2\), and \(ds=dx\). Hence

$$ S=2\pi\int_0^2\frac{x}{\sqrt2}\,dx =\frac{2\pi}{\sqrt2}\left[\frac{x^2}{2}\right]_0^2 =\frac{2\pi}{\sqrt2}\cdot2 =2\sqrt2\,\pi. $$

To check the result geometrically, the radii at the endpoints are \(0\) and \(\sqrt2\), and the segment length is \(2\). The revolved segment forms a cone, whose lateral area is \(\pi(0+\sqrt2)(2)=2\sqrt2\pi\).

Using the Formula in Rotated Coordinates

Sometimes the profile is easiest to describe after the coordinates have been rotated. If it is given as \(v=g(u)\), then \(ds=\sqrt{1+(g'(u))^2}\,du\) on an interval where \(u\) is the independent coordinate. The surface-area integral is then

$$ S=2\pi\int |g(u)|\sqrt{1+\bigl(g'(u)\bigr)^2}\,du. $$

This is the same formula expressed in coordinates aligned with the axis. It is especially useful when the axis is oblique in the original \(x,y\)-plane but the profile has a simple description in \(u,v\).

Worked Example: A Profile Given in the Rotated System

Let the axis be \(y=x\), and describe the profile in coordinates aligned with it by \(u=t,\ v=1+t\), for \(0\leq t\leq1\). Since \(v>0\), its distance from the axis is \(1+t\). The profile length element is

$$ ds=\sqrt{\left(\frac{du}{dt}\right)^2+ \left(\frac{dv}{dt}\right)^2}\,dt =\sqrt{1^2+1^2}\,dt =\sqrt2\,dt. $$

The area is therefore

$$ S=2\pi\int_0^1(1+t)\sqrt2\,dt =2\pi\sqrt2\left[t+\frac{t^2}{2}\right]_0^1 =3\pi\sqrt2. $$

For a Cartesian check, the inverse rotation for \(\theta=\pi/4\) gives \(x=(u-v)/\sqrt2=-1/\sqrt2\) and \(y=(u+v)/\sqrt2=(1+2t)/\sqrt2\). Thus the profile is a straight segment of length \(\sqrt2\), with endpoint radii \(1\) and \(2\). Its frustum area is \(\pi(1+2)\sqrt2=3\pi\sqrt2\), as obtained above.

Axis Crossings and a Common Pitfall

The formula uses distance from the axis, so the radius is \(|v|\), not the signed coordinate \(v\). If a profile crosses the axis, split it at the crossing before applying the formula to its one-sided pieces. This prevents treating a signed distance as a negative radius.

Splitting alone does not guarantee that areas of the pieces can simply be added: distinct profile portions can generate the same patch of surface. For example, consider the segment \(u=0\), \(-1\leq v\leq1\). It crosses the axis, and its two radial portions each sweep out the same disk of radius \(1\). The geometric union is one disk, of area \(\pi\). Adding the two one-sided integrals would count that disk twice.

$$ 2\pi\int_{-1}^{1}|v|\,dv =2\pi\left(\int_{-1}^{0}(-v)\,dv+\int_{0}^{1}v\,dv\right) =2\pi. $$

This value is the area counted with multiplicity by the two sweeps, not the area of their union. Each radial half sweeps the disk once, with area \(2\pi\int_0^1r\,dr=\pi\); because those two disks coincide, the union has area \(\pi\). Thus, the integral gives the geometric surface area directly when the generated patches do not overlap. If they do overlap, identify the overlap and count the union only once.

Common mistake: Applying the frustum lateral-area formula \(\pi(r_1+r_2)\ell\) to a segment that crosses the axis, or adding the areas of coincident sweeps. The frustum formula applies to a segment that stays on one side and forms a frustum band; crossing segments must be split, and any coincident patches must not be counted twice.

Summary of the Setup

To set up surface area about an oblique axis, rotate coordinates so that the axis becomes \(v=0\). Find the signed perpendicular coordinate \(v\), take its absolute value for the radius, and calculate \(ds\) along the original profile. Then integrate circumference times profile arc length, checking that the selected pieces do not duplicate surface patches.

1
Align coordinates with the axis.
Use \(u\) along the axis and \(v\) perpendicular to it; translate first if the axis does not pass through the origin.
2
Find the radius and arc length.
The radius is \(|v|\), and \(ds=\sqrt{du^2+dv^2}\) along the profile.
3
Integrate the band areas.
Use \(S=2\pi\int |v|\,ds\) on one-sided pieces, and check for crossings or coincident sweeps.

Check Your Understanding

Use the rotated-coordinate method and explain the geometric conditions relevant to each setup.

  1. For an axis with direction angle \(\theta\), write the formulas for \(u\) and \(v\) in terms of \(x\) and \(y\). What does each coordinate measure?
  2. A profile is \(y=2x\), \(0\leq x\leq1\), revolved about \(y=x\). Find \(u(x)\), \(v(x)\), and \(ds\), then set up and evaluate its surface-area integral.
  3. Why does a narrow band have area approximately \(2\pi r\,ds\)? State what \(r\) and \(ds\) represent.
  4. What must be checked before adding areas from two profile pieces that cross the axis?
  5. For a profile given as \(v=g(u)\), write the surface-area integral on an interval where \(u\) is the independent coordinate.

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