Computing Radius To An Oblique Axis

Why the Radius Must Be Perpendicular

A surface of revolution is generated by rotating a curve around an axis. At each point of the curve, the radius of the circular path is the shortest distance from that point to the axis. When the axis is oblique—that is, neither horizontal nor vertical—this radius is generally not just a difference between \(x\)- or \(y\)-coordinates.

To find the radius correctly, represent the axis as a line and calculate the perpendicular distance from a point on the profile to that line. The same distance then goes into the surface-area formula \(S=2\pi\int r\,ds\), where \(ds\) is an element of arc length along the profile.

Distance from a Point to an Oblique Line

Let the axis be the line \(ax+by+c=0\), where \(a\) and \(b\) are not both zero. The vector \(\mathbf{n}=(a,b)\) is perpendicular to this line. Its length is \(\sqrt{a^2+b^2}\), so \(\mathbf{n}/\sqrt{a^2+b^2}\) is a unit normal vector.

For a point \(P=(x_0,y_0)\), the expression \(ax_0+by_0+c\) measures its signed displacement from the line in the direction of the normal, scaled by the length of that normal. Dividing by the normal’s length gives the signed distance; taking its absolute value gives the perpendicular distance.

This formula applies to every point in the plane. For a curve that is rotated about the line, evaluate the formula at each profile point to obtain the radius as a function of the curve’s parameter.

Using Distance as the Surface-Area Radius

Suppose a smooth profile is given by \(y=f(x)\) for \(x\) in \([u,v]\). Its arc-length element is \(ds=\sqrt{1+(f'(x))^2}\,dx\). If the axis is \(ax+by+c=0\), the radius at \(x\) is

When the profile stays on one side of the axis and its revolution sweeps out a surface without being counted more than once, the surface area is \(2\pi\int r\,ds\). Combining the radius and arc-length expressions gives

The absolute value ensures a nonnegative radius. If the profile is known to remain on one side of the axis, the sign of \(ax+b f(x)+c\) is constant on the interval, so the absolute value can be removed using that sign. This tutorial uses profiles for which the surface is traced once; a profile that crosses the axis or produces overlapping surfaces needs additional care.

Why the Surface-Area Integral Uses This Radius

A short arc of profile of length approximately \(ds\), at distance \(r\) from the axis, sweeps out a narrow band. If the profile is locally close to a straight segment and the radius changes gradually, the band is approximately a frustum. Its area is \(2\pi\) times the average of the two end radii times the slant length. For a sufficiently short segment, that average approaches \(r\), and its slant length approaches \(ds\). Thus the band’s area approaches \(2\pi r\,ds\). Adding the areas of the bands along the profile leads to the integral.

Worked Example: A Line Segment Revolved About an Oblique Axis

Revolve the line segment \(y=0\), \(1\le x\le 3\), about the oblique axis \(y=x\). Write the axis as \(x-y=0\). A point on the segment is \((x,0)\), so its radius is

Here \(x>0\), so the absolute value becomes \(x\). The segment is horizontal, hence \(ds=dx\). The surface area is therefore

Check: The endpoint radii are \(1/\sqrt{2}\) and \(3/\sqrt{2}\), and the segment’s length is \(2\). The frustum-area calculation gives \(\pi(1/\sqrt{2}+3/\sqrt{2})\cdot 2=4\sqrt{2}\pi\), matching the integral.

Worked Example: A Parabolic Profile Revolved About an Oblique Axis

Revolve \(y=x^2\), \(0\le x\le 1\), about the axis \(x+y+1=0\). The radius at \((x,x^2)\) is

Since \(x^2+x+1>0\) on this interval, the profile remains on one side of the axis. Also, \(f'(x)=2x\), so \(ds=\sqrt{1+4x^2}\,dx\). Thus

To evaluate the integral, split it into three parts:

For the first integral, use \(u=2x\), so \(dx=du/2\) and \(x^2=u^2/4\). The standard antiderivative follows by differentiation:

Consequently, evaluating at \(u=2\) and \(u=0\) gives

For the second integral, substitute \(w=1+4x^2\), so \(dw=8x\,dx\). For the third, use the antiderivative shown below; its derivative is \(\sqrt{1+4x^2}\).

Adding the three contributions and multiplying by \(\sqrt{2}\pi\) yields the surface area:

The area is approximately \(13.03\) square units, rounded to two decimal places. Check: Differentiating the antiderivative for the third integral gives \(\sqrt{1+4x^2}\); the two substitutions above also reverse to their original integrands. The radius is positive throughout the interval, as required for this setup.

A Common Mistake: Using a Coordinate Difference

A frequent error is to treat the radius to an oblique axis as a vertical or horizontal distance. For example, for a point \((2,0)\) and the axis \(y=x\), using the vertical difference gives \(2\). That is not the perpendicular distance: the shortest segment from the point to the line is perpendicular to it.

Correction: Put the axis in the form \(ax+by+c=0\), substitute the point into the numerator, and divide by \(\sqrt{a^2+b^2}\). The denominator accounts for the length of the normal vector; omitting it generally produces the wrong radius.