Parametric Profiles In Surface-Area Integrals

From a Parametrized Profile to a Surface

A profile need not be written as a graph \(y=f(x)\) to generate a surface of revolution. It may be more natural to specify its coordinates using a parameter \(t\), especially when the curve bends back or when neither coordinate is a convenient independent variable. The surface-area calculation still has two geometric ingredients: the circumference traced at each point and the small distance traveled along the profile.

For a parametrized curve, the distance traveled per unit change in \(t\) is its speed. Multiplying that speed by the radius of rotation and integrating gives a direct surface-area formula. The same method applies to circular arcs, straight segments, and other sufficiently smooth profiles.

Key idea: Surface area is accumulated along the profile. At parameter value \(t\), a thin strip contributes approximately its circumference, \(2\pi r(t)\), times its arc-length width.

The Parametric Surface-Area Formula

Let a profile be given by \(x=x(t)\) and \(y=y(t)\), for \(a\le t\le b\), where \(x\) and \(y\) are continuously differentiable. Its speed is \(\sqrt{[x'(t)]^2+[y'(t)]^2}\), so a small change \(dt\) corresponds to arc length \(ds=\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt\).

When the profile is revolved about an axis in its plane, let \(r(t)\) denote the nonnegative distance from \((x(t),y(t))\) to that axis. For example, for the \(x\)-axis, \(r(t)=|y(t)|\); for the \(y\)-axis, \(r(t)=|x(t)|\). The surface-area integral is

S=2\pi\int_a^b r(t)\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt.

The formula assumes the parametrized profile traces the intended portion of the curve and that its rotation does not produce overlapping surface patches of positive area. If a surface is swept more than once, the integral counts that area with multiplicity. A profile that touches or crosses the axis should be treated with care: the radius is a distance, and overlapping generated portions must not be counted as distinct area unless the intended quantity is area with multiplicity.

Definition: The parametric speed is \(v(t)=\sqrt{[x'(t)]^2+[y'(t)]^2}\). The differential of arc length is \(ds=v(t)\,dt\), and the surface-area formula can also be written \(S=2\pi\int r\,ds\).

Why the Formula Works

Consider a short portion of the profile near parameter value \(t\). If its arc length is approximately \(\Delta s\) and its distance from the axis is approximately \(r(t)\), rotating it makes a narrow band. The band’s circumference is approximately \(2\pi r(t)\), so its area is approximately \(2\pi r(t)\Delta s\). Adding these contributions along the curve leads to the integral.

More formally, partition \([a,b]\) into short intervals. On an interval from \(t_{i-1}\) to \(t_i\), let \(\Delta s_i\) be the profile’s arc length and choose a sample point \(t_i^*\). The band approximation is \(2\pi r(t_i^*)\Delta s_i\). As the partition becomes finer, the sum approaches the integral of \(2\pi r\) with respect to arc length. Since \(ds=v(t)\,dt\), that integral becomes the stated formula.

\sum_i 2\pi r(t_i^*)\Delta s_i \longrightarrow 2\pi\int r\,ds = 2\pi\int_a^b r(t)v(t)\,dt.

This also explains why the speed uses both coordinate derivatives: arc length records the curve’s actual motion in the plane, not just its horizontal or vertical change. Replacing \(ds\) with \(dx\) or \(dy\) alone would generally omit part of that motion.

Worked Examples

Worked Example: A Circular Profile Revolved About the Axis

Consider the circle parametrized by \(x(t)=\cos t\), \(y(t)=2+\sin t\), for \(0\le t\le 2\pi\), revolved about the \(x\)-axis. Find the surface area.

The radius is the distance from the profile to the \(x\)-axis. Because \(y(t)=2+\sin t\ge 1\), this distance is \(r(t)=2+\sin t\). The derivatives are \(x'(t)=-\sin t\) and \(y'(t)=\cos t\), so the speed is

v(t)=\sqrt{(-\sin t)^2+(\cos t)^2} =\sqrt{\sin^2 t+\cos^2 t}=1.

Substitute the radius and speed into the formula:

\begin{aligned} S &=2\pi\int_0^{2\pi}(2+\sin t)(1)\,dt\\ &=2\pi\left[2t-\cos t\right]_0^{2\pi}\\ &=2\pi\bigl((4\pi-1)-(0-1)\bigr)\\ &=2\pi(4\pi)=8\pi^2. \end{aligned}

As a check, the profile is a circle of radius \(1\) whose center is \(2\) units from the axis. The generated surface is a torus, whose area is circumference around the central ring, \(2\pi(2)\), times the circumference of the tube, \(2\pi(1)\). Their product is \(8\pi^2\), agreeing with the integral.

Worked Example: A Line Segment Revolved Into a Frustum

Parametrize a line segment by \(x(t)=t\), \(y(t)=1+t\), for \(0\le t\le 2\). Revolve it about the \(x\)-axis and find the curved surface area.

The radius is \(r(t)=1+t\), since \(y(t)\) is positive on this interval. The derivatives are \(x'(t)=1\) and \(y'(t)=1\), so

v(t)=\sqrt{1^2+1^2}=\sqrt{2}.

Therefore,

\begin{aligned} S &=2\pi\int_0^2(1+t)\sqrt{2}\,dt\\ &=2\pi\sqrt{2}\left[t+\frac{t^2}{2}\right]_0^2\\ &=2\pi\sqrt{2}\left((2+2)-0\right)\\ &=8\pi\sqrt{2}. \end{aligned}

The endpoint radii are \(1\) and \(3\), and the segment’s length is \(2\sqrt{2}\). The frustum formula gives \(\pi(1+3)(2\sqrt{2})=8\pi\sqrt{2}\), which confirms the result. The integral includes the curved lateral surface, not the two circular end faces.

Worked Example: A Line Segment Revolved About the Vertical Axis

Let \(x(t)=2+t\), \(y(t)=3t\), for \(0\le t\le 1\), and revolve the segment about the \(y\)-axis. Find the surface area.

The distance from the point to the \(y\)-axis is \(r(t)=|x(t)|=2+t\), which is positive throughout the interval. The coordinate derivatives are \(x'(t)=1\) and \(y'(t)=3\). Thus

v(t)=\sqrt{1^2+3^2}=\sqrt{10}.

The integral evaluates as follows:

\begin{aligned} S &=2\pi\int_0^1(2+t)\sqrt{10}\,dt\\ &=2\pi\sqrt{10}\left[2t+\frac{t^2}{2}\right]_0^1\\ &=2\pi\sqrt{10}\left(2+\frac12\right)\\ &=2\pi\sqrt{10}\cdot\frac52\\ &=5\pi\sqrt{10}. \end{aligned}

For a check, the two endpoint radii are \(2\) and \(3\), while the segment length is \(\sqrt{(3-2)^2+(3-0)^2}=\sqrt{10}\). The frustum calculation is \(\pi(2+3)\sqrt{10}=5\pi\sqrt{10}\), matching the parametrized integral.

Parameter Choice and Common Pitfalls

The parameter is a label for points on the curve; it need not represent distance. Consequently, the derivatives must be combined into the speed. For example, in the vertical-axis example, \(t\) changes by \(1\) over the whole segment, but the segment’s actual length is \(\sqrt{10}\), not \(1\). The speed accounts for that difference at every parameter value.

A common error is to use only one coordinate derivative in the arc-length factor. For the segment \(x(t)=t\), \(y(t)=1+t\), using \(dx=dt\) would give \(2\pi\int_0^2(1+t)\,dt=8\pi\), which is too small. The segment is slanted, so \(ds=\sqrt{(dt)^2+(dt)^2}=\sqrt{2}\,dt\). The correct area is \(8\pi\sqrt{2}\), not \(8\pi\).

Another error is to use a signed coordinate as the radius without checking its sign. A radius is a distance and cannot be negative. For revolution about the \(x\)-axis, it is \(|y(t)|\); for revolution about the \(y\)-axis, it is \(|x(t)|\). If the profile crosses the axis, split the integral where needed and examine whether different parts generate the same surface.

Practical check: Before integrating, identify the axis, write the radius as a nonnegative distance, calculate both coordinate derivatives, and simplify the speed. After evaluating, compare the result with the geometry and dimensions of the generated surface.

When This Representation Is Useful

Parametric profiles are particularly useful when a curve has a vertical tangent, doubles back in one coordinate, or is naturally described by an angle or another parameter. The surface-area formula does not require solving for \(y\) as a function of \(x\), nor does it require \(x\) to increase along the profile. It requires a valid differentiable parametrization, the correct radius, and the speed along that parametrization.

For a regular change of parameter that traces the same profile once, the geometric area is unchanged: the radius is still measured at the same points and the arc-length differential remains the same. Thus the formula depends on the curve and its position relative to the axis, not on the particular parameter used to describe it.

Check Your Understanding

For each question, identify the radius and speed before evaluating or comparing formulas.

For \(x(t)=3\cos t\), \(y(t)=3\sin t\), \(0\le t\le \pi/2\), revolved about the \(x\)-axis, what are \(r(t)\) and \(v(t)\)?
For \(x(t)=1+2t\), \(y(t)=t\), \(0\le t\le 1\), revolved about the \(y\)-axis, write the surface-area integral without evaluating it.
Why is the speed \(\sqrt{[x'(t)]^2+[y'(t)]^2}\), rather than just \(x'(t)\)?
A proposed calculation uses \(y(t)\) as the radius when the profile may lie below the \(x\)-axis. What correction is needed?
For what geometric reason can a parametrized integral overcount the area of a generated surface?