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Surface Area of Revolution · Tutorial 435 of 998

Axis Crossings And Overlapping Surfaces

Use absolute radii to handle profiles on or across an axis, and compare branches to avoid counting the same surface patch more than once.

Intermediate 9 min read

What You'll Learn

  • Explain why a radius is the absolute distance from the axis, even when a profile crosses it.
  • Set up surface-area integrals on intervals where a profile meets or lies on the axis.
  • Identify when two profile branches generate identical patches of a surface of revolution.
  • Distinguish coincident patches from surfaces that meet only along a curve or at a point.
  • Compute geometric surface area without counting overlapped patches more than once.

Why Crossings and Overlaps Need Care

A profile may cross the axis about which it is revolved, touch the axis without crossing, or coincide with the axis for part of its length. In each case, the distance from the profile to the axis is a physical radius and cannot be negative. A second issue arises when a surface is described by multiple profile branches: two different branches can generate the same patch when revolved. Adding their separate areas would count that patch more than once.

The essential checks are therefore distinct. First, use the absolute distance to the axis in the area integrand. Second, compare branches by their axial position and their distance from the axis. A crossing of one profile does not by itself imply any overlap, and two distinct profiles do not necessarily generate distinct surfaces.

Key takeaway: Surface area uses the nonnegative distance to the axis. If separate profile branches generate the same surface patch, the geometric area of their union counts that patch once.

Radius and Surface-Area Formulas

For revolution about the \(x\)-axis, a graph \(y=f(x)\) has radius \(r(x)=|f(x)|\). Provided \(f\) is continuously differentiable on the interval of integration, its generated surface area is

$$ A=2\pi\int_a^b |f(x)|\sqrt{1+\bigl(f'(x)\bigr)^2}\,dx. $$

For revolution about the \(y\)-axis, a profile written as \(x=g(y)\) has radius \(r(y)=|g(y)|\), and its area is

$$ A=2\pi\int_c^d |g(y)|\sqrt{1+\bigl(g'(y)\bigr)^2}\,dy. $$

These formulas use the axial coordinate as the integration variable. They require a suitable smooth graph on the stated interval; if a profile must be split into smooth pieces, apply the formula on each piece. At points where the radius is zero, the integrand is zero. If the profile lies on the axis over an interval, it generates no surface area there.

The same principle applies to an axis shifted from a coordinate axis: the radius is the absolute distance from the profile to the axis. The formulas above are written for coordinate axes so that the role of the absolute value is explicit.

Why the Radius Is an Absolute Value

For a profile \(y=f(x)\) revolved about the \(x\)-axis, a parametrization of the generated surface is

$$ \mathbf{R}(x,\theta) = \bigl(x,\ f(x)\cos\theta,\ f(x)\sin\theta\bigr), \qquad 0\leq\theta\leq 2\pi. $$

At a fixed \(x\), the distance of this circle from the axis is \(|f(x)|\), whether \(f(x)\) is positive or negative. The area element follows by taking the magnitude of the cross product of the two tangent vectors:

$$ \mathbf{R}_x=(1,f'(x)\cos\theta,f'(x)\sin\theta), \qquad \mathbf{R}_\theta=(0,-f(x)\sin\theta,f(x)\cos\theta), $$
$$ \left|\mathbf{R}_x\times\mathbf{R}_\theta\right| = |f(x)|\sqrt{1+\bigl(f'(x)\bigr)^2}. $$

Integrating this element over \(0\leq\theta\leq2\pi\) gives the factor \(2\pi|f(x)|\sqrt{1+(f'(x))^2}\). In particular, replacing \(f(x)\) by \(|f(x)|\) in the radius does not change the surface generated by the profile; it expresses the actual distance to the axis.

Worked Example: A Smooth Profile Crossing the Axis

Worked Example: A Smooth Profile Crossing the Axis

Revolve \(y=(x-1)^3\), for \(0\leq x\leq2\), about the \(x\)-axis. Find the surface area. Assume coordinates are measured in units, so the answer is in square units.

The profile crosses the axis at \(x=1\). Its derivative is \(f'(x)=3(x-1)^2\), so the area formula gives

$$ A=2\pi\int_0^2 |(x-1)^3|\sqrt{1+9(x-1)^4}\,dx. $$

Set \(u=x-1\). The limits become \(-1\) and \(1\), and the integrand is even. Therefore,

$$ A=4\pi\int_0^1 u^3\sqrt{1+9u^4}\,du. $$

Use \(t=1+9u^4\), so \(dt=36u^3\,du\). The limits in \(t\) are \(1\) and \(10\):

$$ \begin{aligned} A &=4\pi\left(\frac{1}{36}\int_1^{10}t^{1/2}\,dt\right)\\ &=\frac{\pi}{9}\left[\frac{2}{3}t^{3/2}\right]_1^{10}\\ &=\frac{2\pi}{27}\left(10\sqrt{10}-1\right). \end{aligned} $$

To check, differentiating \(\frac{1}{54}(1+9u^4)^{3/2}\) gives \(u^3\sqrt{1+9u^4}\). Its change from \(u=0\) to \(u=1\) is \(\frac{10\sqrt{10}-1}{54}\); doubling for the two halves and multiplying by \(2\pi\) gives the same area. The result is positive, as a surface area must be.

Axis Touching and Axis Segments

A profile can meet the axis without changing sides. For instance, \(f(x)=x^2\) touches the \(x\)-axis at \(x=0\) and is nonnegative on both sides of that point. Its radius is still \(|f(x)|\), which here equals \(x^2\). The formula remains valid across the touch point; splitting the integral there is optional if the graph is smooth on the full interval.

If a profile lies on the axis for a whole interval, its radius is zero throughout that interval. Consequently the area integrand is zero there. Including the interval in the integral adds nothing. This is different from a profile that merely meets the axis at one point: that point contributes zero by itself, but the profile on either side may generate surface area.

When a profile changes sign, one may split the integral at its zeros to remove the absolute-value notation. On each resulting interval, determine the sign of the profile before replacing \(|f(x)|\). Splitting is a convenience, not permission to discard the portions where the profile is negative.

When Do Different Branches Overlap?

For rotation about the \(y\)-axis, the axial coordinate is \(y\), while the radius is \(|x|\). A branch \(x=g(y)\) therefore generates a circular patch at each \(y\), with radius \(|g(y)|\). Two branches \(g_1\) and \(g_2\) generate the same circular patch at a given \(y\) precisely when

$$ |g_1(y)|=|g_2(y)|. $$

To have the same patch over a range of axial positions, this equality must hold throughout that range. Such a shared patch contributes once to the area of the geometric union, not once for each description of it.

By contrast, branches with different radii at the same axial coordinate generate different circles. They do not overlap in a surface patch there. Surfaces from separate branches may meet at an isolated circle, curve, or point; such an intersection has zero surface area and does not require subtracting area from the sum of the branch integrals. The concern is a coincident patch, which has positive surface area.

Worked Example: Two Branches Describe an Overlapping Patch

Worked Example: Two Branches Describe an Overlapping Patch

Rotate the branches \(x=y\) for \(0\leq y\leq2\) and \(x=-y\) for \(1\leq y\leq2\) about the \(y\)-axis. Find the area of the surface they generate, counting a coincident patch only once.

For the first branch, the radius is \(|y|=y\), and \(dx/dy=1\). Its area is

$$ A_1=2\pi\int_0^2 y\sqrt{1+1^2}\,dy =2\pi\sqrt{2}\left[\frac{y^2}{2}\right]_0^2 =4\pi\sqrt{2}. $$

For the second branch, the radius is also \(|-y|=y\), and \(dx/dy=-1\). Its area, considered separately, is

$$ A_2=2\pi\int_1^2 y\sqrt{1+(-1)^2}\,dy =2\pi\sqrt{2}\left[\frac{y^2}{2}\right]_1^2 =3\pi\sqrt{2}. $$

At every \(y\) from \(1\) to \(2\), the branches have equal absolute \(x\)-coordinates, so they generate the same circular patch. The second branch therefore adds no new surface beyond the first. The area of the union is \(A_1=4\pi\sqrt{2}\), not \(A_1+A_2=7\pi\sqrt{2}\). As a check, the first branch already covers all axial positions from \(0\) to \(2\), including the full overlap interval.

Worked Example: Different Radii Do Not Make a Shared Patch

Worked Example: Different Radii Do Not Make a Shared Patch

Rotate \(x=y\) and \(x=2y\), both for \(0\leq y\leq1\), about the \(y\)-axis. Find the area of their combined surface.

For \(x=y\), the radius is \(y\) and the derivative is \(1\). Thus

$$ A_1=2\pi\int_0^1 y\sqrt{2}\,dy =2\pi\sqrt{2}\left[\frac{y^2}{2}\right]_0^1 =\pi\sqrt{2}. $$

For \(x=2y\), the radius is \(2y\) and the derivative is \(2\). Thus

$$ A_2=2\pi\int_0^1 2y\sqrt{5}\,dy =4\pi\sqrt{5}\left[\frac{y^2}{2}\right]_0^1 =2\pi\sqrt{5}. $$

For every \(y>0\), the radii \(y\) and \(2y\) differ, so the branches generate different circular patches. They meet at the axis when \(y=0\), but that single point contributes no surface area. The total is therefore

$$ A=A_1+A_2=\pi\sqrt{2}+2\pi\sqrt{5}. $$

The overlap check is decisive: equality of axial coordinates alone is not enough; the absolute radii must also be equal for the patches to coincide.

A Common Mistake

A frequent error is to use the signed profile value as the radius. For the crossing profile in the first example, omitting the absolute value would give

$$ 2\pi\int_0^2 (x-1)^3\sqrt{1+9(x-1)^4}\,dx=0, $$

because the integrand is odd after the substitution \(u=x-1\), and the limits are symmetric. A surface with zero area has not been found: the calculation has incorrectly subtracted the contribution from one side of the axis from the other. The correction is to use \(|(x-1)^3|\), giving the positive area \(\frac{2\pi}{27}(10\sqrt{10}-1)\).

A related error is adding the areas of coincident branches as if they were separate pieces of the geometric surface. Before adding branch integrals, compare their axial coordinates and absolute radii. Count a shared patch once; add distinct patches, even if they meet along a boundary.

Check Your Understanding

For each question, consider both the radius and the patch generated by a profile.

  1. What radius should be used for the graph \(y=f(x)\) revolved about the \(x\)-axis when \(f(x)<0\)?
  2. If a profile lies on its axis of revolution for an interval, what does that interval contribute to surface area?
  3. When do two branches \(x=g_1(y)\) and \(x=g_2(y)\), revolved about the \(y\)-axis, generate the same circular patch at a given \(y\)?
  4. Two branches meet at one point but have different radii at every other shared axial coordinate. Does that meeting point require subtracting area from their summed areas? Explain.
  5. Why is integrating a signed profile value instead of its absolute value incorrect for surface area?

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