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Surface Area of Revolution · Tutorial 441 of 998

Matching Surface Areas From Two Profile Descriptions

See how the derivative and arc-length factors transform together when a profile is rewritten using its inverse.

Intermediate 10 min read

What You'll Learn

  • State conditions that ensure a profile has a differentiable inverse and positive radius from either coordinate axis.
  • Relate the derivatives of a function and its inverse.
  • Transform the arc-length factor and integration differential consistently.
  • Verify equality of the two descriptions’ surface-area integrals about the x-axis.
  • Verify equality of the two descriptions’ surface-area integrals about the y-axis.
  • Identify the error caused by omitting the change-of-variable factor.

Why Compare Two Descriptions?

A curve in the first quadrant can often be described either as \(y=f(x)\) or, if \(f\) is one-to-one, as \(x=f^{-1}(y)\). These are not two different curves: they are two ways of specifying the same points. When we revolve that curve about an axis, each description should give the same surface area, provided we transform the derivative, the arc-length factor, and the integration differential consistently.

The comparison is made about one fixed axis at a time. Equality between the two descriptions does not mean that revolving a curve about the \(x\)-axis gives the same area as revolving it about the \(y\)-axis. Those are generally different surfaces.

Surface-Area Formulas for the Two Descriptions

Let \(f\) be continuously differentiable on \([a,b]\), with \(0<a<b\) and \(f(x)>0\) throughout the interval. Suppose also that \(f\) is strictly monotone and \(f'(x)\ne 0\) throughout \([a,b]\). Then the curve has an inverse function on its image, and its points have positive \(x\)- and \(y\)-coordinates. These positivity conditions ensure that the radii used below are distances from the axes, not signed coordinates.

For a graph \(y=f(x)\), the arc-length element is \(ds=\sqrt{1+(f'(x))^2}\,dx\). Revolving a small arc about an axis produces a surface band whose circumference is \(2\pi\) times the distance from that axis. Thus the surface areas are given by the following formulas.

$$ \begin{aligned} S_x &=2\pi\int_a^b f(x)\sqrt{1+(f'(x))^2}\,dx, \\[4pt] S_y &=2\pi\int_a^b x\sqrt{1+(f'(x))^2}\,dx. \end{aligned} $$

Here \(S_x\) is the area when the curve is revolved about the \(x\)-axis, so the radius is \(y=f(x)\). The quantity \(S_y\) is the area when it is revolved about the \(y\)-axis, so the radius is \(x\).

For the inverse description \(x=f^{-1}(y)\), the corresponding arc-length element is \(\sqrt{1+((f^{-1})'(y))^2}\,dy\). Its surface-area formulas use the same radii, expressed in terms of \(y\): \(y\) for revolution about the \(x\)-axis and \(f^{-1}(y)\) for revolution about the \(y\)-axis.

Key relationship: The inverse function rule gives \((f^{-1})'(y)=1/f'(x)\) when \(y=f(x)\). In changing variables between the descriptions, the differential also changes: \(|dy|=|f'(x)|\,dx\). Both changes are needed.

Why the Integrals Agree

We now prove the equality for either axis. Write \(g=f^{-1}\), and let \(y=f(x)\). Because \(f'\) is continuous and never zero on the interval, it has a constant sign there. The inverse function rule gives \(g'(f(x))=1/f'(x)\). The key identity is

$$ \sqrt{1+\left(\frac{1}{f'(x)}\right)^2}\,|f'(x)| = \sqrt{(f'(x))^2+1} = \sqrt{1+(f'(x))^2}. $$

The absolute value appears because, if \(f\) is decreasing, increasing \(x\) makes \(y\) decrease. In an ordinary integral with increasing limits, the change of variables therefore contributes \(|dy|=|f'(x)|\,dx\). This factor combines with the inverse arc-length factor to recover the original arc-length factor.

Theorem: Suppose \(f\) is continuously differentiable on \([a,b]\), where \(0<a<b\), and \(f(x)>0\) and \(f'(x)\ne 0\) throughout that interval. Then, for the same coordinate axis, the surface-area integral using \(y=f(x)\) equals the integral using \(x=f^{-1}(y)\), with the inverse integral taken over the interval between \(f(a)\) and \(f(b)\).
1
Revolution about the \(x\)-axis.
The inverse description has radius \(y\), so its integral is \(2\pi\int y\sqrt{1+(g'(y))^2}\,dy\), with increasing \(y\)-limits. Substitute \(y=f(x)\); the factor \(y\) becomes \(f(x)\), and the arc-length factor times the changed differential becomes \(\sqrt{1+(f'(x))^2}\,dx\). The result is \(S_x\).
2
Revolution about the \(y\)-axis.
The inverse description has radius \(x=g(y)\), so its integral is \(2\pi\int g(y)\sqrt{1+(g'(y))^2}\,dy\). Under the same substitution, \(g(y)=x\), and the transformed arc-length factor and differential give \(\sqrt{1+(f'(x))^2}\,dx\). The result is \(S_y\).

This argument proves equality for increasing and decreasing profiles alike. In the decreasing case, the inverse limits reverse order when expressed as \(f(a)\) and \(f(b)\); using increasing limits and the absolute change-of-variable factor accounts for that reversal.

Worked Examples

Worked Example: An Increasing Line About Either Axis

Take \(y=2x+1\) for \(1\le x\le 3\). The profile is positive and increasing, and its inverse is \(x=(y-1)/2\), for \(3\le y\le 7\). Since \(dy/dx=2\), the arc-length factor in the \(x\)-description is \(\sqrt{1+2^2}=\sqrt{5}\). The inverse derivative is \(dx/dy=1/2\), whose arc-length factor is \(\sqrt{1+(1/2)^2}=\sqrt{5}/2\).

About the \(x\)-axis, the original description gives

$$ \begin{aligned} S_x &=2\pi\int_1^3(2x+1)\sqrt{5}\,dx\\ &=2\pi\sqrt{5}\left[x^2+x\right]_1^3\\ &=2\pi\sqrt{5}\bigl((9+3)-(1+1)\bigr)\\ &=20\pi\sqrt{5}. \end{aligned} $$

The inverse description gives

$$ \begin{aligned} S_x &=2\pi\int_3^7 y\frac{\sqrt{5}}{2}\,dy\\ &=\pi\sqrt{5}\left[\frac{y^2}{2}\right]_3^7\\ &=\frac{\pi\sqrt{5}}{2}(49-9)\\ &=20\pi\sqrt{5}. \end{aligned} $$

About the \(y\)-axis, the original description gives

$$ \begin{aligned} S_y &=2\pi\int_1^3 x\sqrt{5}\,dx\\ &=2\pi\sqrt{5}\left[\frac{x^2}{2}\right]_1^3\\ &=\pi\sqrt{5}(9-1)\\ &=8\pi\sqrt{5}. \end{aligned} $$

From the inverse description, the radius is \(x=(y-1)/2\), so

$$ \begin{aligned} S_y &=2\pi\int_3^7\frac{y-1}{2}\frac{\sqrt{5}}{2}\,dy\\ &=\frac{\pi\sqrt{5}}{2}\left[\frac{y^2}{2}-y\right]_3^7\\ &=\frac{\pi\sqrt{5}}{2}\bigl((49/2-7)-(9/2-3)\bigr)\\ &=\frac{\pi\sqrt{5}}{2}(16)\\ &=8\pi\sqrt{5}. \end{aligned} $$

The two descriptions agree for each axis. The \(x\)-axis area and the \(y\)-axis area are not equal to one another: their respective values are \(20\pi\sqrt{5}\) and \(8\pi\sqrt{5}\).

Worked Example: A Decreasing Circular Profile

Consider \(y=\sqrt{1-x^2}\) for \(3/5\le x\le 4/5\). Both coordinates are positive, and the derivative \(dy/dx=-x/\sqrt{1-x^2}\) is nonzero on this interval. The inverse is \(x=\sqrt{1-y^2}\). As \(x\) increases from \(3/5\) to \(4/5\), \(y\) decreases from \(4/5\) to \(3/5\), so the inverse integrals use increasing limits \(3/5\le y\le 4/5\).

The original arc-length factor simplifies to

$$ \sqrt{1+\left(\frac{-x}{\sqrt{1-x^2}}\right)^2} = \sqrt{\frac{1}{1-x^2}} = \frac{1}{\sqrt{1-x^2}}, $$

and the inverse arc-length factor is \(1/\sqrt{1-y^2}\). About the \(x\)-axis, the original description gives

$$ \begin{aligned} S_x &=2\pi\int_{3/5}^{4/5}\sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}}\,dx\\ &=2\pi\left[x\right]_{3/5}^{4/5}\\ &=2\pi\left(\frac45-\frac35\right) =\frac{2\pi}{5}. \end{aligned} $$

The inverse description gives

$$ \begin{aligned} S_x &=2\pi\int_{3/5}^{4/5}\frac{y}{\sqrt{1-y^2}}\,dy\\ &=2\pi\left[-\sqrt{1-y^2}\right]_{3/5}^{4/5}\\ &=2\pi\left(-\frac35+\frac45\right) =\frac{2\pi}{5}. \end{aligned} $$

About the \(y\)-axis, the original description gives

$$ \begin{aligned} S_y &=2\pi\int_{3/5}^{4/5}\frac{x}{\sqrt{1-x^2}}\,dx\\ &=2\pi\left[-\sqrt{1-x^2}\right]_{3/5}^{4/5}\\ &=2\pi\left(-\frac35+\frac45\right) =\frac{2\pi}{5}. \end{aligned} $$

The inverse description gives

$$ \begin{aligned} S_y &=2\pi\int_{3/5}^{4/5}\sqrt{1-y^2}\frac{1}{\sqrt{1-y^2}}\,dy\\ &=2\pi\left[y\right]_{3/5}^{4/5}\\ &=\frac{2\pi}{5}. \end{aligned} $$

All four calculations give \(2\pi/5\). In this particular case, the two axis areas also happen to match; that is a feature of this example, not a general consequence of using inverse descriptions.

Worked Example: The Profile \(y=x^2\) and Its Inverse

Let \(y=x^2\) on \(1\le x\le 2\). Its inverse is \(x=\sqrt{y}\) on \(1\le y\le 4\). Both coordinates are positive, and the derivatives are \(dy/dx=2x\) and \(dx/dy=1/(2\sqrt{y})\).

For revolution about the \(y\)-axis, the original description gives

$$ \begin{aligned} S_y &=2\pi\int_1^2 x\sqrt{1+4x^2}\,dx\\ &=\frac{\pi}{4}\int_5^{17}\sqrt{u}\,du \qquad (u=1+4x^2,\;du=8x\,dx)\\ &=\frac{\pi}{4}\left[\frac{2}{3}u^{3/2}\right]_5^{17}\\ &=\frac{\pi}{6}\left(17\sqrt{17}-5\sqrt{5}\right). \end{aligned} $$

The inverse description produces the same value, as the substitution \(y=x^2\) shows directly:

$$ \begin{aligned} S_y &=2\pi\int_1^4 \sqrt{y}\sqrt{1+\frac{1}{4y}}\,dy\\ &=2\pi\int_1^2 x\frac{\sqrt{1+4x^2}}{2x}\,(2x\,dx)\\ &=2\pi\int_1^2x\sqrt{1+4x^2}\,dx\\ &=\frac{\pi}{6}\left(17\sqrt{17}-5\sqrt{5}\right). \end{aligned} $$

The \(x\)-axis integrals also match. Their common value can be written explicitly. An antiderivative is

$$ \int x^2\sqrt{1+4x^2}\,dx = \frac{2x\sqrt{1+4x^2}(8x^2+1)-\operatorname{arsinh}(2x)}{64}. $$

Differentiating the right-hand side gives \(x^2\sqrt{1+4x^2}\). Therefore the original description gives

$$ S_x =2\pi\int_1^2 x^2\sqrt{1+4x^2}\,dx =\frac{\pi}{32}\left(132\sqrt{17}-18\sqrt{5}-\operatorname{arsinh}(4)+\operatorname{arsinh}(2)\right). $$

For the inverse description, substitute \(y=x^2\), so \(dy=2x\,dx\) and \(x=\sqrt{y}\). Its integral becomes

$$ \begin{aligned} S_x &=2\pi\int_1^4 y\sqrt{1+\frac{1}{4y}}\,dy\\ &=2\pi\int_1^2 x^2\sqrt{1+4x^2}\,dx\\ &=\frac{\pi}{32}\left(132\sqrt{17}-18\sqrt{5}-\operatorname{arsinh}(4)+\operatorname{arsinh}(2)\right). \end{aligned} $$

Thus both inverse descriptions reproduce the same area about each axis.

A Common Mistake: Forgetting the Changed Differential

When changing from \(x\) to \(y\), it is not enough to replace \(f'(x)\) by the inverse derivative \(1/f'(x)\). The differential changes too. For example, take the increasing line \(y=2x+1\) from the first example. A mistaken calculation for its \(x\)-axis area might use the inverse arc-length factor but omit \(dx/dy=1/2\):

$$ 2\pi\int_3^7 y\sqrt{1+\left(\frac12\right)^2}\,dy = 2\pi\cdot\frac{\sqrt{5}}{2}\cdot 20 = 20\pi\sqrt{5}. $$

The expression \(2\pi\int_3^7 y\,(\sqrt5/2)\,dy\) is the correct inverse-description integral: \(\int_3^7 y\,dy=\frac{49-9}{2}=20\), so it evaluates to \(20\pi\sqrt5\). The factor \(\sqrt5/2\) here is the inverse arc-length factor \(\sqrt{1+(dx/dy)^2}\), and the integration variable is \(y\), so no additional \(dx/dy\) factor is needed. The extra factor appears only when this integral is converted back to an \(x\)-integral, because then \(dy=2\,dx\). The mistake to avoid is mixing the two descriptions: the inverse arc-length factor must be paired with \(dy\), and the original factor \(\sqrt5\) with \(dx\).

More directly, the inverse arc-length factor alone is not the original arc-length factor: \(\sqrt{1+(1/f')^2}\ne\sqrt{1+(f')^2}\) in general. The correct comparison multiplies by \(|f'|\), from the differential change. For this line, \((\sqrt5/2)\cdot 2=\sqrt5\), exactly the original arc-length factor. Always check the combined factor, not either factor in isolation.

Conditions and Interpretation

Positivity is part of the surface-area setup, not a cosmetic convenience. If a profile lies below the \(x\)-axis, its distance to that axis is \(|y|\), not \(y\); likewise, the distance to the \(y\)-axis is \(|x|\). The theorem above avoids signed-radius errors by requiring \(x>0\) and \(f(x)>0\). If a curve crosses an axis, one must account for absolute radii and consider whether different portions of the curve generate overlapping surfaces; the simple comparison stated here does not resolve those issues.

The proof also requires a genuine inverse with a finite, nonzero derivative on the interval. The assumption \(f'(x)\ne0\) ensures the inverse derivative formula applies throughout. Subject to these conditions, equality is a consequence of the ordinary change-of-variable rule and the geometry of arc length: the curve’s length element does not depend on whether it is recorded using \(x\) or \(y\).

Takeaway: To match surface-area integrals for inverse descriptions, use the same axis and the same geometric radius, replace the derivative by its reciprocal, and include the absolute change-of-variable factor. Together, the inverse arc-length factor and the changed differential recover the original arc-length factor.

Check Your Understanding

Use the definitions and relationships in this tutorial to answer the following questions.

  1. For a curve \(y=f(x)\) in the first quadrant, what radius belongs in the surface-area formula for revolution about the \(y\)-axis?
  2. Why does changing from \(x\) to \(y=f(x)\) contribute \(|f'(x)|\) to an integral with increasing limits?
  3. For a decreasing profile, why must the inverse integral’s limits be arranged in increasing \(y\)-order?
  4. What expression results from multiplying the inverse arc-length factor by the change-of-variable factor?
  5. Does equality of the two descriptions imply equal areas for revolution about different axes? Explain briefly.

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