From an Exponential or Logarithmic Profile to Surface Area
A curve such as \(y=e^x\) or \(y=\ln x\) can describe the same profile in two different ways: one gives \(y\) as a function of \(x\), and the other gives \(x\) as a function of \(y\). When the curve is revolved about a coordinate axis, the surface-area integral must account for both the distance to the axis and the length of the curve.
The exponential and logarithmic functions are especially useful here because they are inverses. Their derivatives also give straightforward expressions for the arc-length factor. The main choices to make are which axis the profile revolves around, which coordinate is the radius, and which variable makes the integral easiest to express.
In each formula, the function value is the radius and the square-root factor accounts for the curve’s arc length. For example, if \(y=f(x)\) is revolved about the \(y\)-axis and \(x\ge 0\), the corresponding formula is \(S=2\pi\int_a^b x\sqrt{1+[f'(x)]^2}\,dx\). The radius is \(x\), not \(f(x)\), because the distance to the \(y\)-axis is measured horizontally.
If a profile lies on the opposite side of an axis, its radius is the absolute value of the relevant coordinate. The examples here use profiles on the positive side, so no absolute-value notation is needed. A profile that crosses an axis requires additional care about what surface is swept out.
Exponential Profiles Revolved About the \(x\)-Axis
For an exponential profile \(y=Ae^{kx}\), where \(A>0\) and \(k\ne 0\), the derivative is \(y'=kAe^{kx}=ky\). If the profile stays above the \(x\)-axis, the surface-area integral is therefore
Sometimes a substitution using the profile height is convenient. With \(u=Ae^{kx}\), we have \(du=ku\,dx\), so \(dx=du/(ku)\). If \(k>0\), the integral becomes
The changed bounds are the heights at the original endpoints. This substitution changes how the area is expressed; it does not change the curve or the surface.
Worked Example: Revolving \(y=e^x\) About the \(x\)-Axis
Find the surface area generated by revolving \(y=e^x\), for \(0\le x\le \ln 2\), about the \(x\)-axis.
First, differentiate the profile: \(y'=e^x\). Its height is positive, so the radius is \(e^x\). Substituting into the formula gives
Set \(u=e^x\). Then \(du=e^x\,dx\). When \(x=0\), \(u=1\); when \(x=\ln 2\), \(u=2\). Thus
Evaluating the endpoints gives
To check the integration, differentiate \(F(u)=\tfrac12\left(u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})\right)\). The product rule gives \(\frac{d}{du}(u\sqrt{1+u^2})=\sqrt{1+u^2}+u^2/\sqrt{1+u^2}\), while the logarithm has derivative \(1/\sqrt{1+u^2}\). Hence \(F'(u)=\sqrt{1+u^2}\), as required. The bounds \(1\) and \(2\) also match the profile heights at the original endpoints.
Logarithmic Profiles Revolved About the \(y\)-Axis
The logarithmic profile \(y=\ln x\), for \(x>0\), can also be written as \(x=e^y\). These are inverse descriptions of the same curve. When using \(y=\ln x\) and integrating with respect to \(x\), its derivative is \(dy/dx=1/x\). If the curve is revolved about the \(y\)-axis and \(x>0\), the radius is \(x\), giving
where the interval must lie within \(x>0\). Alternatively, using \(x=e^y\), the radius is \(e^y\), and the derivative with respect to \(y\) is also \(e^y\). That description gives \(S=2\pi\int e^y\sqrt{1+e^{2y}}\,dy\), with bounds expressed as \(y\)-values.
Worked Example: Revolving \(y=\ln x\) About the \(y\)-Axis
Find the area generated by revolving the part of \(y=\ln x\) from \(x=1\) to \(x=2\) about the \(y\)-axis.
Since \(dy/dx=1/x\) and the radius to the \(y\)-axis is \(x\), the surface-area integral is
An antiderivative of \(\sqrt{x^2+1}\) is \(\tfrac12\left(x\sqrt{x^2+1}+\ln(x+\sqrt{x^2+1})\right)\). Therefore
For a check, let \(G(x)=\tfrac12\left(x\sqrt{x^2+1}+\ln(x+\sqrt{x^2+1})\right)\). Differentiating its first term gives \(\tfrac12\sqrt{x^2+1}+\tfrac{x^2}{2\sqrt{x^2+1}}\); differentiating the logarithm contributes \(1/(2\sqrt{x^2+1})\). Their sum is \(\sqrt{x^2+1}\), the required integrand. The endpoint values \(x=1\) and \(x=2\) give respectively \(\sqrt{2}+\ln(1+\sqrt{2})\) and \(2\sqrt{5}+\ln(2+\sqrt{5})\) inside the evaluated bracket.
Why Equivalent Graph Descriptions Give Equivalent Integrals
The same increasing profile can be described either by \(y=f(x)\) or by its inverse \(x=g(y)\). Suppose \(f\) is differentiable on \([a,b]\), with \(f'(x)>0\), and let \(g=f^{-1}\). The inverse derivative rule gives \(g'(y)=1/f'(g(y))\). For revolution about the \(x\)-axis, the two descriptions give, respectively,
Set \(y=f(x)\) in the second integral. Since \(dy=f'(x)\,dx\), and \(f'(x)>0\), its integrand including the differential becomes
The transformed bounds are \(x=a\) and \(x=b\), so the two integrals are equal. This proves that changing from \(y=f(x)\) to \(x=g(y)\) changes the variable and the appearance of the formula, but not the surface area. The increasing-profile assumption keeps the bounds and the sign of \(dy\) straightforward.
Worked Example: The Same Exponential Profile in Two Variables
Use both \(y=e^x\) and \(x=\ln y\) to set up and evaluate the surface area when the profile is revolved about the \(x\)-axis, from \((0,1)\) to \((\ln 2,2)\).
Using \(y=e^x\), the derivative is \(dy/dx=e^x\), so the area is
For the inverse description \(x=\ln y\), we have \(dx/dy=1/y\). The radius to the \(x\)-axis is \(y\), so the same surface is described by
The first integral becomes the second under \(y=e^x\), because \(dy=e^x\,dx\) and the bounds change from \(x=0,\ln 2\) to \(y=1,2\). Both evaluate to
The antiderivative check is the one used in the first example: differentiating \(\tfrac12\left(y\sqrt{1+y^2}+\ln(y+\sqrt{1+y^2})\right)\) gives \(\sqrt{1+y^2}\). Thus both graph descriptions produce the same verified value.
A Second Exponential Setup
A coefficient in the exponent changes the slope and therefore the square-root factor. It must be differentiated before the formula is formed. For instance, for \(y=3e^{2x}\), the derivative is \(6e^{2x}\), not \(3e^{2x}\). The following example shows how to retain that factor through a change of variable.
Worked Example: Revolving \(y=3e^{2x}\) About the \(x\)-Axis
Take \(0\le x\le \tfrac12\ln 2\). The profile height increases from \(3\) to \(6\), and \(y'=2y\). The area integral is
Set \(u=3e^{2x}\). Then \(du=6e^{2x}\,dx=2u\,dx\), so \(dx=du/(2u)\). The bounds become \(u=3\) and \(u=6\). Hence
An antiderivative is \(H(u)=\tfrac{u}{2}\sqrt{1+4u^2}+\tfrac14\ln(2u+\sqrt{1+4u^2})\). Its derivative is \(\sqrt{1+4u^2}\): differentiating the first term gives \(\tfrac12\sqrt{1+4u^2}+2u^2/\sqrt{1+4u^2}\), and differentiating the logarithmic term gives \(1/(2\sqrt{1+4u^2})\); together these simplify to \(\sqrt{1+4u^2}\). Therefore the checked area is
At \(u=6\), the radical is \(\sqrt{145}\); at \(u=3\), it is \(\sqrt{37}\). Substitution gives the exact value $$ S=\pi\left(3\sqrt{145}+\frac14\ln(12+\sqrt{145}) -\frac32\sqrt{37}-\frac14\ln(6+\sqrt{37})\right). $$ The changed bounds agree with the original endpoint heights, confirming that the substitution covers the intended profile segment.
Common Setup Errors
A frequent mistake is to use the profile’s height as the radius even when the axis is vertical. For \(y=\ln x\) revolved about the \(y\)-axis, the radius is \(x\), not \(\ln x\). The incorrect setup would be
This uses the distance to the \(x\)-axis, rather than the distance to the \(y\)-axis. The correction is to use \(x\) as the radius:
Also check the derivative in the arc-length factor before simplifying. In \(y=3e^{2x}\), the derivative is \(6e^{2x}\), so omitting the factor \(2\) would give the wrong integrand. The axis determines the radius; the chosen graph description determines which derivative and differential appear.
Choosing the Integration Variable
Either graph description is valid when the profile is invertible on the interval of interest. A useful choice is the one that makes the radius, derivative, bounds, or substitution easiest to handle. For exponential and logarithmic profiles, the inverse relation often turns an awkward-looking integral into a simpler one, but it does not remove the need to change the bounds consistently.
| Description and axis | Radius | Arc-length factor |
|---|---|---|
| \(y=f(x)\), about the \(x\)-axis | \(y=f(x)\) | \(\sqrt{1+[f'(x)]^2}\,dx\) |
| \(x=g(y)\), about the \(x\)-axis | \(y\) | \(\sqrt{1+[g'(y)]^2}\,dy\) |
| \(y=f(x)\), about the \(y\)-axis | \(x\) | \(\sqrt{1+[f'(x)]^2}\,dx\) |
| \(x=g(y)\), about the \(y\)-axis | \(x=g(y)\) | \(\sqrt{1+[g'(y)]^2}\,dy\) |
In every row, multiply \(2\pi\) times the radius by the arc-length factor and integrate over the interval belonging to the displayed variable. This keeps the geometry, derivative, and bounds consistent.
Check Your Understanding
For each question, identify the radius and the correct integration variable before evaluating or simplifying.
- Write the surface-area integral for \(y=e^{2x}\), \(0\le x\le 1\), revolved about the \(x\)-axis.
- For \(y=\ln x\), \(1\le x\le e\), revolved about the \(y\)-axis, what is the radius and what is \(dy/dx\)?
- Rewrite the profile \(y=e^x\) as \(x\) in terms of \(y\), and state the corresponding derivative.
- Explain why the bounds in the \(y\)-variable for \(y=e^x\), \(0\le x\le\ln 2\), are \(1\) and \(2\).
- Identify the error in using \(\ln x\) as the radius when \(y=\ln x\) is revolved about the \(y\)-axis.