Paraboloid Surface Area From Power Profiles

From a Power Profile to a Surface

A power profile gives a family of curves whose slopes and distances from the axis can be described directly. When such a curve is revolved about the $x$-axis, its surface area depends on both its height above the axis and how quickly that height changes. The quadratic profile is especially important: its revolution is a paraboloid.

Let $y=kx^p$, where $k>0$, on an interval $0\leq a\leq x\leq b$, with $p>0$. Revolving the graph about the $x$-axis produces a surface with radius $y$ at each $x$. The surface area considered here is the curved area generated by the profile; it does not include any end caps.

Definition: For a nonnegative differentiable profile $y=f(x)$ revolved about the $x$-axis on $[a,b]$, its curved surface area is $$ S=2\pi\int_a^b f(x)\sqrt{1+\bigl(f'(x)\bigr)^2}\,dx, $$ provided the integral exists. If the derivative is unbounded at an endpoint, the integral is interpreted as an improper integral.

The factor $2\pi f(x)$ is the circumference of the circular band at height $f(x)$. The factor $\sqrt{1+(f'(x))^2}\,dx$ is the arc length of a small piece of the profile. Thus each band contributes approximately its circumference multiplied by its slanted width. Integrating adds the contributions along the curve.

The General Power-Profile Setup

For $y=kx^p$, differentiation gives $y'=kp\,x^{p-1}$ for $x>0$. Substituting the profile and its derivative into the surface-area formula gives

S=2\pi k\int_a^b x^p\sqrt{1+k^2p^2x^{2p-2}}\,dx.

This is the standard setup for a power profile. It applies directly on intervals with $a>0$. When the interval includes $x=0$, the endpoint must be checked: for $p>0$, the area integral is finite there, although the slope is unbounded when $0<p<1$. In that case, the integral is understood as a limit from positive $x$.

For example, when $0<p<1$, near zero the square-root factor behaves like a constant times $x^{p-1}$, so the full integrand behaves like a constant times $x^{2p-1}$. The integral of this power is finite at zero because $2p-1>-1$, which is equivalent to $p>0$.

Why the Surface-Area Formula Has These Factors

The circumference factor records the radius of each circular band, while the square root corrects for the fact that the profile is slanted rather than horizontal. The arc length of a small profile segment is $ds=\sqrt{dx^2+dy^2}$. Since $dy=y'\,dx$, this becomes $ds=\sqrt{1+(y')^2}\,dx$. A band of slant width $ds$ and circumference $2\pi y$ has area approximately $2\pi y\,ds$. Adding these contributions along the interval gives the integral.

Find the radius.
At position $x$, the radius to the $x$-axis is $y=f(x)$.

Find the slant-width factor.
The profile contributes arc length $ds=\sqrt{1+(f'(x))^2}\,dx$.

Add the band areas.
Multiply circumference by slant width and integrate: $S=2\pi\int f(x)\sqrt{1+(f'(x))^2}\,dx$.

Worked Example: A Paraboloid from a Quadratic Profile

Worked Example: Surface Area of a Paraboloid

Revolve $y=\frac12x^2$ for $0\leq x\leq2$ about the $x$-axis. Find the curved surface area.

First differentiate the profile:

y=\frac12x^2, \qquad y'=x.

Substitute both expressions into the surface-area formula:

S=2\pi\int_0^2 \frac12x^2\sqrt{1+x^2}\,dx =\pi\int_0^2 x^2\sqrt{1+x^2}\,dx.

An antiderivative is

F(x)=\frac18\left[x\sqrt{1+x^2}(2x^2+1)-\operatorname{asinh}(x)\right],

where $\operatorname{asinh}(x)=\ln\!\left(x+\sqrt{1+x^2}\right)$ for $x\geq0$. Therefore,

\begin{aligned} S &=\pi\bigl(F(2)-F(0)\bigr)\\ &=\frac{\pi}{8}\left[2\sqrt5(9)-\ln(2+\sqrt5)\right]\\ &=\frac{\pi}{8}\left[18\sqrt5-\ln(2+\sqrt5)\right]. \end{aligned}

To check the antiderivative, differentiate its bracket. The derivative of $x\sqrt{1+x^2}(2x^2+1)$, after putting its terms over $\sqrt{1+x^2}$, is $(8x^4+8x^2+1)/\sqrt{1+x^2}$. Subtracting the derivative $1/\sqrt{1+x^2}$ of $\operatorname{asinh}(x)$ leaves $8x^2\sqrt{1+x^2}$. Dividing by $8$ gives $F'(x)=x^2\sqrt{1+x^2}$, as required. Also, $F(0)=0$, so the lower endpoint contributes zero.

When Does the Integral Have an Elementary Antiderivative?

The paraboloid integral has an elementary antiderivative, but a general power profile need not. Two values are immediate: if $p=1$, the profile is a straight line and the square-root factor is constant; if $p=2$, substitution reduces the integral to a standard form involving a square root of a quadratic expression.

For other rational powers, the relevant general result is Chebyshev’s theorem for binomial differentials. In the form needed here, if $m,n,q$ are rational numbers, $n\ne0$, and $A$ and $B$ are nonzero constants, then $\int x^m(A+Bx^n)^q\,dx$ has an elementary antiderivative if and only if at least one of $q$, $(m+1)/n$, and $q+(m+1)/n$ is an integer. For the power-profile area integral with rational $p\ne1$, the corresponding values are $m=p$, $n=2p-2$, and $q=\frac12$. Thus the test is whether at least one of

\frac12, \qquad \frac{p+1}{2p-2}, \qquad \frac12+\frac{p+1}{2p-2}

is an integer. The first value is not an integer, but either of the other two can be. This criterion is for rational $p$ with $p\ne1$; $p=1$ is handled separately. It does not classify irrational exponents.

Takeaway: Every power profile has a surface-area integral, but not every such integral has an elementary antiderivative. For rational $p\ne1$, Chebyshev’s theorem gives an exact test; otherwise, the integral may need to be left in integral form or evaluated numerically.

Worked Example: A Fractional Power with a Finite Tip Area

Worked Example: A Square-Root Profile

Revolve $y=2\sqrt{x}$ on $0\leq x\leq4$ about the $x$-axis. The derivative is unbounded at zero, so evaluate the surface-area integral as an improper integral.

For $x>0$, $y'=x^{-1/2}$. The integrand simplifies as follows:

y\sqrt{1+(y')^2} =2\sqrt{x}\sqrt{1+\frac1x} =2\sqrt{x}\sqrt{\frac{x+1}{x}} =2\sqrt{x+1}.

Thus the area is the limit of the integrals from $\varepsilon$ to $4$:

\begin{aligned} S &=\lim_{\varepsilon\to0^+}4\pi\int_\varepsilon^4\sqrt{x+1}\,dx\\ &=\lim_{\varepsilon\to0^+}\frac{8\pi}{3} \left[(x+1)^{3/2}\right]_\varepsilon^4\\ &=\frac{8\pi}{3}(5\sqrt5-1). \end{aligned}

The area is finite even though the slope is unbounded at the tip. To check the integration, differentiating $\frac23(x+1)^{3/2}$ gives $\sqrt{x+1}$; multiplying by $4\pi$ recovers the area integrand. Taking $\varepsilon\to0^+$ gives the stated endpoint value.

Worked Example: A Linear Power Profile

Worked Example: A Cone from a Straight Profile

Revolve $y=2x$ on $0\leq x\leq3$ about the $x$-axis. Find its curved surface area.

Here $y'=2$, so the surface-area integral is

\begin{aligned} S &=2\pi\int_0^3 2x\sqrt{1+2^2}\,dx\\ &=4\pi\sqrt5\int_0^3x\,dx\\ &=4\pi\sqrt5\left[\frac{x^2}{2}\right]_0^3\\ &=18\pi\sqrt5. \end{aligned}

For a check, the antiderivative $2\pi\sqrt5\,x^2$ differentiates to $4\pi\sqrt5\,x$, the integrand. Evaluating it at $3$ and $0$ gives $18\pi\sqrt5$. This is the lateral surface area of the cone, not its circular base.

A Common Setup Error

A frequent mistake is to keep the arc-length factor but forget the radius factor $y$. For $y=2x$, this incorrect setup would give

2\pi\int_0^3\sqrt{1+2^2}\,dx =6\pi\sqrt5.

This is not the surface area: it treats each band as though its radius were constant and equal to $1$. The radius actually changes with $x$, so the circumference is $2\pi(2x)$. Including it gives the correct integral $2\pi\int_0^3(2x)\sqrt5\,dx=18\pi\sqrt5$.

Interpreting the Result

The power $p$ affects both the radius $kx^p$ and the slope $kp\,x^{p-1}$, so it affects the surface area in two ways. The radius determines the circumference of each band; the slope determines how much slanted profile length the band covers. This is why the surface-area integral is not simply an integral of the profile height.

The paraboloid case is useful because it gives a familiar surface from a simple power profile and produces an elementary antiderivative. Other powers may also be elementary, but the general formula is a setup, not a promise that the integral can be expressed using elementary functions. If no elementary antiderivative is available, the definite integral still precisely describes the area and can be approximated using numerical integration.

Check Your Understanding

Use the radius and arc-length factors to reason through these questions.

Write the surface-area integral for $y=3x^2$ on $0\leq x\leq R$, revolved about the $x$-axis.
For $y=kx^p$, what are the radius and the derivative that enter the surface-area formula?
Why can a profile with an unbounded slope at an endpoint still generate a finite surface area?
State the three integer tests in Chebyshev’s criterion for a rational power-profile integral with $p\ne1$.
What factor is omitted in the incorrect setup $2\pi\int\sqrt{1+(y')^2}\,dx$?