Finding the surface area of a region which is generated by revolving a curve around a line

Question

The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is early on in the book so I would expect / hope any integral would be easy to solve.

Problem:
Find the area of the surface generated by revolving the following curve about the line $y = -1$. The curve is $y = \frac{x^3}{3} + \frac{1}{4x}$ for $1 \leq x \leq 3$.

Answer:

Since we are revolving the curve about $y = -1$, I augment the function by adding $1$ to it and treating it as revolving it around $y = 0$. The format of the integral for surface area revolved around the y-axis is: $$ S = \int_a^b 2\pi x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \,\, dx $$ Now we need to find the bounds on $y$. \begin{align*} y(1) &= \frac{1^3}{3} + \frac{1}{4} = \frac{1}{3} + \frac{1}{4} \\ y &= \frac{7}{12} \\ y(3) &= \frac{3^3}{3} + \frac{1}{4(3)} = 9 + \frac{1}{12} \\ y(3) &= \frac{109}{12} \\ y &= \frac{x^3}{3} + \frac{x^{-1}}{4} \\ \frac{dy}{dx} &= x^2 - \frac{x^{-2}}{4} \\ \frac{dy}{dx} &=\frac{4x^2 - x^{-2}}{4} \\ \frac{dx}{dy} &= \frac{4}{4x^2 - x^{-2}} \\ S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{1 + \left( \frac{4}{4x^2 - x^{-2}} \right)^2} \,\, dx \end{align*} Now we need to integrate. \begin{align*} S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{1 + \frac{16}{\left(4x^2 - x^{-2}\right)^2 }} \,\, dx \\ S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{ \frac{\left(4x^2 - x^{-2}\right)^2 + 16 }{\left(4x^2 - x^{-2}\right)^2 }} \,\, dx \\ \end{align*} This does not seem right to me.

Based upon the comments from the group, I updated my solution.

Since we are revolving the curve about $y = -1$, I augment the function by adding $1$ to it and treating it as revolving it around $y = 0$. Let $S$ be the surface area we are trying to find. \begin{align*} y &= \frac{x^3}{3} + \frac{x^{-1}}{4} \\ y' &= x^2 - \frac{x^{-2}}{4} \\ S &= \int_1^3 2 \pi \left(y+1 \right) \sqrt{1 + \left( x^2 - \frac{x^{-2}}{4} \right) ^2 } \,\, dx \\ S &= \int_1^3 2 \pi \left(\frac{x^3}{3} + \frac{x^{-1}}{4}+1 \right) \sqrt{1 + \left( x^2 - \frac{x^{-2}}{4} \right) ^2 } \,\, dx \\ S &= \int_1^3 2 \pi \left(\frac{x^3}{3} + \frac{x^{-1}}{4}+1 \right) \sqrt{ \frac{16x^4 + 8 + x^{-4}}{16} } \,\, dx \\ \end{align*}

This does not seem right to me. How do I complete this integration?

Answer 1

Consider a scalar function $f(x)$: $R$ $\rightarrow$ $R$. Then the surface area of the solid formed by revoling $f(x)$ about $y=0$ is S=$2\pi$$\int_a^b$$f(x)ds$, where $ds$ represents an infintesimal arclength element of the curve $f(x)$.

To calculate the surface area of the solid formed by revolving $f(x)$ about y=$-1$, you must add 1 to the integrand. So we have that S=$2\pi$$\int_a^b$$(f(x)+1)ds$= $2\pi$$\int_a^b$$g(x)$ds, where $g(x)$=$f(x)+1$=${x^3\over 3}$+${1\over 4x}$+$1$.

Note that $ds$=$\sqrt{1+({dg\over dx})^2}$$dx$=$\sqrt{{1}+(x^2-{{1\over4x^2}}})^2$$dx$.

Now substitute a=1, b=3, and the appropriate values for $g(x)$ and $ds$ into the integral S=$2\pi$$\int_1^3$$g(x)$ds. Calculating it will yield the desired surface area.

Answer 2

The integral formula for $S$ In the post is questionable. It is good for surfaces revolving around $x$, not $y$,

The following expression should be used, instead, to integrate surfaces around $y$, $$2\pi \int_1^3 (y+1)\sqrt{1 + \left( \frac{dy}{dx} \right)^2} \,\, dx $$

Otherwise, the integrand goes to infinity.