2.4 Arc length of a curve and surface area  (Page 4/8)

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the $x\text{-axis}\text{.}$ A representative band is shown in the following figure.

Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select $x_i^{*}\in \left[x_{i-1},x_i\right]$ such that $f^{\prime }(x_i^{*})=\left(\text{Δ}{y}_i\right)\text{/}\text{Δ}x.$ This gives us

Furthermore, since $f(x)$ is continuous, by the Intermediate Value Theorem, there is a point $x_i^{**}\in \left[x_{i-1},x_i\right]$ such that $f(x_i^{**})=\left(1\text{/}2\right)\left[f(x_{i-1})+f(x_i)\right],$ so we get

Then the approximate surface area of the whole surface of revolution is given by

This almost looks like a Riemann sum, except we have functions evaluated at two different points, $x_i^{*}$ and $x_i^{**},$ over the interval $\left[x_{i-1},x_i\right].$ Although we do not examine the details here, it turns out that because $f(x)$ is smooth, if we let $n\to \infty ,$ the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both $x_i^{*}$ and $x_i^{**}$ are in the interval $\left[x_{i-1},x_i\right],$ so it makes sense that as $n\to \infty ,$ both $x_i^{*}$ and $x_i^{**}$ approach $x.$ Those of you who are interested in the details should consult an advanced calculus text.

Taking the limit as $n\to \infty ,$ we get

As with arc length, we can conduct a similar development for functions of $y$ to get a formula for the surface area of surfaces of revolution about the $y\text{-axis}.$ These findings are summarized in the following theorem.

Calculating the surface area of a surface of revolution 1

Let $f(x)=\sqrt{x}$ over the interval $\left[1,4\right].$ Find the surface area of the surface generated by revolving the graph of $f(x)$ around the $x\text{-axis}.$ Round the answer to three decimal places.

The graph of $f(x)$ and the surface of rotation are shown in the following figure.

We have $f(x)=\sqrt{x}.$ Then, $f^{\prime }(x)=1\text{/}\left(2\sqrt{x}\right)$ and ${\left(f^{\prime }(x)\right)}^2=1\text{/}\left(4x\right).$ Then,

$\begin{array}{cc}\hfill \text{Surface Area}& ={\displaystyle {\int }_a^b\left(2\pi f(x)\sqrt{1+{\left(f^{\prime }(x)\right)}^2}\right)dx}\hfill \\ & ={\displaystyle {\int }_1^4\left(2\pi \sqrt{x}\sqrt{1+\frac{1}{4x}}\right)}dx\hfill \\ & ={\displaystyle {\int }_1^4\left(2\pi \sqrt{x+\frac{1}{4}}\right)dx}.\hfill \end{array}$

Let $u=x+1\text{/}4.$ Then, $du=dx.$ When $x=1,$ $u=5\text{/}4,$ and when $x=4,$ $u=17\text{/}4.$ This gives us

$\begin{array}{cc}\hfill {\displaystyle {\int }_0^1\left(2\pi \sqrt{x+\frac{1}{4}}\right)dx}& ={\displaystyle {\int }_{5\text{/}4}^{17\text{/}4}2\pi \sqrt{u}}du\hfill \\ & =2\pi {\left[\frac{2}{3}{u}^{3\text{/}2}\right]|}_{5\text{/}4}^{17\text{/}4}=\frac{\pi }{6}\left[17\sqrt{17}-5\sqrt{5}\right]\approx \mathrm{30.846.}\hfill \end{array}$

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