Applications Surface Area

Problem 240: Find the area of the surface generated by revolving the given curve about the $x$ -axis.

(1) $\begin{equation*} y = 4x, \qquad 0 \le x \le 3 \end{equation*}$

Solution: The surface area by revolving around the $x$ -axis is given by

(2) $\begin{equation*} \text{SA} = 2\pi \int_{a}^{b} R(x) \sqrt{1 + \left[f'(x) \right]^2}\ dx,\end{equation*}$

where $R(x)$ is the distance between the $x$ axis and the given curve, and $a$ and $b$ are the limits of integration. Let start by finding the derivative. That is,

(3) $\begin{align*} f(x) = 4x \implies f'(x) = 4. \end{align*}$

We have

(4) $\begin{align*} \text{SA} & = 2\pi \int_{a}^{b} R(x) \sqrt{1 + \left[f'(x) \right]^2}\ dx \\ & = \text{SA} = 2\pi \int_{0}^{3} (4x-0) \sqrt{1 + \left(4 \right)^2}\ dx \\ & = 2 \pi \sqrt{17} \int_{0}^{3} x \ dx \\ & = 8\sqrt{17} \pi \left(\frac{x^2}{2} \right)_{0}^{3} \\ & = 36\sqrt{17}\pi.\end{align*}$

That is,

(5) $\begin{equation*} \text{SA} = 36\sqrt{17} \pi. \end{equation*}$

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