Applications Surface Area II

Problem 241: Find the area of the surface generated by revolving the given curve about the $y$ -axis.

(1) $\begin{equation*} x = \sqrt{36-y^2}, \qquad -2 \le y \le 2 \end{equation*}$

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

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

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

(3) $\begin{align*} f(x) = \sqrt{36-y^2} \implies f'(y) = \frac{1}{2}(36-y^2)^{-1/2} \cdot 2y = \frac{y}{\sqrt{36-y^2}}. \end{align*}$

We have

(4) $\begin{align*} \text{SA} & = 2\pi \int_{a}^{b} R(x) \sqrt{1 + \left[f'(x) \right]^2}\ dy \\ & = \text{SA} = 2\pi \int_{-2}^{2} \left( \sqrt{36-y^2}\right)\sqrt{1 + \left( \frac{y}{\sqrt{36-y^2}}\right)^2}\ dy \\ & = 2 \pi \int_{-2}^{2}\sqrt{36-y^2} \frac{\sqrt{36-y^2+y^2}}{\sqrt{36-y^2}} \ dy \\ & = 2\pi \int_{-2}^{2} \sqrt{36} \ dy \\ & = 12\pi \int_{-2}^{2} dy \\ & = 12 \pi (y)_{-2}^{2} \\ & = 48 \pi.\end{align*}$

That is,

(5) $\begin{equation*} \text{SA} = 48\pi. \end{equation*}$

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