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# Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the x-axis. $y=\frac{x^{3}}{18}$, $3 \leq x \leq 6$

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To solve we will use the surface of a revolution formula

$S=2 \pi \int_{a}^{b} r(x) \sqrt{1+ [f'(x)]^2}\,dx$

We will use $r(x)=\frac{x^3}{18}$ because it is the distance from the $x$ axis, the radius. When we differentiate $f(x)=\frac{x^3}{18}$ we obtain $f'(x)=\frac{x^2}{6}$.

\begin{align*} S &= 2\pi \int \frac{x^3}{18} \sqrt{1+\frac{x^4}{36}}\,dx \quad \quad \text{let } u=1+\frac{x^4}{36}\\ &= 2 \pi \int \frac{x^3}{18} \sqrt{u} \frac{9}{x^3}\,du\\ &= \pi \int \sqrt{u}\,du\\ &= \frac{2 \pi}{3} u^\frac{3}{2}\\ &= \frac{2 \pi}{3} \left(1+\frac{x^4}{36} \right)^\frac{3}{2} \end{align*}

Now we can plug in the limits of integration

\begin{align*} S &= \frac{2 \pi}{3} \left[\left(1+\frac{x^4}{36} \right)^\frac{3}{2} \right]_{3}^{6}\\ &= \frac{2 \pi}{3} \left(37^\frac{3}{2} - \left(\frac{13}{4} \right)^\frac{3}{2} \right)\\ & \approx 459.068 \end{align*}

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