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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=x318y=\frac{x^{3}}{18}, 3x63 \leq x \leq 6


Answered 2 years ago
Answered 2 years ago
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To solve we will use the surface of a revolution formula

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

We will use r(x)=x318r(x)=\frac{x^3}{18} because it is the distance from the xx axis, the radius. When we differentiate f(x)=x318f(x)=\frac{x^3}{18} we obtain f(x)=x26f'(x)=\frac{x^2}{6}.

S=2πx3181+x436dxlet u=1+x436=2πx318u9x3du=πudu=2π3u32=2π3(1+x436)32\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

S=2π3[(1+x436)32]36=2π3(3732(134)32)459.068\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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