# How do you find the surface area of the solid obtained by rotating about the x-axis the region bounded by y=x^3/6+1/(2x) on the interval 1/2<=x<=1 ?

Nov 15, 2014

$y = \frac{{x}^{3}}{6} + \frac{1}{2 x} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left({x}^{2} - \frac{1}{x} ^ 2\right)$

Let us find the arc length element $\mathrm{ds}$.

$\mathrm{ds} = \sqrt{1 + {\left(\setminus \frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx} = \sqrt{1 + \frac{1}{4} \left({x}^{4} - 2 + \frac{1}{x} ^ 4\right)} \mathrm{dx}$

$= \sqrt{\frac{1}{4} \left({x}^{4} + 2 + \frac{1}{x} ^ 4\right)} \mathrm{dx} = \sqrt{\frac{1}{2} ^ 2 {\left({x}^{2} + \frac{1}{x} ^ 2\right)}^{2}} \mathrm{dx}$

$= \frac{1}{2} \left({x}^{2} + \frac{1}{x} ^ 2\right) \mathrm{dx}$

The surface area $A$ can be found by

$A = 2 \pi {\int}_{\frac{1}{2}}^{1} \left({x}^{3} / 6 + \frac{1}{2 x}\right) \cdot \frac{1}{2} \left({x}^{2} + \frac{1}{x} ^ 2\right) \mathrm{dx}$

$= \frac{\pi}{6} {\int}_{\frac{1}{2}}^{1} \left({x}^{5} + 4 x + 3 {x}^{- 3}\right) \mathrm{dx}$

$= \frac{\pi}{6} {\left[{x}^{6} / 6 + 2 {x}^{2} - \frac{3}{2} {x}^{- 2}\right]}_{\frac{1}{2}}^{1}$

$= \frac{\pi}{36} {\left[{x}^{6} + 12 {x}^{2} - 9 {x}^{- 2}\right]}_{\frac{1}{2}}^{1}$

$= \frac{\pi}{36} \left[1 + 12 - 9 - \left(\frac{1}{64} + 3 - 36\right)\right] = \frac{263 \pi}{256}$

I hope that this was helpful.