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

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

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Answer 1 · 2014-11-15T16:24:16

$y = \frac{x^{3}}{6} + \frac{1}{2 x} \Rightarrow \frac{d y}{d x} = \frac{1}{2} (x^{2} - \frac{1}{x^{2}})$ $y={x^3}/{6}+1/{2x} => {dy}/{dx}=1/2(x^2-1/x^2)$

Let us find the arc length element $d s$ $ds$ .

$d s = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x = \sqrt{1 + \frac{1}{4} (x^{4} - 2 + \frac{1}{x^{4}})} d x$ $ds=sqrt{1+(\frac{dy}{dx})^2}dx=sqrt{1+1/4(x^4-2+1/x^4)}dx$

$= \sqrt{\frac{1}{4} (x^{4} + 2 + \frac{1}{x^{4}})} d x = \sqrt{\frac{1}{2^{2}} {(x^{2} + \frac{1}{x^{2}})}^{2}} d x$ $=sqrt{1/4(x^4+2+1/x^4)}dx=sqrt{1/2^2(x^2+1/x^2)^2}dx$

$= \frac{1}{2} (x^{2} + \frac{1}{x^{2}}) d x$ $=1/2(x^2+1/x^2)dx$

The surface area $A$ $A$ can be found by

$A = 2 π \int_{\frac{1}{2}}^{1} (\frac{x^{3}}{6} + \frac{1}{2 x}) \cdot \frac{1}{2} (x^{2} + \frac{1}{x^{2}}) d x$ $A=2pi int_{1/2}^1(x^3/6+1/{2x})cdot 1/2(x^2+1/x^2)dx$

$= \frac{π}{6} \int_{\frac{1}{2}}^{1} (x^{5} + 4 x + 3 x^{- 3}) d x$ $=pi/6 int_{1/2}^1(x^5+4x+3x^{-3})dx$

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

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

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

I hope that this was helpful.

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

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How do you find the surface area of the solid obtained by rotating about the xxx-axis the region bounded by y=x^3/6+1/(2x)y=x36+12xy=x^3/6+1/(2x) on the interval 1/2<=x<=112≤x≤11/2<=x<=1 ?

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