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 $1/2<=x<=1$ ?

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Answer 1 · 2014-11-15T16:24:16

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

Let us find the arc length element $ds$ .

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

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

$=1/2(x^2+1/x^2)dx$

The surface area $A$ can be found by

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

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

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

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

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

I hope that this was helpful.

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