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.