How do you find the area of the surface generated by rotating the curve about the y-axis #y=1/4x^4+1/8x^-2, 1<=x<=2#?

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Answer 1 · 2017-04-13T20:04:17

#(253pi)/(20)#

#dy/dx=x^3-x^(-3)/4#

So, the arc length element is:

#sqrt{1+(dy/dx)^2}=sqrt(1+(x^3)^2-1/2+(x^(-3)/4)^2) =sqrt((x^3)^2+1/2+(x^(-3)/4)^2)=sqrt((x^3+x^(-3)/4)^2) =x^3+x^(-3)/4#

Hence, the surface area can be expressed as:

#S=2pi int_1^2 x sqrt(1+(dy/dx)^2) dx =2pi int_1^2(x^4+x^(-2)/4) dx#

#=2pi[x^5/5-1/(4x)]_1^2=(253pi)/(20)#