`y = x^3/6 + 1/(2x) , 1<=x<=2` Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

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Area of the surface obtained by revolving the curve `y=f(x)` about `x`-axis between `a leq x leq b` is given by 

`S_x=2pi int_a^b y sqrt(1+y'^2)dx`

Let us therefore, first find `y'.`

`y'=x^2/2-1/(2x^2)=(x^4-1)/(2x^2)`

`y'^2=(x^8-2x^4+1)/(4x^4)`

We can now calculate the surface area.

`S_x=2pi int_1^2 (x^3/6+1/(2x))sqrt(1+(x^8-2x^4+1)/(4x^4))dx=`

` `  `2pi int_1^2(x^3/6+1/(2x))sqrt((x^8+2x^4+1)/(4x^4))=`

`2pi int_1^2(x^3/6+1/(2x))sqrt(((x^4+1)/(2x^2))^2)dx=`

`2pi int_1^2(x^3/6+1/(2x))(x^4+1)/(2x^2)dx=`

Multiplying the terms...

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Area of the surface obtained by revolving the curve `y=f(x)` about `x`-axis between `a leq x leq b` is given by 

`S_x=2pi int_a^b y sqrt(1+y'^2)dx`

Let us therefore, first find `y'.`

`y'=x^2/2-1/(2x^2)=(x^4-1)/(2x^2)`

`y'^2=(x^8-2x^4+1)/(4x^4)`

We can now calculate the surface area.

`S_x=2pi int_1^2 (x^3/6+1/(2x))sqrt(1+(x^8-2x^4+1)/(4x^4))dx=`

` `  `2pi int_1^2(x^3/6+1/(2x))sqrt((x^8+2x^4+1)/(4x^4))=`

`2pi int_1^2(x^3/6+1/(2x))sqrt(((x^4+1)/(2x^2))^2)dx=`

`2pi int_1^2(x^3/6+1/(2x))(x^4+1)/(2x^2)dx=`

Multiplying the terms under integral yields

`2pi int_1^2 (x^5/12+x/12+x/4+1/(4x^3))dx=`

`2pi int_1^2(x^5/12+x/3+1/(4x^3))dx=`

`2pi (x^6/72+x^2/6-1/(8x^2))|_1^2=`

` ` `2pi(64/72+2/3-1/32-1/72-1/6+1/8)=2pi cdot 47/32=(47pi)/16`

` `The area of surface generated by revolving the given curve about `x`-axis is `(47pi)/16.`    

Graphs of the curve and the surface can be seen in the images below.

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