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...
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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