Question

What is the surface area of the solid created by revolving `f(x) = e^(x)/2 , x in [2,7]` around the x axis?

Answer 1

The area is `=pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))`

Explanation:

The volume of a small disc is `dS=piydr`
and `dr=sqrt((dx)^2+(dy)^2)`
`dS=piysqrt((dx)^2+(dy)^2)=piysqrt(1+(dy/dx)^2)dx`
So `y=e^x/2` `=>` `dy/dx=e^x/2`
so `dS=pie^x/2*sqrt(1+e^(2x)/4)dx`
`dS=pie^x/4*sqrt(4+e^(2x))dx`
so `S=pi/4inte^xsqrt(4+e^(2x))dx`
let `u=e^x` `=>` `du=e^xdx`
`S=pi/4*intsqrt(4+u^2)du`
`u=2tanv` `=>` `du= sec^2u (dv)`
and `4tan^2v+4=4sec^2v`
`S=4pi/4intsec^3vdv=pi(1/2intsecvdv+secvtanv/2)`
`pi((ln(tanv+secv)/2)+secvtanv/2)`
`S=pi/8(4(ln(e^(2x)+4)+e^x)+e^xsqrt(e^(2x)+4)))_2^7`
`=pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))`