Question

What is the surface area of the solid created by revolving `f(x) =ln(4-x) , x in [1,3]` around the x axis?

Answer 1

Surface area of the solid created by revolving `f(x)=ln(4-x)`, `x in[1,3]` is `2pi(2+3ln3)`

Explanation:

See the portion of the curve between `x=1` and `x=3`. Let us divide the interval `[1,3}` into say `n` equal sub-intervals each of of width `Deltax`, where `Deltax=(3-1)/n` and as `n->oo`, `Deltax->dx`.

graph{ln(4-x) [-0.99, 4.01, -0.52, 1.98]}

On each of these sub-intervals we can revolve the curve around `x`-axis to form a thin cylinder, whose widt is `dx` and radius is `r=f(x)`.

Then the surface area of this thin cylinder will be `2pirl`, where `l=dx` and `r=ln(4-x)` and adding them up will give surface area of the solid created by revolving the function `ln(4-x)` and in place of adding we can integrate it to get the surface area, which will be

`2piint_1^3ln(4-x)dx`

and as `intln(4-x)dx=-(4-x)ln(4-x)+x`

and `2piint_1^3ln(4-x)dx`

= `2pi[-(4-x)ln(4-x)+x]_1^3`

= `2pi[-ln1+3+3ln3-1]`

= `2pi(2+3ln3)`