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

Mar 23, 2018

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

Explanation:

See the portion of the curve between $x = 1$ and $x = 3$. Let us divide the interval $\left[1 , 3\right\}$ into say $n$ equal sub-intervals each of of width $\Delta x$, where $\Delta x = \frac{3 - 1}{n}$ and as $n \to \infty$, $\Delta x \to \mathrm{dx}$.

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 $\mathrm{dx}$ and radius is $r = f \left(x\right)$.

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

$2 \pi {\int}_{1}^{3} \ln \left(4 - x\right) \mathrm{dx}$

and as $\int \ln \left(4 - x\right) \mathrm{dx} = - \left(4 - x\right) \ln \left(4 - x\right) + x$

and $2 \pi {\int}_{1}^{3} \ln \left(4 - x\right) \mathrm{dx}$

= $2 \pi {\left[- \left(4 - x\right) \ln \left(4 - x\right) + x\right]}_{1}^{3}$

= $2 \pi \left[- \ln 1 + 3 + 3 \ln 3 - 1\right]$

= $2 \pi \left(2 + 3 \ln 3\right)$