# What is the surface area produced by rotating f(x)=x^2lnx, x in [0,3] around the x-axis?

Oct 20, 2016

$\approx 311.4$

#### Explanation:

If we consider a small strip width $\mathrm{dx}$, it will have radius $y \left(x\right)$ as it is revolved about the x axis, and thus circuference $2 \pi y$.

The arc length $\mathrm{ds}$ of the tip of strip $\mathrm{dx}$ is:
$\mathrm{ds} = \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$

With $y ' = x \left(1 + 2 \ln x\right)$#

and so the surface area of the element is

$\mathrm{dS} = 2 \pi y \mathrm{ds}$

$= 2 \pi {x}^{2} \ln x \sqrt{1 + {\left(x \left(1 + 2 \ln x\right)\right)}^{2}} \mathrm{dx}$

For $x \in \left[1 , 3\right]$, the surface area $S$ is therefore:

$S = 2 \pi {\int}_{1}^{3} \setminus {x}^{2} \ln x \sqrt{1 + {\left(x \left(1 + 2 \ln x\right)\right)}^{2}} \mathrm{dx}$

However because $y < 0$ for $x \in \left[1 , 3\right]$, which would generate a negative radius, we need to be sure to place a negative number on the integration.

The surface area in total is therefore

$S = 2 \pi \left({\int}_{1}^{3} \setminus {x}^{2} \ln x \sqrt{1 + {\left(x \left(1 + 2 \ln x\right)\right)}^{2}} \mathrm{dx} - {\int}_{0}^{1} \setminus {x}^{2} \ln x \sqrt{1 + {\left(x \left(1 + 2 \ln x\right)\right)}^{2}} \mathrm{dx}\right)$