# What is the volume of the solid produced by revolving f(x)=e^x-xlnx, x in [1,5] around the x-axis?

Jul 18, 2018

The volume is $29486.554$ ${\text{u}}^{3}$, or the sort of exact result is:

$V = \frac{\pi}{2} \left({e}^{10} - {e}^{2}\right) - 1696.084 \pi + \frac{125 \pi}{3} \left[{\ln}^{2} \left(5\right) - \frac{2}{3} \ln 5 + \frac{2}{9}\right] - \frac{2 \pi}{27}$ ${\text{u}}^{3}$

To do these kinds of solid of revolution problems around the $x$ axis, just remember that you are projecting a circle whose radius varies as $r = f \left(x\right)$.

The volume is then given by:

$V = \pi {\int}_{a}^{b} {f}^{2} \left(x\right) \mathrm{dx}$

The $f \left(x\right)$ serves as the varying radius, and the integral over $\left[a , b\right]$ gives the length of the object, sliced up into discs of thickness $\mathrm{dx}$.

In this case, we have:

$V = \pi {\int}_{1}^{5} {\left({e}^{x} - x \ln x\right)}^{2} \mathrm{dx}$

$= \pi {\int}_{1}^{5} {e}^{2 x} - 2 x {e}^{x} \ln x + {x}^{2} {\ln}^{2} x \mathrm{dx}$

= pi overbrace(int_(1)^(5) e^(2x)dx)^("Integral 1") - 2pi overbrace(int_(1)^(5) xe^xlnxdx)^"Integral 2" + pi overbrace(int_(1)^(5) x^2ln^2xdx)^"Integral 3"

• The first integral is $\frac{1}{2} {e}^{2 x}$ evaluated from $x = 1$ to $5$, which is $\frac{1}{2} \left({e}^{10} - {e}^{2}\right)$.
• The second integral is not possible with elementary functions, and the numerical solution is $848.042$.
• The third integral is hard but doable here.

int x^2ln^2xdx = ?

Let:

• $u = {\ln}^{2} x$
• $\mathrm{dv} = {x}^{2} \mathrm{dx}$
• $\mathrm{du} = \frac{2 \ln x}{x} \mathrm{dx}$
• $v = {x}^{3} / 3$

Thus,

$\int {x}^{2} {\ln}^{2} x \mathrm{dx}$

$= \frac{{x}^{3} {\ln}^{2} x}{3} - \int {x}^{3} / 3 \frac{2 \ln x}{x} \mathrm{dx}$

$= \frac{1}{3} {x}^{3} {\ln}^{2} x - \frac{2}{3} \int {x}^{2} \ln x \mathrm{dx}$

Repeat with:

• $u = \ln x$
• $\mathrm{dv} = {x}^{2} \mathrm{dx}$
• $\mathrm{du} = \frac{1}{x} \mathrm{dx}$
• $v = {x}^{3} / 3$

And we get:

$\implies \frac{1}{3} {x}^{3} {\ln}^{2} x - \frac{2}{3} \left[\frac{1}{3} {x}^{3} \ln x - \int {x}^{3} / 3 \frac{1}{x} \mathrm{dx}\right]$

$= \frac{1}{3} {x}^{3} {\ln}^{2} x - \frac{2}{3} \left[\frac{1}{3} {x}^{3} \ln x - \frac{1}{3} \int {x}^{2} \mathrm{dx}\right]$

$= \frac{1}{3} {x}^{3} {\ln}^{2} x - \left[\frac{2}{9} {x}^{3} \ln x - \frac{2}{9} \int {x}^{2} \mathrm{dx}\right]$

$= \frac{1}{3} {x}^{3} {\ln}^{2} x - \frac{2}{9} {x}^{3} \ln x + \frac{2}{9} \int {x}^{2} \mathrm{dx}$

= {:[1/3 x^3ln^2x - 2/9x^3lnx + 2/9 x^3/3]|:}_(1)^(5)

$= \left[\frac{1}{3} {\left(5\right)}^{3} {\ln}^{2} \left(5\right) - \frac{2}{9} {\left(5\right)}^{3} \ln \left(5\right) + \frac{2}{9} {\left(5\right)}^{3} / 3\right] - \left[{\cancel{\frac{1}{3} {\left(1\right)}^{3} {\ln}^{2} \left(1\right)}}^{0} - {\cancel{\frac{2}{9} {\left(1\right)}^{3} \ln \left(1\right)}}^{0} + \frac{2}{9} {\left(1\right)}^{3} / 3\right]$

$= \left[\frac{1}{3} 125 {\ln}^{2} \left(5\right) - \frac{2}{9} \left(125 \ln 5 - \frac{125}{3}\right)\right] - \frac{2}{9} \left(\frac{1}{3}\right)$

$= \frac{125}{3} \left[{\ln}^{2} \left(5\right) - \frac{2}{3} \ln 5 + \frac{2}{9}\right] - \frac{2}{27}$

So, adding up all the results,

$V = \pi \left\{\frac{1}{2} \left({e}^{10} - {e}^{2}\right)\right\} - 2 \pi \left\{848.042\right\} + \pi \left\{\frac{125}{3} \left[{\ln}^{2} \left(5\right) - \frac{2}{3} \ln 5 + \frac{2}{9}\right] - \frac{2}{27}\right\}$

$= \textcolor{b l u e}{\frac{\pi}{2} \left({e}^{10} - {e}^{2}\right) - 1696.084 \pi + \frac{125 \pi}{3} \left[{\ln}^{2} \left(5\right) - \frac{2}{3} \ln 5 + \frac{2}{9}\right] - \frac{2 \pi}{27}}$

or $29486.554$.