# What is the volume of the solid produced by revolving f(x)=cotx, x in [pi/4,pi/2] around the x-axis?

Mar 9, 2018

$V = \pi - \frac{1}{4} {\pi}^{2}$

#### Explanation:

The formula for finding the volume of a solid produced by revolving a function $f$ around the $x$-axis is

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

So for $f \left(x\right) = \cot x$, the volume of its solid of revolution between $\pi \text{/} 4$ and $\pi \text{/} 2$ is

$V = {\int}_{\pi \text{/"4)^(pi"/"2)pi(cotx)^2dx=piint_(pi"/"4)^(pi"/"2)cot^2xdx=piint_(pi"/"4)^(pi"/"2)csc^2x-1dx=-pi[cotx+x]_(pi"/"4)^(pi"/} 2} = - \pi \left(\left(0 - 1\right) + \left(\frac{\pi}{2} - \frac{\pi}{4}\right)\right) = \pi - \frac{1}{4} {\pi}^{2}$

Mar 9, 2018

$\text{Area of revolution around}$ $x \text{-axis} = 0.674$

#### Explanation:

$\text{Area of revolution around}$ $x \text{-axis} = \pi {\int}_{a}^{b} {\left(f \left(x\right)\right)}^{2} \mathrm{dx}$

$f \left(x\right) = \cot x$
$f {\left(x\right)}^{2} = \cot x$

${\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} {\cot}^{2} x \mathrm{dx} = {\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} {\csc}^{2} x - 1 \mathrm{dx}$
$\textcolor{w h i t e}{{\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} {\cot}^{2} x \mathrm{dx}} = \pi {\left[- \cot x - x\right]}_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$\textcolor{w h i t e}{{\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} {\cot}^{2} x \mathrm{dx}} = \pi \left[\left(- \cot \left(\frac{\pi}{2}\right) - \frac{\pi}{2}\right) - \left(- \cot \left(\frac{\pi}{4}\right) - \frac{\pi}{4}\right)\right]$
$\textcolor{w h i t e}{{\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} {\cot}^{2} x \mathrm{dx}} = \pi \left[\left(- 0 - \frac{\pi}{2}\right) - \left(- 1 - \frac{\pi}{4}\right)\right]$
$\textcolor{w h i t e}{{\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} {\cot}^{2} x \mathrm{dx}} = \pi \left[- \frac{\pi}{2} + 1 + \frac{\pi}{4}\right]$
$\textcolor{w h i t e}{{\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} {\cot}^{2} x \mathrm{dx}} = 0.674$