# What is the volume of the solid produced by revolving f(x)=cscx-cotx, x in [pi/8,pi/3] around the x-axis?

Aug 4, 2016

A pretty ugly answer, but I got:

$V = \frac{\left(48 - 16 \sqrt{3} + 5 \pi\right) \sqrt{2} + 32 \sqrt{3} - 10 \pi - 96 \sqrt{2 - \sqrt{2}}}{24 \left(2 - \sqrt{2}\right)} \pi$

$\approx 0.3216$

Unfortunately, it's the simplest exact numerical solution, apparently, we can't make it look any nicer. :)

DISCLAIMER: The integral is not that hard, but this answer requires a lot of simplification work!

First, let's see how this graph looks.

graph{(cscx - cotx) [0.3927, 1.047, 0, 0.8]}

Here we can see it's a simple curve within this interval. Along the x-axis, the easiest way to do this is to form discs that are perpendicular to the x-axis.

The integral in general is:

$\setminus m a t h b f \left(V = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} \pi r {\left(x\right)}^{2} \mathrm{dx}\right)$

where $r \left(x\right) = \csc x - \cot x$ is the function by which the radius varies for each disc we stack of width $\mathrm{dx}$ and area $\pi r {\left(x\right)}^{2}$ along the x-axis.

$= \pi {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\left(\csc x - \cot x\right)}^{2} \mathrm{dx}$

$= \pi {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\csc}^{2} x - 2 \csc x \cot x + {\cot}^{2} x \mathrm{dx}$

$= \pi \left[{\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\csc}^{2} x \mathrm{dx} - 2 {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} \csc x \cot x \mathrm{dx} + {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\cot}^{2} x \mathrm{dx}\right]$

If you recall...

• $\textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left[\csc x\right] = - \csc x \cot x}$, similar to how $\frac{d}{\mathrm{dx}} \left[\sec x\right] = \sec x \tan x$. So let's adjust the middle integral to make it easier to integrate.
• $\frac{d}{\mathrm{dx}} \left[\tan x\right] = {\sec}^{2} x$, so $\textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left[\cot x\right] = - {\csc}^{2} x}$. So let's also adjust the first integral.
• Since $1 + {\tan}^{2} x = {\sec}^{2} x$, $\textcolor{g r e e n}{1 + {\cot}^{2} x = {\csc}^{2} x}$, and so, let's adjust the third integral as well.

$= \pi \left[- {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} - {\csc}^{2} x \mathrm{dx} + 2 {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} - \csc x \cot x \mathrm{dx} - {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} 1 + \left(- {\csc}^{2} x\right) \mathrm{dx}\right]$

Now we're ready to evaluate each one.

$= \pi \left[- \cot x + 2 \csc x - x - \cot x\right]$

$= \pi | \left[2 \csc x - 2 \cot x - x\right] {|}_{\pi \text{/"8)^(pi"/} 3}$

$= \pi \left[\left(2 \csc \left(\frac{\pi}{3}\right) - 2 \cot \left(\frac{\pi}{3}\right) - \frac{\pi}{3}\right) - \left(2 \csc \left(\frac{\pi}{8}\right) - 2 \cot \left(\frac{\pi}{8}\right) - \frac{\pi}{8}\right)\right]$

At this point, either use your calculator, or work out the second half using half-angle formulas (since $\frac{\pi}{8} = {22.5}^{\circ}$). I just used Wolfram Alpha to save time on $\csc \left(\frac{\pi}{8}\right)$ and $\cot \left(\frac{\pi}{8}\right)$.

$= \pi \left[\left(2 \cdot \frac{2}{\sqrt{3}} - 2 \frac{\sqrt{3}}{3} - \frac{\pi}{3}\right) - \left(2 \cdot \frac{2}{\sqrt{2 - \sqrt{2}}} - 2 \left(1 + \sqrt{2}\right) - \frac{\pi}{8}\right)\right]$

Use common denominators to merge some fractions, distributing negative signs over parentheses carefully.

$= \pi \left[\left(\frac{4 \sqrt{3}}{3} - \frac{2 \sqrt{3}}{3} - \frac{\pi}{3}\right) - \left(\frac{4}{\sqrt{2 - \sqrt{2}}} - 2 - 2 \sqrt{2} - \frac{\pi}{8}\right)\right]$

$= \pi \left[\frac{2 \sqrt{3}}{3} - \frac{\pi}{3} - \frac{4}{\sqrt{2 - \sqrt{2}}} + 2 + 2 \sqrt{2} + \frac{\pi}{8}\right]$

$= \pi \left[\frac{16 \sqrt{3}}{24} - \frac{8 \pi}{24} + \frac{48}{24} + \frac{48 \sqrt{2}}{24} + \frac{3 \pi}{24} - \frac{4}{\sqrt{2 - \sqrt{2}}}\right]$

Here we multiply by a unit fraction to get rid of the outer radical in the denominator.

$= \pi \left[\frac{16 \sqrt{3} - 5 \pi + 48 + 48 \sqrt{2}}{24} - \frac{4}{\sqrt{2 - \sqrt{2}}} \cdot \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 - \sqrt{2}}}\right]$

Cross-multiply to merge fractions again.

$= \pi \left[\frac{\left(16 \sqrt{3} - 5 \pi + 48 + 48 \sqrt{2}\right) \left(2 - \sqrt{2}\right)}{24 \left(2 - \sqrt{2}\right)} - \frac{96 \sqrt{2 - \sqrt{2}}}{24 \left(2 - \sqrt{2}\right)}\right]$

$= \pi \left[\frac{\left(16 \sqrt{3} - 5 \pi + 48 + 48 \sqrt{2}\right) \left(2 - \sqrt{2}\right) - 96 \sqrt{2 - \sqrt{2}}}{24 \left(2 - \sqrt{2}\right)}\right]$

Expand, and then cancel out anything you can.

$= \pi \left[\frac{32 \sqrt{3} - 10 \pi \cancel{+ 96} + 96 \sqrt{2} - 16 \sqrt{6} + 5 \pi \sqrt{2} - 48 \sqrt{2} \cancel{- 96} - 96 \sqrt{2 - \sqrt{2}}}{24 \left(2 - \sqrt{2}\right)}\right]$

$= \pi \left[\frac{48 \sqrt{2} - 16 \sqrt{6} + 5 \pi \sqrt{2} + 32 \sqrt{3} - 10 \pi - 96 \sqrt{2 - \sqrt{2}}}{24 \left(2 - \sqrt{2}\right)}\right]$

Lastly, I found terms that I could factor $\sqrt{2}$ out of. Since $\sqrt{6} = \sqrt{3} \sqrt{2}$, make sure you don't accidentally write $16$ instead of $16 \sqrt{3}$!

$= \textcolor{b l u e}{\frac{\left(48 - 16 \sqrt{3} + 5 \pi\right) \sqrt{2} + 32 \sqrt{3} - 10 \pi - 96 \sqrt{2 - \sqrt{2}}}{24 \left(2 - \sqrt{2}\right)} \pi}$

$\approx 0.1024 \pi$

$\approx \textcolor{b l u e}{0.3216}$