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

2 Answers
Sep 2, 2016

$\frac{\pi}{48} \left(10 \pi - 12 \left(1 + \sqrt{2}\right)\right) = 0.16005$ cubic units, nearly.

Explanation:

I supplied the missing the factor 1.2 in the integral of sin 2x now. I

am sorry for missing it, in my earlier version..

$\left[\frac{\pi}{8} , \frac{\pi}{3}\right] \in \left(0 , \frac{\pi}{2}\right)$.

Both sin x and cos x are > 0 in (0, $\frac{\pi}{2}$). So,

$| \sin x | - | \cos x | = \sin x - \cos x , x \in \left[\frac{\pi}{8} , \frac{\pi}{3}\right]$.

Now, the volume of solid of revolution is

$\pi \int {\left(\sin x - \cos x\right)}^{2} d x$, between the limits $\frac{\pi}{8} \mathmr{and} \frac{\pi}{3}$

$= \pi \int \left({\sin}^{2} x + {\cos}^{2} x - 2 \sin x \cos x\right) d x$, between the limits

$= \pi \int \left(1 - \sin 2 x\right) d x$, between the limits

$= \pi \left[x + \frac{1}{2} \cos 2 x\right]$, between the limits

=pi((pi/3-pi/8)+1/2(cos (2/3pi)-cos(pi/4))

$= \pi \left(\frac{5}{24} \pi + \frac{1}{2} \left(- \frac{1}{2} - \frac{1}{\sqrt{2}}\right)\right)$

$= \frac{\pi}{48} \left(10 \pi - 12 \left(1 + \sqrt{2}\right)\right)$ cubic units.

Sep 2, 2016

Interesting fact.

Explanation:

Comparing $f \left(x\right) = \sin x - \cos x$ for $x \in \left[\frac{\pi}{8} , \frac{\pi}{3}\right]$ with
$y = {y}_{1} + \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \left(x - {x}_{1}\right)$ in which

${x}_{1} = \frac{\pi}{8} , {y}_{1} = f \left({x}_{1}\right) = \sin \left(\frac{\pi}{8}\right) - \cos \left(\frac{\pi}{8}\right)$
${x}_{2} = \frac{\pi}{3} , {y}_{2} = f \left({x}_{2}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2}$

we observe a maximum deviation of $0.007$ nearly

Attached a plot with $f \left(x\right)$ and $y$

The difference between the computed volumes using either function is in the order of $0.0032$ units