# Volume of Solids of Revolution

• Jul 27th 2005, 05:01 AM
mactev
Volume of Solids of Revolution
This question has been taxing me for some time. I am a first year Mathematics student in London and for some reason am struggling with the substitution needed to answer this question.

Q.

Find the volume of the solid of revolution obtained when the region undr the graph of f(x)= (1/x)sec(1/x), from x=3/pi to x=4/pi, is rotated about the x-axis. Give your answers to 2 decimal places.

Chris
• Jul 27th 2005, 07:30 AM
Rebesques
Quote:

Find the volume of the solid of revolution obtained when the region undr the graph of f(x)= (1/x)sec(1/x), from x=3/pi to x=4/pi, is rotated about the x-axis. Give your answers to 2 decimal places.

Lets see... When you have an integrable function $\displaystyle f(x), x \ \in [a,b]$, then the volume of the solid obtained by rotation about the x-axis is given by

$\displaystyle V=\pi \int_a^b f^2(x)dx.$ (1)

In this case, you will need to find the value of $\displaystyle \pi\int_{\frac{3}{\pi}}^{\frac{4}{\pi}} \frac{1}{x^2 \cos^2(\frac{1}{x})} \ dx.$

There are math packages than can save you the bother, but I guess they ask of you, to calculate by numerical integration. If this is the case, search your textbook for the appropriate method, or ask again if I talk crap. :( :confused:

Note: (1) holds true. (BO-RING... :( )

Consider a partition $\displaystyle (x_i), 1\leq i\leq n$ of [a,b], and intermediate points $\displaystyle x_{i-1}\leq\xi_i\leq x_i$. In revolving the function, we obtain n-1 elementary volumes of cylinders, with height $\displaystyle x_i-x_{i-1}$, radious $\displaystyle f(\xi_i)$ and -therefore- volume $\displaystyle \pi f^2(\xi_i)(x_i-x_{i-1})$. Adding up, we get the Riemmanian sum
$\displaystyle \sum_{i=1}^{n}\pi f^2(\xi_i)(x_i-x_{i-1}) \rightarrow \int_a^b\pi f^2(x)dx$ as $\displaystyle n\rightarrow\infty$, due to integrability.
• Jul 27th 2005, 10:34 AM
mactev
Thank you very much for your timely response. I can see the need to substitute (1/x) in the integrand to get pi(-tan(1/x)dx as my answer.

Cheers.
• Jul 27th 2005, 11:39 AM
Rebesques
oh!
You are right!

$\displaystyle \int_{\frac{3}{\pi}}^{\frac{4}{\pi}} \frac{1}{x^2 \cos^2(\frac{1}{x})} \ dx = -\int_{\frac{3}{\pi}}^{\frac{4}{\pi}} d\left\{\tan (\frac{1}{x})\right\} \ dx,$

and childsplay henceforth... :cool: :eek: