# volume of solid of revolution

• Jul 3rd 2011, 03:56 AM
vidhi96
volume of solid of revolution
Hi I have to find volume of solids for f(x)= (1/x) cos(1/x) from x= 2/3pi to x= 2/pi and so far this is what I have done attached here as I don't know how to type all fancy stuff- so please have a look and any guidance in completeing this will be great

• Jul 3rd 2011, 03:59 AM
Prove It
Re: volume of solid of revolution
Is this the volume of the region rotated about the x axis?
• Jul 3rd 2011, 04:16 AM
vidhi96
Re: volume of solid of revolution
yes - sorry I forgot to mention- it is rotated about x-axis
• Jul 3rd 2011, 04:35 AM
Prove It
Re: volume of solid of revolution
First of all, your integral has been set up wrong... the volume of a region rotated about the $\displaystyle x$ axis is given by $\displaystyle \pi\int_a^b{[f(x)]^2\,dx}$, so in this case

\displaystyle \begin{align*} V &= \pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{\left[\frac{1}{x}\cos{\left(\frac{1}{x}\right)}\right]^2\,dx} \\ &= \pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{\frac{1}{ x^2}\cos^2{\left(\frac{1}{x}\right)}\,dx} \\ &= -\pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{-\frac{1}{x^2}\cos^2{\left(\frac{1}{x}\right)}\,dx} \end{align*}

Now make the substitution $\displaystyle u = \frac{1}{x} \implies du = -\frac{1}{x^2}\,dx$, and note that when $\displaystyle u\left(\frac{2}{3\pi}\right) = \frac{3\pi}{2}$ and $\displaystyle u\left(\frac{2}{\pi}\right) = \frac{\pi}{2}$ so that the integral becomes

\displaystyle \begin{align*} -\pi\int_{\frac{3\pi}{2}}^{\frac{\pi}{2}}{\cos^2{(u )}\,du} &= \pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\cos^2{(u )}\,du} \\ &= \pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\frac{1}{ 2} + \frac{1}{2}\cos{(2u)}\,du} \end{align*}

Go from here.
• Jul 3rd 2011, 04:41 AM
vidhi96
Re: volume of solid of revolution
thanks- I can see my error- Can I ask- how do you get this Math text on here as I alwats have to PDF and attached
• Jul 3rd 2011, 04:56 AM
Prove It
Re: volume of solid of revolution
It's called LaTeX. You can learn more in the LaTeX subforum. Once you know the code, put it in tex tags.
• Jul 3rd 2011, 04:57 AM
vidhi96
Re: volume of solid of revolution
Hey I am clear up until -1/x^2 disapperars and replaced by 1/2+1/2(cos(2u)- can you please just clarify that please

Quote:

Originally Posted by Prove It
First of all, your integral has been set up wrong... the volume of a region rotated about the $\displaystyle x$ axis is given by $\displaystyle \pi\int_a^b{[f(x)]^2\,dx}$, so in this case

\displaystyle \begin{align*} V &= \pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{\left[\frac{1}{x}\cos{\left(\frac{1}{x}\right)}\right]^2\,dx} \\ &= \pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{\frac{1}{ x^2}\cos^2{\left(\frac{1}{x}\right)}\,dx} \\ &= -\pi\int_{\frac{2}{3\pi}}^{\frac{2}{\pi}}{-\frac{1}{x^2}\cos^2{\left(\frac{1}{x}\right)}\,dx} \end{align*}

Now make the substitution $\displaystyle u = \frac{1}{x} \implies du = -\frac{1}{x^2}\,dx$, and note that when $\displaystyle u\left(\frac{2}{3\pi}\right) = \frac{3\pi}{2}$ and $\displaystyle u\left(\frac{2}{\pi}\right) = \frac{\pi}{2}$ so that the integral becomes

\displaystyle \begin{align*} -\pi\int_{\frac{3\pi}{2}}^{\frac{\pi}{2}}{\cos^2{(u )}\,du} &= \pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\cos^2{(u )}\,du} \\ &= \pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\frac{1}{ 2} + \frac{1}{2}\cos{(2u)}\,du} \end{align*}

Go from here.

• Jul 3rd 2011, 05:04 AM
Prove It
Re: volume of solid of revolution
$\displaystyle \frac{du}{dx} = -\frac{1}{x^2} \implies du = -\frac{1}{x^2}\,dx$. Can you see where $\displaystyle -\frac{1}{x^2}\,dx$ is in your original integral? Replace it with $\displaystyle du$.

Also, you should know that $\displaystyle \cos{(2\theta)} \equiv \cos^2{(\theta)} - \sin^2{(\theta)}$. Get everything in terms of cosines and solve it for $\displaystyle \cos^2{(\theta)}$.
• Jul 3rd 2011, 05:20 AM
vidhi96
Re: volume of solid of revolution
Yes I understand the identity replacement where you got 1/2(cos(2u) but you alos got extra 1/2 i.e. 1/2 +1/2c0s(2u) du that part is unclear- sorry about this but I am so new to all these
• Jul 3rd 2011, 05:25 AM
Prove It
Re: volume of solid of revolution
\displaystyle \begin{align*} \cos{(2\theta)} &\equiv \cos^2{(\theta)} - \sin^2{(\theta)} \\ \cos{(2\theta)} &\equiv \cos^2{(\theta)} - [1 - \cos^2{(\theta)}] \\ \cos{(2\theta)} &\equiv \cos^2{(\theta)} - 1 + \cos^2{(\theta)} \\ \cos{(2\theta)} &\equiv 2\cos^2{(\theta)} - 1 \\ 1 + \cos{(2\theta)} &\equiv 2\cos^2{\theta} \\ \frac{1}{2} + \frac{1}{2}\cos{(2\theta)} &\equiv \cos^2{(\theta)}\end{align*}
• Jul 3rd 2011, 05:36 AM
vidhi96
Re: volume of solid of revolution
thanks for this- so now I am clear- where you say go from here- I need to find the integral of 1/2+1/2 cos(2u) right?
• Jul 3rd 2011, 05:42 AM
Prove It
Re: volume of solid of revolution
Yes.
• Jul 3rd 2011, 06:33 AM
vidhi96
Re: volume of solid of revolution
Quote:

Originally Posted by Prove It
Yes.