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Math Help - volume of solid of revolution

  1. #1
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    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


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  2. #2
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    Re: volume of solid of revolution

    Is this the volume of the region rotated about the x axis?
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    Re: volume of solid of revolution

    yes - sorry I forgot to mention- it is rotated about x-axis
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    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.
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    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
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  6. #6
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    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.
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    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 View Post
    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.
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    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)}.
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    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
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    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*}
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    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?
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  12. #12
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    Re: volume of solid of revolution

    Yes.
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    Smile Re: volume of solid of revolution

    Quote Originally Posted by Prove It View Post
    Yes.
    Many thanks for every help you offered I managed to find my answer and complete the task
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