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Math Help - rotating a solid about the y axis

  1. #1
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    rotating a solid about the y axis

    so embarrassed that this is my second thread and in both i'm asking for basic help.

    i have taken calc 1 but i have not retained the material very well. i'm now in a class that asks us to solve problems using software and i'm a mess.

    i wish i could contribute by helping people here, but i am not very advanced at all [EDIT TO ADD: I did finally find some posts to answer, so I feel better ! ]

    the problem is as follows:
    Find the volume of the solid that is created by evaluating the function cos(x) from π/2 to π/2 and then rotating the figure about the y axis.


    now to show effort towards a solution:
    i believe that this would be a cylinder, and possibly dy instead of dx as it rotates around the y axis. the way i'd imagine it is by
    π *(integral of x= cos y) from π/2 to π/2 which comes out to 2π

    am i on the right track?


    thanks in advance for any hints!
    Last edited by headabovewater; September 5th 2011 at 09:10 PM.
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  2. #2
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    Re: rotating a solid about the y axis

    You're right about the integral using dy instead of dx, but you forgot about the squared in the solid of revolution.

    Remember that the volume is given by: \text{V} = \pi \int^b_a [f(y)]^2 dy

    Therefore in your case it would be: \text{V} = \pi \int^\frac{\pi}{2}_{-\frac{\pi}{2}} [cos(y)]^2 dy
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  3. #3
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    Re: rotating a solid about the y axis

    Yikes! (pi)*r squared! Oops! Thanks for the catch.
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  4. #4
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    Re: rotating a solid about the y axis

    This is a trickier on than it looks
    First, rearrange for x=arcos(y), and find the volume generated by rotating the area from terminals x=0 to x=pi/2 i.e y=0 to y=1 around the axis using Educated's integral with f(y)=arcos(y). The area you are rotating from -pi/2 to pi/2 will give twice volume required (try picturing this in your head)
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  5. #5
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    Re: rotating a solid about the y axis

    Quote Originally Posted by Istafa View Post
    This is a trickier on than it looks
    First, rearrange for x=arcos(y), and find the volume generated by rotating the area from terminals x=0 to x=pi/2 i.e y=0 to y=1 around the axis using Educated's integral with f(y)=arcos(y).
    Wow, thanks for this response. I did not translate the function for y.

    The area you are rotating from -pi/2 to pi/2 will give twice volume required (try picturing this in your head)
    Do you mean to say that the volume will be twice the result of integrating from y= 0 to y= 1?
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  6. #6
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    Re: rotating a solid about the y axis

    Thats precisely correct
    Think about it this way:
    You are rotating two of the same area 360 degrees about the axis, this will give you twice the volume
    You need only half the total area to be rotated around for the solid required
    Sketch the graph, labelling co-ordinates. It helps to visualise and to switch between the axis with dy or dx more easily!!
    The integration also looks quite tricky, you probably need extra help on this. I'm not exaclty sure how to integrate arcos^2(y)???
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  7. #7
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    Re: rotating a solid about the y axis

    Quote Originally Posted by Istafa View Post
    Thats precisely correct
    Think about it this way:
    You are rotating two of the same area 360 degrees about the axis, this will give you twice the volume
    You need only half the total area to be rotated around for the solid required
    Sketch the graph, labelling co-ordinates. It helps to visualise and to switch between the axis with dy or dx more easily!!
    The integration also looks quite tricky, you probably need extra help on this. I'm not exaclty sure how to integrate arcos^2(y)???
    I wondered about that, as I often see integrals from -x to 0 or 0 to x but never -x to x. So your logic makes sense to me.

    As for integrating arccos--neither do I , but luckily this is a computing course. I input this into Maple and get:
    \text{V} = \pi \int^1_0 [arccos(y)]^2 dy = \pi (\pi -2)

    I tried Maple's Volume of Revolution wizard but it gave me a different answer....However, it wouldnt let me take the integral with respect to y so that may account for the difference.
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