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Disk and washer method

  1. Aug 11, 2011 #1
    Consider the region bounded by the curves y=x^2+1 and y=3-x^2

    a) using the disk/washer method, find the volume of the solid obtained by rotating this region about the x axis

    This was very straight forward
    v=int(1, -1) pi((3-x^2)^2)-(x^2+1)^2))dx
    I finished the problem with 32pi/3 which I think is correct.

    However the next part I have no idea how to set up using the disk washer method.

    b) Set up the integral for finding the volume of the solid obtained by rotating about the y-axis.

    I know that the integration will have to be done in parts but I don't know where to split it into parts. If someone could help me set up the question that would be amazing.

    Thank you,
  2. jcsd
  3. Aug 11, 2011 #2


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    the second part looks much simpler than the first! You have already determined, for the first part, that the two graphs intersect at x= -1 and 1. Those are, of course, at [itex]y= (-1)^2+ 1= (1)^2+ 1= 3- (-1)^2= 3- (1)^2= 2[/itex]. That is, you need to do y= 1 to 2 and y= 2 to 3 separately.
    Last edited by a moderator: Aug 11, 2011
  4. Aug 11, 2011 #3


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    More like y= 1 to 2 and y= 2 to 3. (Oh! HoF already corrected this.)

    Also, if you do the disk/washer method, only include the part of the region for x ≥ 0 , or x ≤ 0, otherwise you will have twice the actual volume.
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