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Homework Help: Solid volume rotation around y-axis

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    We have to trace a pumpkin on graph paper and then find it volumne when rotated around the y-axis. Upon doing so we have 2 pieces we can do. One is a semi circle and the other just a rectangle. Refer to this image:

    r5AZB.png

    Where we can see the diameter of the circle is 13.5 and the full width of the shape is 9.

    2. Relevant equations

    (x-h)^2 + (y-k)^2 = r^2

    v = integral from a to b pi r^2 times thickness (dx or dy)

    3. The attempt at a solution


    Know the diameter is 13.5, we know the radium is (13.5 / 2) or 6.75. Using that we can pinpoint the origin of the semi-circle to (2.25,6.75), where 2.25 is the width of the whole shape (9) minus the radius (6.75).

    We use the formula of a circle with center h,k:
    (x-h)^2 + (y-k)^2 = r^2

    Thus:

    (x-2.25)^2 + (y-6.75)^2 = (6.75)^2.

    We know to be with respect to y there we solve for x:

    x = sqrt( (6.75)^2 - (y-6.75)^2) + 2.25



    Now we get into rotating around the y-axis. The semi-circle is shifted 2.25 from the y-axis so you add 2.25 to the equation, giving you a radius of sqrt( (6.75)^2 - (y-6.75)^2) + 4.5 thus:


    v= pi * integral from 0 to 13.5 (sqrt( (6.75)^2 - (y-6.75)^2) + 4.5)^2 dy
    which is about 1327.56 pi.

    Finally just add to that the retangle rotated around y-axis which is radium 2.25:
    v = pi * integral from 0 to 13.5 (2.25)^2 dy
    which is about 68.3438 pi.

    Answer being about 1395.9 pi.

    I know this is wrong because if I just do a large rectangle of width 9 and rotate it around I get

    v = pi * intefral from 0 to 13.5 (9)^2 dy
    or 1093.5 pi. This should be larger that my previous answer, and it's not.
     
  2. jcsd
  3. Oct 11, 2011 #2

    This will give you the total volume



    I don't know why you are doing the above one ,
    And you need not find the volume for the rotation of rectangle separately because as per the formula its the volume generated when the AREA UNDER THE CURVE x = sqrt( (6.75)^2 - (y-6.75)^2) + 2.25 from 0 to 13.5 is rotated
     
  4. Oct 11, 2011 #3
    It's my understanding that because the rectangle create a 2.25 by 13.5 gap from the y-axis that when you rotate it around the y-axis it will have a center out hole, thats why you need to add the rectangle to fill that piece.

    Also I add 2.25 to the equation because that's the distance from the y-axis.

    I believe part of my issue is that I need the semi circle to have a maximum all the way to the right, basically the top of a circle but rotated 90 degrees clockwise. I thought switching x and y around would do that but I am unsure now.
     
  5. Oct 11, 2011 #4
    Basically what you do is find the volume of the "pumpkin" and subtract the rectange from it.
     
  6. Oct 11, 2011 #5
    The volume of the rectange
     
  7. Oct 11, 2011 #6
    Why would you subtract the rectangle from it? It's apart of the pumpkin.
     
  8. Oct 11, 2011 #7

    SammyS

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    You're rotating about the y-axis (x=0) not about x = -2.25.


    Look at the expression, sqrt( (6.75)^2 - (y-6.75)^2) + 2.25 for y = 6.75. you will get 9, which agrees with you diagram. as the radius of the largest horizontal cross-section.
     
  9. Oct 11, 2011 #8
    Ohhh. It makes sense, sorry I am new to the solid volume rotation, but I understand now. You're right.

    it would be:

    int((sqrt(45.5625-(y-6.75)^2)+2.25)^2, y = 0 .. 13.5) and thus the answer is 800, which makes it fit. Thanks for the explanation.
     
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