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Homework Help: Solids of revolution

  1. Feb 8, 2009 #1
    y=x^2 ;
    y=4;
    rotated around x=2

    im seeing a washer cross section with r=2-y^(1/2);
    im unclear on how to get R to calculate the area it seems to be 2r but this produces incorrect results.
     
  2. jcsd
  3. Feb 8, 2009 #2
    y=x^2 is a parabola and y=4 is a line. Rotate these about the x=2 axis and the object is a paraboloid.
     
  4. Feb 8, 2009 #3
    im clear on the shape of the solid , but how do you get an equation for the large radius of the washer
     
  5. Feb 8, 2009 #4

    Mark44

    Staff: Mentor

    When you rotate the region of the parabola y = x^2 around the line x = 2, the vertical cross-section of this solid looks like two parabolic sections. The part on the right has its vertex at (4, 0) and intersects the line y = 4 at (2, 4) and (6, 4). The equation of this translated parabola is y = (x - 4)^2.

    The large radius of a washer is the x-value on the parabola on the right, minus 2, or sqrt(y) + 4 - 2. You can also get this dimension by taking the x value on the parabola on the left, and you'll get the same value.

    You mentioned that you had calculated the large radius as 2 - sqrt(y). That actually gets you the inner radius. Using the parabolic region on the right, I get an inner radius of 4 - sqrt(y) - 2, which is what you had for the outer radius.
     
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