AP Calculus Help (Last Time xD)

Discussion in 'Random Topic Center' started by OLD_SCHOOL_PLAYER, Jun 6, 2009.

  1. OLD_SCHOOL_PLAYER

    OLD_SCHOOL_PLAYER New Member

    I just need help with two more AP Calculus problems, and then I'll be done for the year.


    Find volume of region generated by rotating shaded area about the x-axis:

    y = 3 - (6/x)

    x = 5

    Window:
    x min: 0
    x max: 6
    y min: 0
    y max: 2

    This is what I did for this problem, not completely sure if I did this correctly:
    y = 3 - (6/x), x = 5
    nr^2
    pi INTEGRAL(from 2 to 5) [ (3-(6/x))^2 ]
    =15.122

    The second problem:

    The base of a solid is the region between the line y = 4 and parabola y = x^2. The cross sections perpendicular to x-axis are semicircles, find volume of solid.



    Thank you for taking the time to read this,
    OSP
     
  2. Diaz

    Diaz New Member

    Its been a while, so someone should correct my mistakes but...
    For the first question, the shape starts at 2 (where y=0) and ends at x=5. Area of a circle is piR^2 so I'm thinking it would be the integral from 2 to 5 of pi((3 - (6/x))^2). You can evaluate that.

    Some one else check this. The second question is similar.
     
  3. Magnechu

    Magnechu Active Member

    First one looks right.

    Second one:

    PI[integral(-2 to 2) of (x^2/2)^2]


    I did this all in my head but that should be right.

    edit: yeah Frankie that's what he had for the 1st one
     
  4. ShadowTogetic

    ShadowTogetic New Member

    Yeah, Frankie. Mathematica gave me the same thing for 1
     

Share This Page