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Find the volume

  1. Nov 1, 2010 #1
    I'm pretty lost with this problem. Can someone please help me with this question? Thanks!!

    Find the volume of the solid generated by revolving the region about the given line

    The region in the first quadrant bounded above by the line y=1, below by the curve y=√(sin6x), and on the left by the y-axis, about the line y=1

    a) pi/6 + 6
    b)pi/12 - 1/6
    c)pi^2/12 - pi/6
    d)pi^2/12 + pi/6

    I set up two possible integrals

    V=pi∫((arcsin y^2)/6)^2 dy from y=0 to y=1

    V=pi∫(√ (sin6x))^2 dx from x=0 to x=pi

    Im not getting any of the possible answers and I dont know what else to do. Please help!
     
  2. jcsd
  3. Nov 1, 2010 #2
    The volume by rotating a function f around the X-AXIS, is

    [tex] \pi\int_0^2{f(x)^2dx} [/tex]

    But you're rotating around the axis y=1. You have to slide down the function first. So the function you should be rotating around the X-axis is [tex]f(x)=\sqrt{\sin(6x)}-1[/tex].
     
  4. Nov 1, 2010 #3
    Thanks! Is that the function i need to use? When I put that in my calculator from 0 to 2 still doesnt give me any of the answers.. :-( what am i doing wrong??
     
  5. Nov 1, 2010 #4
    No, the ranges in my formula are wrong. You don't have to do it from 0 to 2. You need to do it from 0 to when f becomes 0 (sketch the function, thatll make it easier)
     
  6. Nov 1, 2010 #5
    I was wondering where did you get 2 from.. Thanks! I'll try that..
     
  7. Nov 1, 2010 #6
    0 to pi/12?
     
  8. Nov 1, 2010 #7
    Yes, that seems right.
     
  9. Nov 1, 2010 #8
    thank you so much!
     
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