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Finding Volume of a curve

  1. Oct 22, 2009 #1
    I'm suppose to rotate around y-axis...

    x=8y;x=y^3 and y> and equal to 0

    I was told to integrate (x^1/3-0)^2-(1/8x-0)^2

    so.... pi integral 0 to sqrt(512) x^2/3-1/64x^2 dx

    I end up with 151.65 but that's incorrect can someone tell me what i did wrong and how to get the right answer?
     
  2. jcsd
  3. Oct 22, 2009 #2

    HallsofIvy

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    Who told you this? This would be the volume of the region rotated around the x- axis. You said you wanted the volume of the region rotated around the y- axis. That would be the integral of [itex]\pi((8y)^2- (y^2^2) dy[/itex]

    [itex]\sqrt{512}= 16\sqrt{2}[/itex]

     
    Last edited: Oct 22, 2009
  4. Oct 22, 2009 #3
    Oh wait so it should be integral 64y^2-y^6 dy?

    when i integrate that way it comes out negative
     
    Last edited: Oct 22, 2009
  5. Oct 22, 2009 #4

    HallsofIvy

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    You are right. If forgot the squares (I have edited my response so I can pretend I didn't make thata mistake!). The area of a disk is [itex]\pi r^2[/itex] and you are taking the areas of two disks and subtracting them. The volume you want is [itex]\int_0^{2\sqrt{2}} 64y^2- y^4 dy[/itex].
     
  6. Oct 22, 2009 #5
    Ahhh i'm confused ok I started with

    x=8y
    x=y^3
    y>0

    the limits of y are 0 to sqrt(512) and the limits of x is 0 to sqrt(8)

    SO! Integral should look like 64y^2-y^6 dy Oh gosh nvm THANKS SO MUCH :D umm if i am correct its Rout-Rin right?
     
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