1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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 by a moderator: 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


    User Avatar
    Science Advisor

    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


    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook