Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Volume

  1. Nov 30, 2003 #1
    Can someone plz help me w/ these problems below:
    Find the volume generated by revolving the area abounded by x=y^2 and x=4 about
    a)the line y=2
    b)the line x= -1
    ***I tried to write out the integral not sure if it's correct:
    a) V=pi* int. of (sqrt(x)^2-(2-sqrt(x))^2) dx
    **integral form 0->4 ??
    b) V=pi* int. from -2->2 of (4-y^2)-1^2 dy ??
    If the integrals is wrong plz fix it for me...Thanks!
     
  2. jcsd
  3. Dec 1, 2003 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    For both of these think of a "cross section" as being the area between two circles- a "washer".

    For (a) (rotated around y=2), you are measuring the radius in the y direction from 2 to y= √(x) and from 2 to y= -&radic(x): that is, the two circles have radius 2-x1/2 and 2+x1/2. The larger circle has area π(2+x1/2)2= π(4+ 2x1/2+x). The smaller circle has radius π(2-x1/2)2= π(4- 2x1/2+x).
    The difference between those two areas is 4π x1/2. That's what you need to integrate.

    For (b) (rotated around x= -1), the radius is measured in the x direction from -1 to x= y2 and from -1 to x= 4. The two circles have radius y2+1 and 4+1= 5. The area of the larger circle is 25π and the area of the smaller circle is π(y2+1)2= &pi(y4+ 2y2+ 1).
    The difference between those two areas is π(24- y4- 2y2). That's what you need to integrate.
     
  4. Dec 1, 2003 #3
    I am new to volumes too, what is this cross section?

    And can all general objects' volumes be regarded as Washer and another category?(forgot what the other one was).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding Volume
Loading...