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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


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    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).
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