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Volumes of solids of revolution with infinity

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the infinite region in the first quadrant between the curve y=e^-5x and the x-axis.
    Find area= 1/5 (got this part)
    Compute the volume of the solid generated by revolving the region about the x-axis:
    Compute the volume of the solid generated by revolving the region about the y-axis:

    2. The attempt at a solution
    I do not understand how to do the last two parts. I tried to do
    dV=base*height
    dV=pi*x^2*dy

    because area of circle is

    base= pi*x^2
    leaving me with
    dv= pi*x^2*(-1/5e^(-5x)
    and taking limit from 0 to inifity by taking limit from 0 to b, but i am not understanding how to get the answer. Please advise. I am so lost. Thank you in advance.
     
  2. jcsd
  3. Mar 23, 2014 #2

    BvU

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    For the volume of the solid generated by revolving the region about the x-axis, you now integrate perpendicular to the axis of revolution (dy). It's not all that easy to express ##x^2## in y, but you do need it then.

    What would you write down if you integrated parallel to the axis of revolution (i.e. integrate over dx) ?
     
  4. Mar 23, 2014 #3

    LCKurtz

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    You have the choice of disks or shells and also the choice of dx or dy elements. That should be your first decision. Since ##y## is given in terms of ##x##, you will find using ##dx## elements to be most convenient. So, first question:

    Revolving about the ##x## axis using ##dx## element, would you be using disks or shells? What is the corresponding dV?

    Same questions for about the ##y## axis using a ##dx## element.

    Show us those dV elements. The rest will be easy.
     
  5. Mar 23, 2014 #4

    BvU

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    Over to Lynn (bedtime for me...)
     
  6. Mar 23, 2014 #5
    for rotation about the x-axis, would it be
    v=∫ (0 to infinity) of (2∏*x*e^-5x)dx?
    since im using washers, since the vertical stick is horizontal to axis of rotation?

    im getting the answer as 2/25 pi which, for some reason is the answer to the rotation about the y axis, and not the x? Can you tell me why?

    i still cannot get the x-axis one, and i dont understand why the y axis one is correct.
     
    Last edited: Mar 23, 2014
  7. Mar 23, 2014 #6

    LCKurtz

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    The ##x## factor in your integral is the distance of the vertical element from the ##y## axis. So ##2\pi x## is the distance around the ##y## axis and the ##e^{-5x}## is the height of the element. ##dx## is the thickness. So you have height*circumference*thickness of the shell revolving about the ##y## axis. You are getting the ##y## axis number because that is the integral you set up.

    Now, going around the ##x## axis, using disks, what do you get?
     
  8. Mar 23, 2014 #7
    is is simply v=integral(pi*e^-5x)dx to set up?
     
  9. Mar 23, 2014 #8
    or would i have to square the inside of the integral making the inside e^-10x?

    got answer as pi/10, but honestly I guessed. Did I set my integral up correctly? I still dont understand if I had to square it, or if I squared correctly. (If the integral I wrote on this post is correct, then I understand)
     
    Last edited: Mar 23, 2014
  10. Mar 23, 2014 #9

    LCKurtz

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    Remembering that the area of a disk of radius ##r## is ##\pi r^2##, does that give you a hint about your guess? So ##dV## would be ##(\pi r^2 \cdot \text{thickness})##.
     
  11. Mar 23, 2014 #10
    i think im getting confused with powers for some reason. could you clear something up? when you have something like e^x^2. Is that the same as e^2x? Well, I guess it's not. But is there some way you can rewrite the e^x^2 as an easier power?
     
  12. Mar 24, 2014 #11

    BvU

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    Like ##(4^3)^2 = 4^3 \times 4^3 = 4^{3+3}## you write ## (e^y)^2 = e^y \times e^y = e^{y+y} = e^{2y}##
     
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