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Volume by washer or shell help

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the area given by y=x^2, y=4, x=0 and revolved about y = -2

    Use either the washer or shell method


    2. Relevant equations
    Shell; integral from a to b of 2∏x(f(x))dx

    3. The attempt at a solution
    Alright so I tried to do this is terms of dy so since we are revolving about y=-2 which would be the horizonal axis. I got x= sqrt(y) and then I set up the integral from 0 to 4 of 2pi(y-2)(sqrt(y))dy

    but this seems to be giving me a volume of 0 which means ive got to be doing something wrong!

    I also attempted it with a dx and I decided to use the integral from 0 to 2 of 2 pi(x-2)(4-x^2)dx but recieved a value of 0 as well! I feel like this means that I am doing something consistently wrong which is kind of good, but I would like to fix that.
     
    Last edited: Sep 10, 2014
  2. jcsd
  3. Sep 10, 2014 #2
    Ok so now I am really confused. I mean integral from 0 to 4 of 2pi(y+2)(sqrt(y)) and when I enter that in wolfram I don't get the correct value, but when I move the 2pi out of the integral and then multiply I get the right answer? I thought it wasn't supposed to matter....
     
  4. Sep 10, 2014 #3

    haruspex

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    I'd guess you are entering it into Wolfram wrongly. Maybe a problem with the parentheses?
     
  5. Sep 10, 2014 #4

    HallsofIvy

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    Using the washer method, a vertical line from y= -2 through the figure, at fixed x, crosses it at [itex](x, x^2)[/itex] and [itex](x, 4)[/itex] so the inner and outer boundaries are at distance [itex]x^2+ 2[/itex] and [itex]4+ 2= 6[/itex]. Those are the inner and outer radii of the "washer". The area of such a "washer" will be [itex]\pi(x^2+ 2)^2- \pi(6)^2= \pi(x^4+ 4x^2- 32)[/itex]. The thickness of such a washer will be "dx" so the volume [itex]\pi(x^4+ 4x^2- 32)dx[/itex]. Integrate that from x= 0 to x= 1.

    Using the "shell method", at a given y, the "shell" will have length [itex]x- 0= \sqrt{y}[/itex] and will rotate around a radius [itex]y- (-2)= y+ 2[/itex], describig a circle of circumference [itex]2\pi(y+ 2)[/itex]. The area of such a shell will be [itex]2\pi(y+ 2)\sqrt{y}[/itex]. The thickness will be "dy" so the volume of [itex]2\pi(y+ 2)\sqrt{y}dy[/itex]. Integrate that from y= 0 to y= 4.
     
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