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Finding Volumes by using the Disc and Washer Method

  1. Jul 6, 2010 #1
    Theres a few key concepts about the disc and washer method that I can't quite grasp and I was hoping if I could get a bit of clarification.

    1) How do you find your outer and inner radius? I can provide an example if needed.

    2) If a problem has its function, for example f(x)= sec x, g(x)=tan x, x=0, and x=1.. what do the values x=0 and x=1 mean in the question. Do they represent some sort of asymptote?

  2. jcsd
  3. Jul 6, 2010 #2
    The [itex]x=0,x=1[/itex] are also bounding functions for your region of interest.

    The best way to determine bounds for these things is the actually draw the picture (grpah) of what is going on...

    In this instance, the region is a vaguely rectangular thing with [itex]x=0[/itex] on the left, [itex]x=1[/itex] on the right, [itex]\sec(x)[/itex] on top, and [itex]\tan(x)[/itex] on bottom - this is all viewing it basically with respect to the [itex]x[/itex] variable (in the traditional way).

    Viewing it with respect to the [itex]y[/itex] variable (on it's side), then it is much messier. I'll leave that to you.
  4. Jul 6, 2010 #3


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    Did you try drawing the graphs, y= sec(x), y= tan(x), y= 0, and y= 1? They are the boundaries of the region referred to.

  5. Jul 6, 2010 #4
    Yes, after graphing it I can see that the area inbetween x=0 and x=1 is the "area" wanted to find the volume. For the radii, would f(x)=sec x be the outer (or upper)? and g(x)=tan x the inner (or lower)?
  6. Jul 6, 2010 #5


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    It should be evident from your graph. Which of those two functions is greater in the interval 0<x<1?
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