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Volume Integral

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Sketch the region R bounded by the curves y = x, x = 2 - y^2 and y = 0.

    This is the initial part of an integral problem and I'm just curious about the method here.
    2. Relevant equations



    3. The attempt at a solution

    So, would it be proper to take the x = 2 - y^2 function and write it in terms of y as the dependent variable? ie, y = sqrt(2-x)

    I believe the graph looks the same but for setting up the integral, would it change anything?

    Thanks!
    Jim
     
  2. jcsd
  3. May 29, 2013 #2

    Simon Bridge

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    You should sketch out ##x=2-y^2## just to be on the safe side.
    i.e. ... what happens to x for negative values of y?
    In ##y=\sqrt{2-x}## can y ever be negative?
    If x > 2, y is imaginary - but in the original form, x cannot be bigger than 2.
    When you change over like that these are things you have to think about.

    It may not matter this time, since your other bound is y=x - but note that it is easier to change that one to x=y.

    I suspect the objective is to find an area - in which case it is reasonable to change to any representation that has the same area and a good idea if it gives you easier math. I don't think the surd does that for you, but swapping the roles of x and y will ... just do ##\int f(y)dy##.

    Put it another way - is the area enclosed by x=2-y^2 and y=x the same as the area enclosed by y=2-x^2 and x=y?
     
  4. May 29, 2013 #3
    Yep, I see exactly what you're talking about. One is a horizontal parabola and the other is just half of one. I redid my work, keeping it in the original form. It actually made it seem easier too. Any chance you could proof my work Simon, or anyone else? Thanks for your help btw!

    The second part says: Indicate the method you use to set up the integrals (do not integrate) that give the volume of the solid generated by rotating the region R around:

    i: the x-axis
    ii: the y-axis
    iii: the line x = -2
    iv: the line y = 1

    For i:

    Using cylindrical shells,
    C = 2∏y
    A = 2∏y((2-y2) - y)
    v = 2∏∫(2y-y3-y2)dy for y = [0,1]

    For ii:

    Using the disc method,
    A = ∏(2-y2)2 - y2)
    v = ∏∫(2-y2)2 - y2)dy for y = [0,1]

    for iii,

    Using the disc method,
    A = ∏((2-y2 + 2)2 - (y+2)2)
    V = ∏∫((4-y2)2 - (y+2)2)dy for y = [0,1]

    for iv,

    Using shells,

    Same as i:
    C = 2∏y
    A = 2∏y((2-y2) - y)
    v = 2∏∫(2y-y3-y2)dy for y = [0,1]
     
  5. May 29, 2013 #4

    Simon Bridge

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    When you get over the idea that the x and y axis have to have a particular use, it does often get easier.

    I will advise you to learn LaTeX for writing equations ;)

    Your approach appears to be that of applying definitions and equations.
    That's not how I would construct an integral - but you know your course.

    The bounded region is the positive part of the parabola with a wedge-thing cut out of it.
    For 1 - shell method check.

    ##x=y## intersects ##x=2-y^2## at (x,y)=(1,1)
    ##x=2-y^2## intersects ##y=0## at (x,y)=(2,0)
    so 0<y<1 and 0<x<2.

    a shell radius ##y##, thickness ##dy## and height ##h(y)## has volume ##dV=2\pi y h(y) dy##
    ##h=x_2-x_1## is bounded above by ##x_2=2-y^2## and below by ##x_1=y##
    ##\Rightarrow h(y) = 2-y^2 - y##

    so the volume is $$V=2\pi \int_0^1 y(2-y^2-y)\; dy$$

    Looks like what you have ... notice how I have more detail about how I got things.
     
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