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Volume of revolution: shell method

  1. Oct 31, 2009 #1
    (EDITED)
    1. Use the shell method to find the volume of the solid generated by revolving about the y-axis. [tex]x=y^2, x=y+2[/tex]

    2. same as #1, except change y and x for the two equations and revolve about x-axis.

    I tried doing [tex]2pi\int_{x=0}^4(x)(\sqrt{x}-x+2dx[/tex] but the answer is off for #1.

    I tried doing [tex]2pi\int_{y=-1}^4(y)(\sqrt{y}-y+2dy[/tex], well it's wrong.

    Is there a problem with how I established the shell height?

    Can anyone explain shell method in a better way than my textbook? I learned that shell height and shell radius are something quite useful to know when doing shell method.

    for #2, do I need to use 2 integrals perhaps? since the revolution would provide two different volumes with two different radius?

    Hope somebody can aid me.

    EDIT:

    [tex]2pi\int_{x=0}^4(x)(\sqrt{x}-x+2)dx[/tex]
    [tex]2pi\int_{y=-1}^4(y)(\sqrt{y}-y+2)dy[/tex]

    the answer for these two questions are supposed to give me 72pi/5.
     
    Last edited: Oct 31, 2009
  2. jcsd
  3. Oct 31, 2009 #2

    HallsofIvy

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    Draw an accurate graph. You are missing the part between x= 0 and x= 1. You need to do two parts. If [itex]0\le x\le 1[/itex], the "shell " rotated around around the y-axis has height going form [itex]y= -\sqrt{x}[/itex] to [itex]y= \sqrt{x}[/itex]. For [itex]1\le x\e 4[/itex] (NOT 2), it is from y= x- 2 to [itex]y= \sqrt{x}[/itex].

    Well, of course, it is! You can't have an "x" when you are integrating with respect to y!
    To do this by integrating with respect to y, you need to use the "washer" method or, same thing, find the volume of the region rotating the line x=y+2 around the y-axis and then subtract the volume of the region rotationg the parabola [itex]x= y^2[/itex] around the y-axis. That is the same because those two volumes will be [itex]\pi \int x_1^2dx[/itex] and [itex]\pi \int x_2^2 dx[/itex], respectively. But [itex]\pi\int x_1^2 dx- \pi\int x_2^2 dx= \pi \int (x_1^2- x_2^2)dx[/itex]. Since here "[itex]x_1[/itex]" is x= y+ 2 and "[itex]x_2[/itex]" is [itex]x= y^2[/itex], that would be [itex]\int ((y+2)^2- (y^2)^2)dy[/itex]. And your limits of integration are wrong for that integral also. The two curves intersect at (1, -1) and (4, 2), you integrate from y= -1 to y= 2.

    Well, you got the wrong answer! Since you did not say "how" you got the shell height, I don't know if that was the problem.

    I don't know how your textbook explained it so I don't know if I can do a better job. I think of it like this: since you are rotating around the y-axis, draw a vertical line at a representative distance from that axis. That vertical line, rotated around the axis of rotation will be the radius of the cylinder formed- since the rotation is around the y-axis, the radius is just "x" so you have that right. The height of the cylinder will be the difference between the y-values of the curves forming top and bottom. Since different shells are being taken at different x-values, the thickness of the shell will be "dx" and you integrate with respect to x from the lowest to the highest x-values in the figure.

    ??I don't see a #2. Did you mean the integration by the "washer" method? No, for that you do not need two integrals. You need it for the first integral!
     
  4. Oct 31, 2009 #3

    I made many typos.... for number 2, it is actually rotated about the x-axis...
    for the second integral, it's not x there, it's supposed to be y...
    I edited the original post
     
    Last edited: Oct 31, 2009
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