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Math Help - Shell Method Setup

  1. #1
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    Shell Method Setup

    I am having trouble setting up the 3rd equation in this question. It asks me to use the shell method to find the volume of the region bounded by y=x^2 and y=x+2 revolved around

    x=2
    x=-1
    x axis How?
    y=4

    I know i have the limits of integration correct but i cant figure out what the correct radius or shell height are. Any guidance is appreciated.
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  2. #2
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    Re: Shell Method Setup

    Quote Originally Posted by petenice View Post
    I am having trouble setting up the 3rd equation in this question. It asks me to use the shell method to find the volume of the region bounded by y=x^2 and y=x+2 revolved around
    I presume you have drawn a graph? y= x^2 and y= x+ 2 intersect at (-1, 1) and (2, 4).

    x=2
    Using the shell method, you would rotate a vertical line, at a given "x" around x= 2 so its radius would be 2- x. The shell height will be x+2- x^2.

    x=-1
    This is on the left side so the radius is x- (-1)= x+ 1.

    x axis How?
    Now you are rotating around a horizontal axis (y= 0) so your variable will be y, not x. The radius will be y. As for the length of the "shell", you will need to do it in two parts. from y= 0 up to y= 1, it will be \sqrt{x}- (-\sqrt{x})= 2\sqrt{x} and from y= 1 up to y= 4, it will be [tex]\sqrt{y}- (y- 2)= \sqrt{y}- y+ 2[tex].

    y=4
    Again, this is a horizontal axis so your variable will be y. The radius will by 4- y.

    I know i have the limits of integration correct but i cant figure out what the correct radius or shell height are. Any guidance is appreciated.
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  3. #3
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    Re: Shell Method Setup

    Quote Originally Posted by HallsofIvy View Post
    Now you are rotating around a horizontal axis (y= 0) so your variable will be y, not x. The radius will be y. As for the length of the "shell", you will need to do it in two parts. from y= 0 up to y= 1, it will be \sqrt{x}- (-\sqrt{x})= 2\sqrt{x} and from y= 1 up to y= 4, it will be \sqrt{y}- (y- 2)= \sqrt{y}- y+ 2.

    Im assuming thats supposed to be \sqrt{y}- (-\sqrt{y})= 2\sqrt{y} correct? I think i got it. Took me a while to understand where you got that 2\sqrt{y} but i understand now! Thanks so much.
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