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Math Help - Shell method, can someone explain this problem?

  1. #1
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    Shell method, can someone explain this problem?

    I know how to do very basic and simple shell method problems, but this one is difficult for me. Every problem that I have had to do so far, the graphs always stay on one side of the x and y axis. Meaning they never cross it. This one does, however, and I don't understand how to work with that. So the information given is:
    x= y^4/4 -y^2/2

    x=y^2/2

    There is a graph of this shown, and the graphs intersect at (0,0) and (2,2).
    But the graph of y^4/4 - y^2/2 goes over to the left side of the y axis for a bit, and then goes back to the right. I need to find the volume of the region that would be formed by rotating the graph about the x axis. So how do I find the radius and the height for this? Maybe I'm making it more complicated than it is...
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  2. #2
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    Hi

    I just want to be sure to have correctly understood
    You want to evaluate the volume of the solid bounded by the two graphs and rotating around x-axis right ?

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  3. #3
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    Yeah, that's what I meant. Sorry I'm a little slow on the reply. So yeah, it's being rotated about the x axis. But in order to use the shell method, I need to know the radius and the height, right? The fact that the graph crosses over to the left side of the y axis is making me unsure of what the radius and height would be.
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  4. #4
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    First you need to define the integration boundaries in x. You can show that it goes from x=-1/4 to x=2

    Then you need to separate the calculation into 2 parts :
    - from x=-1/4 to x=0 the curve is bounded only by x= y^4/4 -y^2/2. This means that you need to inverse this expression in order to get y function of x.
    You will get 2 functions : one (f(x)) for the lower part and one (g(x)) for the upper part
    Due to the fact that the rotation is around x axis, the radius is y and the height is x
    The volume is
    V_1 = \int_{-\frac{1}{4}}^{0} \:\int_{0}^{2\pi}\:\int_{f(x)}^{g(x)} y\:dy\:d{\theta}\:dx

    - from x=0 to x=2 the curve is bounded in the lower part by x=y^2/2 and in the upper part by x= y^4/4 -y^2/2. You also need to inverse these expressions to calculate the volume V2
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  5. #5
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    x_1 = \frac{y^2}{2}

    x_2 = \frac{y^4}{4} - \frac{y^2}{2}

    V = 2\pi \int_0^2 y(x_1 - x_2) \, dy
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