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Math Help - Volume

  1. #1
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    Volume

    Please see attachment.
    My workings are -
    I am stuck on Volume part. (b) and (c)
    Please help.
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  2. #2
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    Easiest way to do this is to use shell method (I find).

    2\pi\int\limits_{a}^{b}(y + 1)Rdy

    Where b and a are the y values where the two functions intersect, and R is the equation for the height of the shell. Basically f(y) - g(y).
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  3. #3
    MHF Contributor Calculus26's Avatar
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    For Part c see attachment -- I answered a similar problem yesterday
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  4. #4
    MHF Contributor Calculus26's Avatar
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    If you use the shell method you'd need 2 integrals

    and you'd have to invert y = 2^x to get x = ln(y)

    and y =3 -x^2 to get x = (3-y)^1/2

    Don't know where defleurer got his results

    The easiesies way is to use washers

    outer radius 2^x+1 and inner radius 4-x^2
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  5. #5
    MHF Contributor Calculus26's Avatar
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    Aah! Defluerer didn't see the R in the integral -- I see what you are saying but it would still take 2 integrals and inverting the fns
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  6. #6
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    Ah, sorry. Shouldn't have rushed that answer, I suppose.

    But, with shell method, the idea is to integrate 2\pi rh (surface area) is it not? And your height would be f(y) - g(y) (by which I did mean inverting the functions). Your radius is then the distance you are from the axis of rotation. Were that axis of rotation simply y=0, than the radius would just be y. In this case, our axis of rotation is y=-1 and thus our radius should be y + 1. Thus, i'm still thinking this should work

    2\pi\int\limits_{a}^{b}(y + 1)(f(y) - g(y))dy
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  7. #7
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    Quote Originally Posted by Calculus26 View Post
    If you use the shell method you'd need 2 integrals

    and you'd have to invert y = 2^x to get x = ln(y)

    and y =3 -x^2 to get x = (3-y)^1/2

    Don't know where defleurer got his results

    The easiesies way is to use washers

    outer radius 2^x+1 and inner radius 4-x^2
    Is the part (b) like this?

    {\text{Volume }} = \pi \int\limits_{ - \sqrt 3 }^1 {\left( {3 - x^2  + 1} \right)^2 dx}  - \pi \int\limits_0^1 {\left( {2^x  + 1} \right)^2 } dx
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  8. #8
    MHF Contributor Calculus26's Avatar
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    Certainly the shell methid works

    but it will take 2 integrals one from a to 2 and one from 2 t0 3 as f and g change at the line y = 2
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  9. #9
    MHF Contributor Calculus26's Avatar
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    No you have to integrate over the same interval a to b which as defleurer pointed are the intersection point of the 2 curves
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  10. #10
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    Quote Originally Posted by Calculus26 View Post
    Certainly the shell methid works

    but it will take 2 integrals one from a to 2 and one from 2 t0 3 as f and g change at the line y = 2
    Ah, yes, I see now. Should've have caught that one sooner.

    Nevermind what I said. Play it safe and stick with washer. =p
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