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Math Help - Integration help - Shell Method

  1. #1
    Junior Member xfriendsonfirex's Avatar
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    Integration help - Shell Method

    Hey guys. Any help integrating this function would be much appreciated, so I can apply the shell method.

    Area bound by the curves:

    y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}
    y=0
    x=0
    x=1

    I just don't know where to begin integrating that first function. Thanks!
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by xfriendsonfirex View Post
    Hey guys. Any help integrating this function would be much appreciated, so I can apply the shell method.

    Area bound by the curves:

    y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}
    y=0
    x=0
    x=1

    I just don't know where to begin integrating that first function. Thanks!
    You want to know how to integrate \int \frac{ e^{\frac{-x^2}{2}}}{\sqrt{2\pi}} \,\ dx ?
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  3. #3
    Junior Member xfriendsonfirex's Avatar
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    Yes please.
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  4. #4
    Super Member General's Avatar
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    You can not integrate it in terms of functions we know.
    see :
    integrate ( e^[ (-x^2)/2 ] ) - Wolfram|Alpha

    Post the whole question, please.
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  5. #5
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    Quote Originally Posted by xfriendsonfirex View Post
    Hey guys. Any help integrating this function would be much appreciated, so I can apply the shell method.

    Area bound by the curves:

    y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}
    y=0
    x=0
    x=1

    I just don't know where to begin integrating that first function. Thanks!
    Hi,
    Do you want to find the area of the function between the 3 lines,
    or do you want to rotate the area about an axis (shells) ?
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  6. #6
    Junior Member xfriendsonfirex's Avatar
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    Quote Originally Posted by General View Post
    Post the whole question, please.
    Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

    y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}, y=0, x=0, x=1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfriendsonfirex View Post
    Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

    y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}, y=0, x=0, x=1
    ok, then that function is not what you would integrate. what would you need to integrate using the Shell formula?
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  8. #8
    Super Member General's Avatar
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    Quote Originally Posted by xfriendsonfirex View Post
    Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

    y=\frac{e^{{-x^2}/2}}{\sqrt{2\pi}}, y=0, x=0, x=1
    So you will not integrate what i asked you.
    Do you know the steps of solving suck problem?
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  9. #9
    Junior Member xfriendsonfirex's Avatar
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    What would I need to integrate then? Maybe I can make more sense of that.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfriendsonfirex View Post
    What would I need to integrate then? Maybe I can make more sense of that.
    volume by Shells is given by

    V = 2 \pi \int_a^b (\text{radius})(\text{height})~dx

    have you seen this formula before? have you practiced setting up other problems like this?
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  11. #11
    Junior Member xfriendsonfirex's Avatar
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    V=\int_0^1 x(\frac{e^{{-x^2}/2}}{\sqrt{2\pi}})dx

    Is this what I would be integrating? The x as the width, and the function as the height?

    Sorry, I'm really slow with LaTeX. Never used it before.
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfriendsonfirex View Post
    V=\int_0^1 x(\frac{e^{{-x^2}/2}}{\sqrt{2\pi}})dx

    Is this what I would be integrating? The x as the width, and the function as the height?

    Sorry, I'm really slow with LaTeX. Never used it before.
    actually, you need to multiply that by 2 \pi

    now, as for integrating. do a substitution, u = \frac {-x^2}2
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  13. #13
    Junior Member xfriendsonfirex's Avatar
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    Okay, so I've gotten as far as this:

    y=-2\pi\int_0^1(\frac{e^u}{\sqrt{2\pi}})du

    Can I bring the (2\pi)^{-1/2} in front of the integral, as it is a constant? If so, I can integrate the rest easily.
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfriendsonfirex View Post
    Okay, so I've gotten as far as this:

    y=-2\pi\int_0^1(\frac{e^u}{\sqrt{2\pi}})du

    Can I bring the (2\pi)^{-1/2} in front of the integral, as it is a constant? If so, I can integrate the rest easily.
    yes, you can bring it out. and your limits would change actually. your integral becomes

    V = - \sqrt {2 \pi} \int_0^{- \frac 12}e^u~du

    Now finish up
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  15. #15
    Junior Member xfriendsonfirex's Avatar
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    Well, after integrating, I am getting the following:

    -\sqrt{2\pi}(1-e^{-1/8})

    However, my book key says I should be getting this:

    \sqrt{2\pi}(1-\frac{1}{\sqrt{e}})

    Any suggestions? Thanks for all the help so far.
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