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Math Help - Integrals using cylindrical shells when rotated on non-axis line

  1. #1
    Ife
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    Integrals using cylindrical shells when rotated on non-axis line

    I am getting a bit confused here.. the question asks to find the volume of the solid that results from rotating y^2=8x, x=2 around the line x=4.

    My setup is this: 2 \pi \int_{-4}^4(2-x)(\sqrt{8x}) dx

    but when i integrate and substitute i cant get values for the negative lower limit.. what's my error?
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  2. #2
    MHF Contributor
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    Try integratinf from 0-4 and multiply by another constant of 2.
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  3. #3
    Ife
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    i dont understand??
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  4. #4
    MHF Contributor Amer's Avatar
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    we want the volume the green volume as the picture shows us


    the function is y^2=8x

    y = \sqrt{8x}

    Integrals using cylindrical shells when rotated on non-axis line-1st.jpg

    to make it easier let new x=x+4 to make the volume around the y-axis

    so the new question will be

    y^2 = 8(x+4) and x=-2 around the y-axis

    so the picture will be like this

    Integrals using cylindrical shells when rotated on non-axis line-2cd.jpg

    and we want the volume revolution around y-axis
    note that y = \sqrt{8(x+4)} is not all the curve if we calculate the volume for y it will be half of what we need since it is symmetric
    see this
    Integrals using cylindrical shells when rotated on non-axis line-22cd.jpg

    the volume of this around y-axis

    \int _{-4}^{-2} 2\pi x \sqrt{8(x+4)} dx

    multiply it with 2 to get what you need

    any questions
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  5. #5
    MHF Contributor Amer's Avatar
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    or in other way it will equal

    \int_{-4}^{4} \pi \left( \frac{y^2}{8} -4 \right)^2 - 4 dy
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  6. #6
    Ife
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    Quote Originally Posted by Amer View Post
    we want the volume the green volume as the picture shows us


    the function is y^2=8x

    y = \sqrt{8x}

    Click image for larger version. 

Name:	1st.JPG 
Views:	10 
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ID:	16247

    to make it easier let new x=x+4 to make the volume around the y-axis

    so the new question will be

    y^2 = 8(x+4) and x=-2 around the y-axis

    so the picture will be like this

    Click image for larger version. 

Name:	2cd.JPG 
Views:	17 
Size:	25.7 KB 
ID:	16248

    and we want the volume revolution around y-axis
    note that y = \sqrt{8(x+4)} is not all the curve if we calculate the volume for y it will be half of what we need since it is symmetric
    see this
    Click image for larger version. 

Name:	22cd.JPG 
Views:	14 
Size:	22.7 KB 
ID:	16249

    the volume of this around y-axis

    \int _{-4}^{-2} 2\pi x \sqrt{8(x+4)} dx

    multiply it with 2 to get what you need

    any questions
    Thanks, but how did u get those limits??
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  7. #7
    MHF Contributor Amer's Avatar
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    limits is x=-4 and x=-2 since the region is bounded by this two lines

    Integrals using cylindrical shells when rotated on non-axis line-22cd.jpg

    Cylindrical shell says the volume generated by revolving a region bounded above by a continuous function and from the left by the line x=a and from the right by the line x=b given by

    v= \int_{a}^{b} \pi x f(x) dx

    in the second way I used washers
    look to the picture

    Click image for larger version. 

Name:	22cd.JPG 
Views:	15 
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ID:	16257

    if we look to the two curves

    y^2 = 8(x+4) \Rightarrow x_1 = \frac{y^2}{8} - 4

    and

    x_2=-2

    want the volume bounded by this to curves revolving around y-axis

    find the point of intersection

    -2 =  \frac{y^2}{8} - 4 \Rightarrow y=\mp 4

    so the volume will be

    \pi \int_{-4}^{4}  (x_1)^2 - (x_2)^2 dy

    \pi \int_{-4}^{4} \left( \frac{y^2}{8} -4 \right)^2 - (-2)^2 dy
    Last edited by Amer; April 7th 2010 at 07:23 PM.
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