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Math Help - Spinning around the y-axis

  1. #1
    Member Jones's Avatar
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    Spinning around the y-axis

    lo'

    The area that is limited by the curve y=x^-2 and y=1, y=e gets to rotate around the y axis detirmine the volume of the figure.

    The problem: How do i get the integration values? since it rotates around the y-axis i can't use the x values, right?
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  2. #2
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    Hello, Jones!

    The area that is limited by the curve y = \frac{1}{x^2},\;y = 1,\;y = e is rotated about the y-axis.
    Detirmine the volume of the figure.

    The problem: How do i get the integration values?
    Since it rotates around the y-axis, i can't use the x values, right?
    You can, but it's tricky.
    Code:
                  |
                  |*
                  |
                  | *
                e +  *
                  |::::*
                  |::::::::*
                1 + - - - - - - -*
                  |                           *
            - - - + - - - - - - - - - - - - - - - -
                  |

    We can integrate with respect to y\!:\;\;V \;=\;\pi\int^e_1x^2\,dy

    Since y = \frac{1}{x^2}, we have: . x = \frac{1}{\sqrt{y}}

    Then: . V \;=\;\pi\int^e_1\left(\frac{1}{\sqrt{y}}\right)^2d  y \;=\;\pi\int^e_1\frac{dy}{y} \;= \;\pi\ln(y)\,\bigg]^e_1<br />

    Therefore: . V \;=\;\pi\ln(e) - \pi\ln(1)\;=\;\pi(1) - \pi(0) \;= \;\boxed{\pi}

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  3. #3
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    You could also use the 'shells' method.

    2{\pi}\int_{e^{\frac{-1}{2}}}^{1}\frac{1}{x}dx=\boxed{\pi}
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  4. #4
    Member Jones's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Jones!

    We can integrate with respect to y\!:\;\;V \;=\;\pi\int^e_1x^2\,dy
    why  x^2 ?
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  5. #5
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    Quote Originally Posted by Jones
    why  x^2 ?

    The same reason that a region revolved about the x-axis is:

    . . V\;=\;\pi \int^b_a y^2\,dx

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  6. #6
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    Quote Originally Posted by Jones View Post
    why  x^2 ?

    Hello,

    the volume of a solid which was generated(?) by rotation around the y-axis could be considered as assembled of small cylindrical slices. (See attachment)
    The Volume of a cylinder is V=\pi \cdot r^2 \cdot h. In this case r correspond with x and the height of the cylinder correspond with y. A small part of the complete volume could be calculated:

    \Delta V=\pi \cdot x^2 \cdot \Delta y. Divide by \Delta y and you'll get:

    \frac{\Delta V}{\Delta y}=\pi \cdot x^2. Now let \Delta y approach to zero to get very, very thin slices:

    \lim_{\Delta y \rightarrow \infty}{\frac{\Delta V}{\Delta y}}=\frac{dV}{dy}=\pi \cdot x^2.This is the rate of change of the volume with respect of y.

    To get the complete volume you have to run up all thin cylinders which have the volume dV. You get this term by multiplying the rate of change by dy:

    dV=\pi \cdot x^2 \cdot dy. Running up means you have to calculate the sum. Use the big S to add all cylinders:

    \int (dV) \cdot dy = V. Thus:

    V=\int \pi \cdot x^2 \cdot dy=\pi \int   x^2 \cdot dy

    and here we are!

    EB

    PS.: This is not a rigorous proof or something like that. This post should sho, where the x comes from - and maybe it helps a little bit to understand what integrating means (in the beginning).
    Attached Thumbnails Attached Thumbnails Spinning around the y-axis-rot_vol1.gif  
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  7. #7
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    Quote Originally Posted by earboth View Post
    PS.: This is not a rigorous proof or something like that. This post should sho, where the x comes from - and maybe it helps a little bit to understand what integrating means (in the beginning).
    I seem to have a strange effect on people. Whenever, I am around people always have to mention word "not rigorous". Both my math professors mention that and always look at me when they say that. Even my engineering professor when he makes some argument in class always looks at me and say "this is not a rigorous explaination, but...". To remind you there can be no rigorous explanation to this problem. This is some applied problem. To be rigorous we need a defintion of volume. But we do not. So whenever you deal with areas, volumes, centroids.... All these things do not have proofs. Because they cannot be formally defined.
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  8. #8
    Member Jones's Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    the volume of a solid which was generated(?) by rotation around the y-axis could be considered as assembled of small cylindrical slices. (See attachment)
    The Volume of a cylinder is V=\pi \cdot r^2 \cdot h. In this case r correspond with x and the height of the cylinder correspond with y. A small part of the complete volume could be calculated:

    \Delta V=\pi \cdot x^2 \cdot \Delta y. Divide by \Delta y and you'll get:

    \frac{\Delta V}{\Delta y}=\pi \cdot x^2. Now let \Delta y approach to zero to get very, very thin slices:

    \lim_{\Delta y \rightarrow \infty}{\frac{\Delta V}{\Delta y}}=\frac{dV}{dy}=\pi \cdot x^2.This is the rate of change of the volume with respect of y.

    To get the complete volume you have to run up all thin cylinders which have the volume dV. You get this term by multiplying the rate of change by dy:

    dV=\pi \cdot x^2 \cdot dy. Running up means you have to calculate the sum. Use the big S to add all cylinders:

    \int (dV) \cdot dy = V. Thus:

    V=\int \pi \cdot x^2 \cdot dy=\pi \int   x^2 \cdot dy

    and here we are!
    This was a very good explanation, why can't they give similar explanations in the book?

    Just to make sure i have understood this i'll try to solve this one.

    The area limited by the x and y axis and the line x+2y-4=0 rotates around the x axis. Determine the volume.

    so my suggestion is that we start of by substituting for x.
    x=-2y+4
    V= pi \int^0_{-4} (-2y+4)^2 dy

     <==> (-2/3y+4/3)^3
    Last edited by Jones; December 4th 2006 at 03:42 PM.
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  9. #9
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    Quote Originally Posted by Jones View Post
    This was a very good explanation, why can't they give similar explanations in the book?
    Thank you. I like this kind of response.



    Quote Originally Posted by Jones View Post
    Just to make sure i have understood this i'll try to solve this one.

    The area limited by the x and y axis and the line x+2y-4=0 rotates around the x axis. Determine the volume.

    so my suggestion is that we start of by substituting for x.
    x=-2y+4

    V= pi \int^0_-4 (-2y+4)^2 dy
     <==> (-2/3y+4/3)^3
    Hello,

    the axis of rotation is the x-axis, which is perpendicular to the base area of the solid. Thus the radius of the circle (= base area) is y and the "thickness" of the cylinders is dx. So you get:

    \int_{0}^{4} \pi \cdot y^2\cdot dx=\int_{0}^{4} \pi \cdot (\frac{1}{2}x+2)^2\cdot dx=\pi \int_{0}^{4} (-\frac{1}{4}x^2-2x+4)^2\cdot dx

    I'll leave the rest for you.

    EB

    PS.: With this problem it isn't necessary to use integration:
    You've got a right triangle which rotates around one leg. This will give a cone.
    The radius of the base is the y-intercept of the straight line: (0, 2)
    The height of the cone is the zero of the straight line: (4, 0)
    The volume of a cone can be calculated by:
    V_{cone}=\frac{1}{3}\cdot \pi \cdot r^2 \cdot h. Now plug in all values you know:

    V_{cone}=\frac{1}{3}\cdot \pi \cdot (2)^2 \cdot 4=\frac{1}{3}\cdot \pi \cdot 4 \cdot 4=\frac{16}{3}\cdot \pi
    That's the result, you should get with the integration too.
    Last edited by earboth; December 4th 2006 at 11:08 AM. Reason: additional remarks
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