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Math Help - Spinning around x axis

  1. #1
    Member Jones's Avatar
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    Spinning around x axis

    Hi,

    Im having trouble solving a really simple integration problem.

    R is the finite region bounded by x=y, x=4y-y^2 are rotated around the x axis. Determine the volume of the solids.

    Okay
    The line and the curve interect at Y=0 and Y=4 so these are our upper and lower limits of integration.

    A=\pi\int_{4}^0 y(4y-y^2-y) \rightarrow \pi\int_{4}^0 4y^2-y^3-y^2


    =\frac{4y^3}{3}-\frac{y^4}{4}-\frac{y^3}{3}

    But the answer is zero and it's wrong
    What am i doing wrong?
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi,

    Im having trouble solving a really simple integration problem.

    R is the finite region bounded by x=y, x=4y-y^2 are rotated around the x axis. Determine the volume of the solids.

    Okay
    The line and the curve interect at Y=0 and Y=4 so these are our upper and lower limits of integration.

    A=\pi\int_{4}^0 y(4y-y^2-y) \rightarrow \pi\int_{4}^0 4y^2-y^3-y^2


    =\frac{4y^3}{3}-\frac{y^4}{4}-\frac{y^3}{3}

    But the answer is zero and it's wrong
    What am i doing wrong?
    Not sure how you got an intersection at 4, nor the

    (4y-y^2-y) (Edit: no, this bit is fine, silly me!)

    ... maybe the two are connected? (Edit: no.)

    Surely,

    x = y and x = 4y - y^2

    give you y = 0, 3

    ? ...
    Last edited by tom@ballooncalculus; July 6th 2009 at 09:05 AM.
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  3. #3
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    I guess you are using shells to solve the porblem. I dont know how to use that method. However, tom@ballooncalculus is correct. The intersection points are:

    (0,0) and (3,3)

    because:

    1. x = y
    2. x = 4y-y^2
    y = 4y-y^2
    y = y(4-y)
    1 = 4-y
     <br />
y = 3<br />
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  4. #4
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    Quote Originally Posted by Jones View Post
    Hi,

    Im having trouble solving a really simple integration problem.

    R is the finite region bounded by x=y, x=4y-y^2 are rotated around the x axis. Determine the volume of the solids.

    Okay
    The line and the curve interect at Y=0 and Y=4 so these are our upper and lower limits of integration.

    A=\pi\int_{4}^0 y(4y-y^2-y) \rightarrow \pi\int_{4}^0 4y^2-y^3-y^2


    =\frac{4y^3}{3}-\frac{y^4}{4}-\frac{y^3}{3}

    But the answer is zero and it's wrong
    What am i doing wrong?
    You're using the disks method to evaluate this integral.

    First you need to note that the graphs intersect at y = 0 and y = 3. So your LOWER limit is 0 and your UPPER limit is 3.

    You also need to note that 4y - y^2 \geq y in the range 0 \leq y \leq 3.


    So the integral you need to set up is...

    \pi \int_0^3{(4y - y^2)^2 - y^2\,dy}

     = \pi \int_0^3{16y^2 - 8y^3 + y^4 - y^2\,dy}

     = \pi \int_0^3{y^4 - 8y^3 + 15y^2\,dy}

     = \pi \left[\frac{1}{5}y^5 - 2y^4 + 5y^3\right]_0^3

     = \pi \left[\frac{243}{5} - 162 + 135\right]

     = \pi \left[\frac{108}{5}\right]

     = \frac{108\pi}{5}\,\textrm{units}^3
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  5. #5
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    Quote Originally Posted by Prove It View Post
    You're using the disks method to evaluate this integral.
    I didn't think s/he was... Seems to me we're integrating perpendicular to the axis of revolution... Solid of revolution - Wikipedia, the free encyclopedia


    ____________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus Forum
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  6. #6
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    Hello, Jones!

    R is the region bounded by x\,=\,y,\;x\,=\,4y-y^2 is rotated about the x axis.
    Determine the volume of the solid.
    Code:
              |
              *               *
              |       *     *
              |           o(3,3)
              |         *::::*
              |       *::::::::o(4,2)
              |     *::::::::* 
              |   *:::::::*
              | *:::::* 
          - - o - - - - - - - - - - - - - -
              |


    Intersection: . y \:=\:4y - y^2 \quad\Rightarrow\quad y^2-3y\:=\:0 \quad\Rightarrow\quad y(y-3)\:=\:0

    . . The graphs intersect at (0,0) and (3,3).


    "Shells" Formula: . V \;=\;2\pi\int^b_a y\left(x_{\text{right}} - x_{\text{left}}\right)\,dy

    We have: . V \;=\;2\pi\int^3_0y\bigg[(4y-y^2) - y\bigg]\,dy \;=\;2\pi\int^3_0\bigg[3y^2 - y^3\bigg]\,dy


    Now finish up . . .

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  7. #7
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    Quote Originally Posted by Prove It View Post
    You're using the disks method to evaluate this integral.

    First you need to note that the graphs intersect at y = 0 and y = 3. So your LOWER limit is 0 and your UPPER limit is 3.

    You also need to note that 4y - y^2 \geq y in the range 0 \leq y \leq 3.


    So the integral you need to set up is...

    \pi \int_0^3{(4y - y^2)^2 - y^2\,dy}

     = \pi \int_0^3{16y^2 - 8y^3 + y^4 - y^2\,dy}

     = \pi \int_0^3{y^4 - 8y^3 + 15y^2\,dy}

     = \pi \left[\frac{1}{5}y^5 - 2y^4 + 5y^3\right]_0^3

     = \pi \left[\frac{243}{5} - 162 + 135\right]

     = \pi \left[\frac{108}{5}\right]

     = \frac{108\pi}{5}\,\textrm{units}^3
    But he is rotating about the x-axis, not the y-axis, so, how can you solve it with the disks method and still integrate along the y-axis?

    Soroban's solution looks correct.
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