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Thread: Need help setting up a shells problem

  1. #1
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    Need help setting up a shells problem




    EDIT: Please excuse my typo, I meant integrate from 1 to 5.
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  2. #2
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    Since you are rotating around the $\displaystyle \displaystyle y$ axis, this will be a $\displaystyle \displaystyle dy$ integral.

    The graphs intersect where the equations are equal...

    $\displaystyle \displaystyle x = 1 + (x-3)^2$

    $\displaystyle \displaystyle x = 1 + x^2 - 6x + 9$

    $\displaystyle \displaystyle 0 = x^2 - 7x + 10$

    $\displaystyle \displaystyle 0 = (x - 2)(x - 5)$

    $\displaystyle \displaystyle x = 2$ or $\displaystyle \displaystyle x = 5$.

    When $\displaystyle \displaystyle x = 5, y = 2$, so they intersect at $\displaystyle \displaystyle (x,y) = (5, 2)$.


    The volume is calculated by rotating the area around the $\displaystyle \displaystyle y$ axis.

    This area needs to be thought of as a series of rectangles, which when rotated, become cylinders.

    The length of each rectangle is $\displaystyle \displaystyle (y + 3) - (1 + y^2) = 2 + y - y^2$ and the width is $\displaystyle \displaystyle dy$, a small change in $\displaystyle \displaystyle y$.

    When rotated, each cylinder has a radius the same as the length of each rectangle, so the circular cross-sectional area is $\displaystyle \displaystyle \pi r^2 = \pi (2 + y - y^2)^2$. The height of each cylinder is the same as the width of each rectangle, $\displaystyle \displaystyle dy$.

    Therefore the volume of each cylinder is $\displaystyle \displaystyle \pi (2 + y - y^2)^2\,dy$.

    So the entire volume can be approximated by $\displaystyle \displaystyle \sum{\pi(2+y-y^2)^2\,dy}$, and when you increase the number of cylinders, this sum converges on an integral and the approximation becomes exact.

    So $\displaystyle \displaystyle V = \int_0^2{\pi (2 + y - y^2)^2\,dy}$.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Since you are rotating around the $\displaystyle \displaystyle y$ axis, this will be a $\displaystyle \displaystyle dy$ integral.

    The graphs intersect where the equations are equal...

    $\displaystyle \displaystyle x = 1 + (x-3)^2$

    $\displaystyle \displaystyle x = 1 + x^2 - 6x + 9$

    $\displaystyle \displaystyle 0 = x^2 - 7x + 10$

    $\displaystyle \displaystyle 0 = (x - 2)(x - 5)$

    $\displaystyle \displaystyle x = 2$ or $\displaystyle \displaystyle x = 5$.

    When $\displaystyle \displaystyle x = 5, y = 2$, so they intersect at $\displaystyle \displaystyle (x,y) = (5, 2)$.


    The volume is calculated by rotating the area around the $\displaystyle \displaystyle y$ axis.

    This area needs to be thought of as a series of rectangles, which when rotated, become cylinders.

    The length of each rectangle is $\displaystyle \displaystyle (y + 3) - (1 + y^2) = 2 + y - y^2$ and the width is $\displaystyle \displaystyle dy$, a small change in $\displaystyle \displaystyle y$.

    When rotated, each cylinder has a radius the same as the length of each rectangle, so the circular cross-sectional area is $\displaystyle \displaystyle \pi r^2 = \pi (2 + y - y^2)^2$. The height of each cylinder is the same as the width of each rectangle, $\displaystyle \displaystyle dy$.

    Therefore the volume of each cylinder is $\displaystyle \displaystyle \pi (2 + y - y^2)^2\,dy$.

    So the entire volume can be approximated by $\displaystyle \displaystyle \sum{\pi(2+y-y^2)^2\,dy}$, and when you increase the number of cylinders, this sum converges on an integral and the approximation becomes exact.

    So $\displaystyle \displaystyle V = \int_0^2{\pi (2 + y - y^2)^2\,dy}$.

    Thank you again for helping me with this type of problem. I'm glad you didn't get confused at the beginning where it said x = 1 + y^2 y = x - 3. The comma I typed in somehow became a superscript so it didn't separate the two equations. The corrected way should be x = 1 + y^2, y = x - 3 but it doesn't matter because you got the equation right anyways. Kudos to you.
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  4. #4
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    Wait a minute, are you sure that is the correct setup? I thought you had to set it up like this(disk method),
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  5. #5
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    Quote Originally Posted by florx View Post
    Wait a minute, are you sure that is the correct setup? I thought you had to set it up like this(disk method),
    Read what I wrote. Are there any mistakes in my logic? (Seriously, it helps having someone else proof-read).

    Don't just blindly apply formulae, logic is always more powerful.
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