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Math Help - Rotation around y and x axis + shifting the fuction

  1. #1
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    Rotation around y and x axis + shifting the fuction

    The first problem states to set up a formula for the volume which u get when the area limited by y=X+1 and y=5 rotates around y=5.

    The second problem states to set up a formula for the volume when the area limited by X=2 and y=(8X)^(1/2 rotates around x=2.

    My progress on the first part is F(x)=pi*f(x)^2 integrated from -2 to 2 and I realize that I somehow should change the formula of f(x) so it rotates around y=5. And this is where im stuck. I know the answer but I have problems understanding it entirely. I tought that I would just add + 5 in f(x) which is wrong

    On the second part I basically arrive at the same problem. I change it to y variable and get the answer F(y)=pi*f(y)^2 integrated from 0 to 4.
    . And then I somehow need to change f(y) so it rotates around 2. And I tought that I just add -2 to the formula in y=(8x)^(1/2) which is incorrect.

    I appreciate all theoretical help that I can recive. The reason why I have come to my conclusions is because that if y=X and I want to shift it two steps to the right then I write y=(X-2) and if I want to shift it 5 steps up I add +5 so I get y=X+5. But when I tried to apply this to the volume rotation problems I arrive at the wrong answer.

    ps. I was trying to add my integrals from a word pad document but it did not work. How do you guys add youre integrals?

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  2. #2
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    Quote Originally Posted by Zamzen View Post
    The first problem states to set up a formula for the volume which u get when the area limited by y=X+1 and y=5 rotates around y=5.
    method of disks, taking advantage of symmetry ...

    \displaystyle V = 2\pi \int_0^2 [5-(x^2+1)]^2 \, dx

    The second problem states to set up a formula for the volume when the area limited by X=2 and y=(8X)^(1/2 rotates around x=2.
    method of disks w/r to y ...

    \displaystyle V = \pi \int_0^4 \left(2-\frac{y^2}{8}\right)^2 \, dy

    method of cylindrical shells ...

    \displaystyle V = 2\pi \int_0^2 (2-x) \cdot \sqrt{8x} \, dx
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  3. #3
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    Thank you. I can use the cylindrical shell and disc method up until the point where i have to rotate them around a certain point. For example the first one that you solved. When i add the 5 in the function, is it moved upwards with 5 units so it rotates around y = 5?

    And on the second problem my textbook says that it should be (y^2)/8 -2. And the cylindrical method you used is very nice.

    But i still dont understand the theory behind rotating it around a certain point to 100%. In ex b, u managed to change it so it rotates around x=2 by adding a 2 in the function. And the cylindrical shell method you used with (2-x) suprises me even more. If you can explain the theory how you shift to a point or recomend a website where i can read about it i would be really thankful.
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  4. #4
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    note the red horizontal distance = radius of rotation of the vertical blue strip to form the cylindrical shell ...

    what is the distance of that radius in terms of x ?
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    r=2-x
    x=1
    r=1
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  6. #6
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    Quote Originally Posted by Zamzen View Post
    r=2-x
    x=1
    r=1
    see where the (2-x) came from now?
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  7. #7
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    yep.
    Is it possible to see some problems about rotations around an axis involving around a certain point? i dont have so many in my text book. Or a website that describes it more clearly. Ive been introduced to it recently and i want to understand more about it.
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  8. #8
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    recommend you perform a "google" search ...

    calculus volumes of revolution
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