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    I have this question which is rather interesting and quite hard IMO.

    I have to find the volumes of revolution generated by the regions A and B when they are each rotated about: the x axis and the y axis

    If the equation was y=x^n from a to b,then it means I need to find the volume of revolution for:

    • x = a to b, rotated around the x axis, which would be: \displaystyle\int^b_a \pi y^2 \, dx \Rightarrow \displaystyle\int^b_a \pi x^2^n \, dx
    • y = a to b, roated around the y axis, which would be: \displaystyle\int^b_a \pi x^2 \, dx \Rightarrow \displaystyle\int^b_a \pi y^\frac {2}{n}\, dx


    are those right?

    I also have to find this though and this is what I need help with:

    • x = a to b, rotated around the y axis
    • y = a to b, roated around the x axis


    How would you find those?
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    For the last two, if you have the y co-ordinates and need x co-ordinates, just plug the y co-ordinates into the equation to find the relevant x's. (or vice versa)
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    (Original post by TomLeigh)
    For the last two, if you need y co-ordinates and need x co-ordinates, just plug the y co-ordinates into the equation to find the relevant x's.
    I have the coordinates, I have to rotate it around the opposite axis. For example, if the function was y=x^2, most questions would ask for the volume of revolution from  x = 1 to  x = 3 , rotated around the x axis.

    My question is asking the exact same thing, except rotated around the y axis
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    Yes so find out what y is at x=1 and x=3 (y=1 and y=9) and use those y co-ordinates to rotate around y.
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    I can see the sense in that, and I don't want to undermine your math skills, but are you sure that is right?

    Because my book doesn't have answers so by doing this I wouldn't know if I get the right answer
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    Ah well I could easily be wrong don't worry. I've only just done this chapter but I just used my common sense mixed with the rules we learned in class. If you think about 2 points on an axis say, 3 and 2, with th graph say y=x^2, then you're not going to use the points (3,0) (2,0) as these would make an area of 0, and the only co-ordinates that you can use with these are the x,y co-ordinates, being (3,9) (2,4) If you get me
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    Anyone else?
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    you have a formula for y in terms of x. Presuming you've done integration by substitution you should be able to make this work by making a substitution for y or x depending on the situation. There may well be a quicker way that someone knows of but I believe this should lead you to the answer.
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    (Original post by JoMo1)
    you have a formula for y in terms of x. Presuming you've done integration by substitution you should be able to make this work by making a substitution for y or x depending on the situation. There may well be a quicker way that someone knows of but I believe this should lead you to the answer.

    I dont understand
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    For rotating around the y-axis, it would be \pi\displaystyle\int^b_a \pi y^\frac {2}{n}\, dx

    not

    \pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx.
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    (Original post by Rainfaery)
    For rotating around the y-axis, it would be \pi\displaystyle\int^b_a \pi y^\frac {2}{n}\, dx

    not

    \pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx.
    thats what i thought but if y=x^2 then y^\frac {1}{2}=x

    \pi\displaystyle\int^b_a y^2\, dx = \pi\displaystyle\int^b_a y^\frac {1}{2}^2\, dx = y^\frac {1}{4}

    hence,

    \pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx
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    (Original post by G O D I V A)
    thats what i thought but if y=x^2 then y^\frac {1}{2}=x

    \pi\displaystyle\int^b_a y^2\, dx = \pi\displaystyle\int^b_a y^\frac {1}{2}^2\, dx = y^\frac {1}{4}

    hence,

    \pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx

    No, when you have a power to a power, they are multiplied together. y^\frac {1}{2}=x If you square that, you just get y, because (1/2)^2 is just (2/2), which is one. This is just a basic property of indices.
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    Oh thanks for that.

    I still have to find though,

    * x = a to b, rotated around the y axis
    * y = a to b, roated around the x axis
 
 
 
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