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    (Original post by ABUSHOMAL///M3)
    plz if a Q. asks for volume of a curve rotated 360 or 180 , what do we use
    pie or 2pie for anyone plz answer me any1 now
    and thanks in advance
    rotated 360 degrees around x axis = π ∫ y² ∂x (relates to πr² )

    and 180 degrees I assume is half of that
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    plz if a Q. asks for volume of a curve rotated 360 or 180 , what do we use
    pie or 2pie for anyone plz answer me any1 now
    and thanks in advance


    plz very essential
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    (Original post by ABUSHOMAL///M3)
    plz if a Q. asks for volume of a curve rotated 360 or 180 , what do we use
    pie or 2pie for anyone plz answer me any1 now
    and thanks in advance
    pi for 360.

    pi/2 for 180 (although I have never seen it on a paper).

    Make sure you integrate the function squared, and if it is parametric this will involve: ∫y²(dx/dt)dt.

    Also if it is round the y axis be sure to swap all x's with y's and vice versa. So the limits will have to y coordinates, and you'll have to integrate with respect to y etc.
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    (Original post by Widowmaker)
    Well, i've just finished partial fractions, so I can help people with those, lol.
    Although it's a very short/easy chapter so noone will need my help.
    Easy but always on the exam papers for 5+ marks!
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    (Original post by samdavyson)
    pi for 360.

    pi/2 for 180 (although I have never seen it on a paper).

    Make sure you integrate the function squared, and if it is parametric this will involve: ∫y²(dx/dt)dt.

    Also if it is round the y axis be sure to swap all x's with y's and vice versa. So the limits will have to y coordinates, and you'll have to integrate with respect to y etc.
    many thanks man>
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    (Original post by samdavyson)
    pi for 360.

    pi/2 for 180 (although I have never seen it on a paper).

    Make sure you integrate the function squared, and if it is parametric this will involve: ∫y²(dx/dt)dt.

    Also if it is round the y axis be sure to swap all x's with y's and vice versa. So the limits will have to y coordinates, and you'll have to integrate with respect to y etc.
    if around y axis then x^2 dy ?
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    (Original post by ABUSHOMAL///M3)
    if around y axis then x^2 dy ?
    Yeah, there is a question on Edexcel's 2nd practice paper if you want a go.
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    (Original post by samdavyson)
    Remember: am/n = nth root of a to the power of m.

    So:

    ∫cosθ/(cos²θ )^3/2 dθ = ∫cosθ/(cos³θ) dθ = ∫1/(cos²θ) dθ

    => ∫sec²θ dθ => Standard Integral. ( => tanθ + C)
    Thanks Sam.
    So for the cos²θ^3/2 is it 2 x 3/2 to get 3? To make it cos³x.. Like multiplying the powers.. right?
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    (Original post by violet)
    Thanks Sam.
    So for the cos²θ^3/2 is it 2 x 3/2 to get 3? To make it cos³x.. Like multiplying the powers.. right?
    Yeah exactly that.

    Or you can think about it as the 2 in the 3/2 square roots the expression and the 2 in the cos²θ squares it. So overall ==> the twos cancel.

    => cos³x
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    (Original post by samdavyson)
    Yeah exactly that.

    Or you can think about it as the 2 in the 3/2 square roots the expression and the 2 in the cos²θ squares it. So overall ==> the twos cancel.

    => cos³x
    I think I'll use the multiplying powers way... It sounds easier
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    (Original post by violet)
    I think I'll use the multiplying powers way... It sounds easier
    Fair enough. They are both the same thing.
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    (Original post by samdavyson)
    Fair enough. They are both the same thing.
    Oh, thanks anyway.

    And also..... For Q5 on that paper, I can't get from the 2nd to last step to the last step in the answers.

    I get V = pi[(8 + 1/4 ln4) + 3)] but how do you make that pi[5 + 1/2 ln2] ?
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    (Original post by violet)
    Oh, thanks anyway.

    And also..... For Q5 on that paper, I can't get from the 2nd to last step to the last step in the answers.

    I get V = pi[(8 + 1/4 ln4) + 3)] but how do you make that pi[5 + 1/2 ln2] ?
    Your error is: it is minus 3, not add 3.

    When you have at minus 3:

    V = pi[(8 + 1/4 ln4) - 3]

    => V = pi[(8 + 1/4ln2²) - 3]

    => V = pi[(8 + 1/2ln2) - 3]

    => V = pi(5 + 1/2ln2)

    How was C3 by the way?
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    (Original post by samdavyson)
    Your error is: it is minus 3, not add 3.

    When you have at minus 3:

    V = pi[(8 + 1/4 ln4) - 3]

    => V = pi[(8 - 1/4ln2²) - 3]

    => V = pi[(8 - 1/2ln2) - 3]

    => V = pi(5 - 1/2ln2)

    How was C3 by the way?
    How come you have to change from +1/4 to -1/4

    It was ok i guess. I just scribbled out a lot and hope the examiner can still read it! How was it for you?
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    (Original post by violet)
    How come you have to change from +1/4 to -1/4

    It was ok i guess. I just scribbled out a lot and hope the examiner can still read it! How was it for you?
    I thought the paper was alright.

    I'd hope for an A on it.

    You have a point there with the minus. I integrated it to give a minus, which is wrong. I will edit it in a second.
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    I think now I just need to learn my integrals (http://153rd.com/sam/as/C4/).

    Then I should be ok.

    Do you know them?
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    (Original post by samdavyson)
    I think now I just need to learn my integrals (http://153rd.com/sam/as/C4/).

    Then I should be ok.

    Do you know them?
    Do we need to know all the ones on that page?
    I haven't done much C4 recently.. I did it ages ago and now I've forgotten it all and it suddenly seems so much harder!
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    (Original post by violet)
    Do we need to know all the ones on that page?
    I haven't done much C4 recently.. I did it ages ago and now I've forgotten it all and it suddenly seems so much harder!
    Just the ones that aren't marked with the red line.

    i.e. the first 6 lines.
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    (Original post by samdavyson)
    Just the ones that aren't marked with the red line.

    i.e. the first 6 lines.
    I don't ever remember learning the first one ever :confused:
    Lol, what is it?
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    (Original post by violet)
    I don't ever remember learning the first one ever :confused:
    Lol, what is it?
    Add one to the power, and divide by the new power!
 
 
 
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