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Solid with polar graph as base

  1. Nov 13, 2016 #1
    1. The problem statement, all variables and given/known data

    In this video , why for the xy plane projection , it's a circle with center = y = 1 , i can understand the r = 2sin theta ? why we cant ∬ r dr dtheta where , r = 1 , and with theta = 0 to 2 pi ?

    They are the same , right ? since volume = integration of area with z a-xis in this case...

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 13, 2016 #2

    Simon Bridge

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    because geometry ... that is the projection of the region being integrated over.
    The lecturer derives the equation on the board for you.

    Where did he lose you?
  4. Nov 14, 2016 #3
    i did in in another way , but didnt get the ans ....

    My ans is :

    Attached Files:

  5. Nov 14, 2016 #4

    Simon Bridge

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    Sure ... but not relevant.
    If you will not answer questions I cannot help you.

    note: hardly anyone reads pictures - I'm cutting you lots of slack just watching the video.
    I did look at the pic - it does not mean anything to me.

    In order for a bunch of symbols to mean anything, you have to explain the reasoning you used for doing that in the first place. If you will not explain your reasoning, nobody can help you.

    Have another go.
  6. Nov 17, 2016 #5
    which part you dont understand ?

    In the pictures , i calculate the area first m then i integrate the area over the height , so for the working at the left hand side , it's the area , for the right hand side , it's the integration of area over the height
  7. Nov 18, 2016 #6

    Ray Vickson

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    How about typing out a clear statement of the problem? I, for one, will not look at the video; I figure that if you want help you will take the time to explain your problem. (BTW: I am by no means alone here; most helpers will not look at your submission, and you are just plain lucky to have found one who did.)
    Last edited: Nov 18, 2016
  8. Nov 18, 2016 #7

    Simon Bridge

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    pretty much all of it. I can see you did some integrals... all integrals are areas. I dont know which areas you are calculating or what for. OR how it is relevant. Don't get me wrong, they are nice integrals, pretty, look good on camera. Not everyone can take photos of a sheet of maths and have them come out so nice, but I have to be psychic to know what you intended them to do or why so I can have any way to evaluate whether what you did was at all valid, never mind correct.

    When doing maths, also use words.

    ... the words need to explain your reasoning too.

    Start again from the top, annotate your working as you go... or just answer the question in post #2... then, maybe, some will be able to help.
  9. Nov 20, 2016 #8
    cant i just calculate the area first ? then i integrate the area of the base with the height to get volume , since this is a cylinder , so , the area of the 'circle ' is constant for every plane throughout the whole length ?

    What i mean is imagine a a cylinder is made of many thin planes (circle)stacked together to form a solid cylinder .... So, i calculate the area of base first , then i only multiply with the quantity of the thin planes(circle) , which is the height here

    So, area of thin planes (circle ) = pi(1^2) = pi , then integrate with the height , which if from z = 2y to z = (x^2) + (y^2) , is it wrong to do so ?
  10. Nov 20, 2016 #9

    Simon Bridge

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    Depends, which area?
    Are we looking at the same problem? vis: the volume to be calculated in post #1 is not a cylinder.

    I'm sorry - you have consistently failed to answer questions or follow sugestions - I cannot help you.
    Good luck.
  11. Nov 20, 2016 #10
    sorry , here's the question

    I'm sorry that i have posted the wrong question .
  12. Nov 21, 2016 #11

    Ray Vickson

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    You really do not want help, so you? You have been advised to NOT show the problem as a video or picture---just type it all out---but you have refused every time. Helpers will not bother to help you if you cannot even be bothered to type the problem. The choice is up to you.
  13. Nov 21, 2016 #12
    refer to the question here
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