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Double integrals using polar co-ordinates

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data

    fp7dlf.jpg



    Step 1) I put the following into polar coordinates
    √(16-x2-y2)=√16-r2

    Where r≤4

    step 2 I solved for y in the original problem which is in the link

    y≤√(4-x2)

    step 3. I graphed the above function

    step 4. I put the above function in polar coordinates getting

    r≤2

    So I already had an idea that the graph would be from

    0≤r≤2 dr However my d(theta) was incorrect I got that it would be 0≤theta≤∏ as my interpretation of the graph shows. The answer however shows 0≤theta≤2∏

    Which does't make much sense to me. Where did I got wrong?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 3, 2012 #2

    tiny-tim

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    what is the actual question? :confused:
     
  4. Nov 3, 2012 #3
    Upper left hand corner of the graph, the problem statement says Use a multiple integral and a convenient coordinate system to find the volume of the solid. I forgot to type it up I apologize.
     
  5. Nov 3, 2012 #4

    tiny-tim

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    i still don't get it

    x2+ y2 ≤ 4 isn't a circle, it's a solid cylinder

    anyway, show us how you found the line of interection​
     
  6. Nov 3, 2012 #5
    Honestly I don't know I solved as if it was a circle.
     
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