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nice

  • 5 years ago

Double integrals in polar coordinates?? I Know nothing about it :S

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    Sometimes integrals are easier to evaluate using polar coordinates (r, θ), where x = r cos θ and y = r sin θ. that is the basic idea behind it

  3. nice
    • 5 years ago
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    aha,, and how I can apply it ? can you give me an example ?

  4. anonymous
    • 5 years ago
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    I am not that good with it, just looking at my lecture notes, but I can try

  5. anonymous
    • 5 years ago
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    The height of the hemisphere about the x-y-plane is given by z =sqrt(1 − x^2-y^2) Find the volume under the unit hemisphere with positive z-coordinate and positive y-coordinate.

  6. anonymous
    • 5 years ago
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    region of integration is 1> x^2+y^2 (as it has to be + under sqrt) and y>=0

  7. anonymous
    • 5 years ago
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    now first challenge is to change the region of integration

  8. nice
    • 5 years ago
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    aha...

  9. anonymous
    • 5 years ago
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    x^2+y^2 in polar is =r^2 (as sin^2 +cos^2=1)

  10. anonymous
    • 5 years ago
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    so r has to be between 0 and 1

  11. anonymous
    • 5 years ago
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    y>0 implies that rsinθ>0 r is always >0 here (0 to 1) so sinθ has to be >0 implies θ is from 0 to pi clear so far?

  12. nice
    • 5 years ago
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    aha ..

  13. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{\pi}\int\limits_{0}^{1}\sqrt{1-r ^{2}} drd \theta\]

  14. anonymous
    • 5 years ago
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    so this is what we have to do now

  15. anonymous
    • 5 years ago
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    it is not, because of the change of integral dxdy becomes r drdθ

  16. anonymous
    • 5 years ago
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    \[rsqrt{(1-r ^{2})} \] this is what you have to integrate

  17. anonymous
    • 5 years ago
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    do you know how?

  18. nice
    • 5 years ago
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    one minute, hoe it comes r from 0 to 1 ,, since r^2<1

  19. nice
    • 5 years ago
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    how**

  20. anonymous
    • 5 years ago
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    1-x^2-y^2 has to be + as it is under sqrt x^2+y^2= (rcosθ)^2+(rsinθ)^2=r^2 so 1-r^2 has to be +

  21. nice
    • 5 years ago
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    1-r^2>0 1>r^2 -1<r<1 ??

  22. anonymous
    • 5 years ago
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    u are right but the example says: positive z-coordinate and positive y-coordinate.

  23. nice
    • 5 years ago
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    ahaaa I get it

  24. anonymous
    • 5 years ago
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    the answer is 1/3 pi

  25. nice
    • 5 years ago
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    Thanks alot I understood the idea

  26. anonymous
    • 5 years ago
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    cool!

  27. nice
    • 5 years ago
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    I have another ques. ?

  28. anonymous
    • 5 years ago
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    do you or do you not? :)

  29. anonymous
    • 5 years ago
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    go on

  30. nice
    • 5 years ago
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    in another topic

  31. anonymous
    • 5 years ago
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    Im just a 1st year maths student, so my knowledge has limits :)

  32. anonymous
    • 5 years ago
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    but I can try

  33. nice
    • 5 years ago
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    how to find the projection of a=(1,3) on b=(4,0,2) ??

  34. anonymous
    • 5 years ago
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    no clue

  35. anonymous
    • 5 years ago
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    post it as a different question

  36. nice
    • 5 years ago
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    ok, thanks for trying :)

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