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Polar coor

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Use polar coords to evaluate the double integral x3 + xy2dydx from y = -(9-x2)1/2 to (9-x2)1/2, and x = 0 to 3

    2. Relevant equations



    3. The attempt at a solution

    So the region is a half circle of radius 3, centered @ the origin, with only the possitive x side, (right side of circle)

    x = rcosQ, y = rsinQ

    the integral is r3cos3Q + r3cosQsinQ drdQ

    after integration with respect to r i get and pluging in the limits

    84/4 cos^3 Q + 81/4 cosQsinQ dQ

    i looked up the integrals of cos^3Q in the back of my book and it is quite complex. Have i made any mistakes thus far, and is there an easier way to evaluate it
     
  2. jcsd
  3. May 6, 2009 #2

    dx

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    x³ + xy² = r² cosθ. Check your algebra.
     
  4. May 6, 2009 #3

    Cyosis

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    Also if you go from Cartesian to polar coordinates [itex]dxdy \Rightarrow r dr d\theta[/itex].
    While not relevant to this problem [itex]\cos^3x[/itex] is easily integrated by writing it as [itex]\cos x (1-\sin^2x)[/itex] and substituting [itex]u=\sin x[/itex].
     
  5. May 8, 2009 #4
    i think i messed up somehwere

    [tex]\int r^3cosQdrdQ[/tex] = .25r^4cosQ from 0 to 3
    =[tex]\int81/4 cosQ dQ[/tex] from 0 to pi = 81/4 sinQ from 0 to pi
    sin(0) = 0, sin (pi) = 0
    i got an answer of 0
     
  6. May 8, 2009 #5

    dx

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    Are you sure θ goes from 0 to π?
     
  7. May 13, 2009 #6

    Cyosis

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    Re: Where my mistake Polar coordinates

    You forgot that when you transform from Cartesian to polar coordinates you have to add the Jacobian r ([itex]dx dy \Rightarrow r dr d\theta[/itex]). Secondly your limits for [itex]\theta[/itex] are off. You are integrating over the upper semi circle, but that is not the right region to integrate over. Draw the circle and add the Cartesian limits, over what section should you be integrating?
     
  8. May 13, 2009 #7

    dx

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    Re: Where my mistake Polar coordinates

    Q should go from [tex] -\frac{\pi}{2} [/tex] to [tex] \frac{\pi}{2} [/tex].
     
  9. May 13, 2009 #8

    HallsofIvy

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    Re: Where my mistake Polar coordinates

    Actually, he did include the "r" but you are correct that his limits of integration on [itex]\theta[/itex] are wrong.
     
  10. May 13, 2009 #9

    Cyosis

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    Re: Where my mistake Polar coordinates

    You're right I completely missed the r.
     
  11. May 13, 2009 #10
    uhhh huh, ok i didnt realize that mattered, i figured as long as it was 180 degrees it would be the same

    but i think i did mess up the r

    x(x2 + y2) = rcosQ* r2 = r3cosQ


    then i add the jacobian r and my integral is for r4cosQdrdQ

    the only thing that changes is the 81/4 becomes 243/5

    so the final answer is

    243/5 + 243/5 = 486/5
     
  12. May 13, 2009 #11

    Cyosis

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    That's correct.
     
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