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Vector calculus for ellipse in polar coordinates

  1. May 21, 2013 #1
    Hello =]
    I'm having trouble with this question, can somebody please help me with it! I'll thanks/like your comment if help me =)

    ![Question][1]


    YGKhm.png

    I know that for a ellipse the parametric is x=a sin t , b= b cos t t:0 to 2pi (?)

    for part a) I drew up the graph but not sure if it's right. the circle have 1 radius and for the ellipse I able to find the x= 2
    (x-1)^2 =9
    sqrt(x-1)^2=sqrt9
    x-1=3
    x=4

    for y=2sqrt2
    y^2=8
    sqrty=sqrt8
    y=sqrt4 sqrt 2
    y= 2sqrt2 or 2.8

    So the region should be the circle? since the ellipse like cover the whole circle?
    Thank you very much for helping!
    Cheers.
     
  2. jcsd
  3. May 21, 2013 #2

    CAF123

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    That would be valid if the ellipse was centered at the origin, which is not the case here. (but only a slight adjustment needs to be made).

    There exists two solutions to each of these equations.

    The region is not exactly a circle, but the unit circle is completely contained within the ellipse.
     
  4. May 21, 2013 #3
    so for x it's 4 and -4 and for y its 2sqrt2 and -2sqrt 2 right?
    i did a sketch of it, by any chance you know how to do this question? =(


    Mod note: The image was way too large. Please scale the image down to about 900 x 600 pixels and edit this post to include it again, or add it in a new post in this thread.
     
    Last edited by a moderator: May 21, 2013
  5. May 21, 2013 #4

    CAF123

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    No, for the first equation, you solve ##(x-1)^2 = 9## so ## x-1 = \pm 3## What are the two solutions?
    Similarly for the other equation.

    Your sketch shows the ellipse concentric with the unit circle, but this is incorrect since the ellipse has a centre (1,0). When you find the correct solutions to the above, it should fix things.
     
  6. May 21, 2013 #5

    LCKurtz

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    @dragonxhell: To get started on this problem you need to substitute the polar coordinate values for ##x## and ##y## into the equation of the ellipse and solve it for ##r##. That should lead you to the answer for part b. Then you will use the form$$
    \int_0^{2\pi} \int_{r_{inner}}^{r_{outer}} \frac 1 {r^3}r\,drd\theta$$
     
  7. May 21, 2013 #6
    o so the centre is (1,0) and the equation you show me I have already obtained
    But don't you move the 1 over? To just simply get x by it self? So it should be 4 and -3 (-4+1?)
     
  8. May 21, 2013 #7
    Is this right? Can somebody double check this for me? It seem right to me...


    (x−1)²/9 + y²/8 = 1
    8(x−1)² + 9y² = 72
    8x² − 16x + 8 + 9y² = 72
    9(x²+y²) − x² − 16x + 8 = 72
    9(x²+y²) = x² + 16x + 64
    9(x²+y²) = (x + 8)²
    9r² = (r cosθ + 8)²3r = r cosθ + 88 + r cosθ = 3r
    a = 8, b = 3
    c)R is region outside circle: r = 1*and inside ellipse: r = 8/(3−cosθ)
    Limits:*1 ≤ r ≤ 8/(3−cosθ)
    0 ≤ θ ≤ 2πƒ
    = 1/(x²+y²)^(3/2)
    = 1/(r²)^(3/2)
    = 1/r³∫∫ ƒ dA = ∫ [0 to 2π] ∫ [1 to 8/(3−cosθ)] 1/r³ * r dr dθR. . . . .
    = ∫ [0 to 2π] ∫ [1 to 8/(3−cosθ)] 1/r² dr dθ. . . . .
    = ∫ [0 to 2π] (−1/r) | [1 to 8/(3−cosθ)] dθ. . . . .
    = ∫ [0 to 2π] (−(3−cosθ)/8 − (−1)) dθ. . . . . = ∫ [0 to 2π] 1/8 (cosθ + 5) dθ. . . . .
    = 1/8 (sinθ + 5θ) | [0 to 2π]*. . . . .
    = 1/8 [(0 + 10π) − (0 + 0)]. . . . .
    = 5π/4

    Also for part a) the R region it is the region between the ellipse to the circle right?
     
  9. May 21, 2013 #8

    LCKurtz

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    Yes. Very nice.
     
  10. May 22, 2013 #9
    Can you please check this question for me because my answer is different to my friends and they say its wrong...
    Given the equation: G= (x^3 -3xy^2)I +(y^3-3x^2y)j +2k is conservative find the potential function for the fields that are conservative
    My answer
    f:grad f=F
    fx=x^3-3xy^2
    fy=y^3-3x^2y
    fz=z

    integral (x^3-3xy^2)dx
    =x^4/4-3x^2y^2/2 +h(y, z)
    fy or f'y= -3x^2 y +hy(y, z)
    fy=y^3-3x^2y= -3x^2y+hy(y, z)
    hy(y,z)=y^3

    Integral y^3 dy = y^4/4 +gz
    fz=0 because no z term ^
    Integral 0dz =c
    F= x^4/4-3x^2y^2/2 +y^4/4+c <-potential function
     
  11. May 22, 2013 #10

    LCKurtz

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    There is no F anywhere. Don't you mean grad f = G?

    Is ##f_z = z## or ##f_z = 2##? Look at G.
    Poorly written. At this point you have$$
    f = \frac {x^4} 4 - \frac {3x^2y^2} 2 +\frac{y^4} 4 + h(z)$$
    Now you need to take ##f_z## and set it equal to the third term, whether it is a ##2## or a ##z##. You don't get ##f_z=0## either way.
     
  12. May 22, 2013 #11
    sorry I type it all up on my phone and when I submit it it's so messy....
    so you are saying
    my integral y^4/4 must equal to fz which is z
    So the final answer is
    F=x^4/4 -3x^2y^2/2 +y^4/4 +z^2/2
     
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