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Ellipse sector area

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    i want to derive a formula for an ellipse sector. ellipse is not rotated and its center is in the origin. its semimajor and semiminor axis are a and b, respectively, and angle of the sector begins with t1 and ends with t2.

    it's just a simple surface integral in polar coordinates, i did that already and got the result.
    but then i found this:
    http://mathforum.org/library/drmath/view/53635.html"
    i also have noticed that at t1 = 0 and t2 = 2pi, the area calculated from that formula does not match A = pi*a*b.

    there is a correction for the answer given by doctor sam just below the next post, but i have no clue where did that dTheta come from, i can't see a reason why a normal integration wouldn't give correct results.
    what did i miss?

    for calculating the area of the whole ellipse, we used the following transformation at lessons:
    x = a r cos(fi)
    y = b r sin(fi)
    J (jacobi determinant) = abr
    this thing works only for 0 < fi < 2*pi.
    why?
    why none of our professors mentioned any of these problems?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Nov 9, 2009 #2

    Dick

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    As they said on mathforum, in the formula r^2*d(theta), theta has to be the angle in the ellipse. Using x=A*cos(theta), y=B*sin(theta), theta is the angle in the circle you are mapping from, not the angle in the ellipse. The angles in the ellipse are distorted. To find the angle in the ellipse (call it Theta) you use tan(Theta)=y/x. That's tan(Theta)=B*sin(theta)/(A*cos(theta)) or Theta=arctan(B*tan(theta)/A). Now you find dTheta by differentiating. You'll find the expression they gave is wrong. It should be (sec(theta)^2*B/A)/(tan(theta)^2*B^2/A^2+1). You can now find the area in polar coordinates by integrating (A^2*cos(theta)^2+B^2*sin(theta)^2)*dTheta. It looks complicated but the trig function parts actually cancel.
     
  4. Nov 9, 2009 #3
    ok so that's why integration over 0 < t < 2*pi works with x = a r cos(t) etc. am i right?

    but just why the "angles are distorted", angle is an angle, only r(t) changes, but that has been taken into account... hum... or not. ?

    why the angles aren't distorted in other cases, such as this:
    http://valjhun.fmf.uni-lj.si/~mihael/fs/mat2/pdfizpiti/280809.pdf": see page 3, (it's not in english, but anyway), it's first part is transformation from G (x,y) to G (r, fi) and then integrating to get area of G. how can i tell when to use the simple x = a r cos(fi) and when i need to do what you said? or is just an ellipse a special case?
    thanks for your reply.
     
    Last edited by a moderator: Apr 24, 2017
  5. Nov 9, 2009 #4
    oh yes, and where did the jacobi determinant go in the mathforum case?
     
  6. Nov 9, 2009 #5

    Dick

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    If you are really working in true polar coordinates, you don't need an explicit jacobian, it's built into the formula. Also notice not all angles are distorted, 0 and 2pi are still in the right place (as are pi/2, pi, etc). But take the point at theta=pi/4. If A is not equal to B, then that's not sitting at polar angle pi/4 in the ellipse. Draw a picture to convince yourself.
     
  7. Nov 9, 2009 #6
    how does the jacobian looks if i wanted to calculate this for any angle? it's not any different than abr, why would it be?
    i guess i need to read some stuff on this topic, all this is really confusing.

    thanks for your replies.
     
  8. Nov 9, 2009 #7

    Dick

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    In this case the jacobian doesn't depend on theta or Theta. But that's just this problem. Yes, read up and practice some more examples.
     
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