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Area of ellipse integral.

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    The area of the ellipse [tex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1[/tex] is given by
    [tex]\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx[/tex] . Compute the integral.


    2. Relevant equations



    3. The attempt at a solution
    I know I have to let [tex]x=asin\theta[/tex] and then [tex]dx=acos\theta[/tex] . Then I plug in x. Do I need to solve for a and plug that in, or am I really over thinking this?
    Thanks in advance.
     
  2. jcsd
  3. Oct 13, 2009 #2

    Mark44

    Staff: Mentor

    [tex]dx=acos\theta d\theta[/tex]
    Don't omit the differential.

    You don't "plug in x"; you replace x and dx in your original integral with the expressions that involve theta and its differential. Maybe that's what you meant, but it isn't what you said.
    You can't solve for a, so yes, you are overthinking this.
     
  4. Oct 13, 2009 #3

    zcd

    User Avatar

    a is a constant (semimajor axis), so there's no need to solve for it.
     
  5. Oct 13, 2009 #4
    Oops I had the [tex]d\theta[/tex] written down but forgot to type it.
    Yes, that is what I meant, but since this is a message board I didn't think it'd be a big deal. Thanks for setting me straight about thinking too much :).

    Ok I worked it out, but I'm not really sure what kind of answer I'm supposed to be getting.
    Here's the work(if anything needs to be explained better please say so).

    [tex]Let x=asin\theta[/tex] [tex]Then dx=acos\theta d\theta[/tex]
    [tex]\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-(asin\theta) ^{2}} acos\theta d\theta[/tex]
    [tex]=\frac{4b}{a}\int_{0}^{a}\sqrt{1-sin^{2} \theta} acos\theta d\theta[/tex]
    Pythagorean ID, square root and then multiply
    [tex]=\frac{4b}{a}\int_{0}^{a}a^{2}cos^{2}\theta d\theta[/tex]
    [tex]=\frac{4ba^{2}}{a}\int_{0}^{a}cos^{2}\theta d\theta[/tex]
    Half-Angle
    [tex]=4ab \int_{0}^{a} \frac{1}{2} \left ( 1+cos2\theta \right ) d\theta[/tex]
    Used substitution.
    [tex]=\left [ 2ab\theta \right ]_{0}^{a}+[absin2\theta]_{0}^{a}[/tex]

    [tex]sin\theta=\frac{x}{a}[/tex] and [tex]\theta=arcsin\frac{x}{a}[/tex]

    [tex]=\left [ 2ab*arcsin\frac{x}{a} \right ]_{0}^{a}+[ab\frac{x}{a}]_{0}^{a}[/tex]

    Since [tex]arcsin(1)=\frac{\pi }{2}[/tex] and [tex]arcsin(0)=0[/tex]
    [tex]=2ab\frac{\pi }{2}+ab=\pi ab+ab[/tex]
     
  6. Oct 13, 2009 #5

    Mark44

    Staff: Mentor

    It looks like you did fine right up to the end where you evaluated your antiderivative at 0 and a. After finding the antiderivative (a function of theta), you need to undo your substitution, and get the antiderivative back in terms of the original variable, x.

    Alternatively, you can change the limits of integration from values of x to values of theta. x = 0 ==> theta = 0, so that one is easy, x = a ==> theta = sin-1(a/a) = sin-1(1) = pi/2.

    Either way should give you the same result.
     
  7. Oct 13, 2009 #6
    Hmm...I thought I did. I think it should've been [tex]sin2\theta=\frac{2x}{a}[/tex] , but I don't get where else I messed up. Is that it?

    I have
    [tex][2ab*arcsin\frac{x}{a}]_{0}^{a}+[ab\frac{2x}{a}]_{0}^{a} = \pi ab+2ab[/tex]

    Thanks for the patience.
     
  8. Oct 13, 2009 #7

    Mark44

    Staff: Mentor

    No, your original substitution was x = a sin(theta), so x/a = sin(theta), so theta = sin-1(x/a).

    You're coming out with the wrong answer for the ellipse's area.
     
  9. Oct 13, 2009 #8
    Sorry, you lost me. Yes, that was my original substitution. Then when I'm going back to x doesn't [tex]sin2\theta=\frac{2x}{a}[/tex] ?

    I'm still getting the same answer [tex]\pi ab + 2ab[/tex]
     
  10. Oct 13, 2009 #9

    Mark44

    Staff: Mentor

    The result you're getting is incorrect.

    From post 4, you have
    [tex]=4ab \int_{x = 0}^{x = a} \frac{1}{2} \left ( 1+cos2\theta \right ) d\theta[/tex]
    [tex]=\left [ 2ab\theta \right ]_{x = 0}^{x = a}+[absin2\theta]_{0}^{a}[/tex]

    I added "x = " in your limits of integration, for emphasis and as a reminder that your antiderivative is in terms of theta, not x. In carrying out this integration, you did another substitution, u = 2theta, du = 2d(theta), but you did the integration and undid that substitution.

    You still need to undo your original substitution, x = a sin(theta) OR change your limits of integration. If you undo your first substitution, you'll need to replace theta in 2ab*theta and in ab*sin(2theta).

    x = a sin(theta) <==> theta = sin-1(x/a)
     
  11. Oct 13, 2009 #10

    Mark44

    Staff: Mentor

    To answer your question, no. sin(2*theta) = sin(2*sin-1(x/a))
     
  12. Oct 13, 2009 #11
    I believe this was my mistake all along. I was going back to x but had the sin(2theta) messed up.
    I'm now getting [tex]\pi ab[/tex] for the answer.
     
  13. Oct 13, 2009 #12

    Mark44

    Staff: Mentor

    And that's the right answer.
     
  14. Oct 13, 2009 #13
    It's always the little mistakes. :bugeye:
    Thanks for the help.
     
  15. Oct 13, 2009 #14

    Mark44

    Staff: Mentor

    Sure, you're welcome.

    Being good at calculus is partly about attention to detail.
     
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