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Stuck on Area of Ellipse Proof

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the area of an ellipse with equation

    [tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]

    is [itex]A=\pi ab[/itex].


    2. Relevant equations



    3. The attempt at a solution
    I solved for y, set up the integral for area with lower limit -a and upper limit a, did u substitution, factored out constants, used an integration table and have arrived at:

    [tex]\frac{-2b}{a}\frac{1}{a}\ln{\frac{a + x}{\sqrt{a^2 - x^2}}[/tex]

    from -a to a but substituting either gives an undefined quantity inside the natural log
     
    Last edited by a moderator: Mar 17, 2009
  2. jcsd
  3. Mar 17, 2009 #2

    Tom Mattson

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    EE, your LaTeX is terrible :tongue: I've cleaned it up for you.

    That's wrong. There should be no logarithm in the antiderivative. You'll have to show your work in order for me to spot the wrong turn.
     
  4. Mar 17, 2009 #3
    [tex]y=\frac{b}{a}\sqrt{a^2 - x^2} y=\frac{-b}{a}\sqrt{a^2 - x^2}[/tex]
    1st one it the top curve, 2nd one is bottom
    They intersect at (-a,0) and (a,0)
    Therefore [tex]\int{\frac{2b}{a}\sqrt{a^2 - x^2}[/tex] from -a to a
    So [tex]\frac{2b}{a}\int\sqrt{a^2 - x^2}[/tex] from -a to a
    Let [tex]\sqrt{a^2 - x^2} = u[/tex]
    [tex]u^2 = a^2 - x^2[/tex]
    [tex]2u\frac{du}{dx} = -2x[/tex]
    [tex]dx = \frac{-u}{x}du[/tex]
    [tex]x = \sqrt{a^2 - u^2}[/tex]
    Substitute: [tex]\frac{2b}{a}\int{u\frac{-u}{\sqrt{a^2 - u^2}}}[/tex]
    [tex]\frac{-2b}{a}\int{\frac{u^2}{\sqrt{a^2 - u^2}}}[/tex]
     
    Last edited: Mar 17, 2009
  5. Mar 17, 2009 #4

    Dick

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    You need a trig substitution to do that integral. Try x=a*sin(u).
     
  6. Mar 17, 2009 #5
    Ok, I copied the wrong Integral from the table. With the correct one, I've arrived at:

    [tex]\frac{b}{a} [x\sqrt{a^2 - x^2} + a^2\arcsin{\frac{\sqrt{a^2 - x^2}}{a}}][/tex]

    Again with the lower limit of -a and upper limit of a but I believe this yields zero, as [tex]\arcsin{0}=0[/tex] and [tex]\sqrt{0}=0[/tex] as well.
     
    Last edited: Mar 17, 2009
  7. Mar 17, 2009 #6

    Mark44

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    You can make your integral a little simpler by exploiting the symmetry of this ellipse. The area of the entire ellipse is four times the area within the first quadrant.
     
  8. Mar 17, 2009 #7

    Dick

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    That's what comes of copying stuff from tables. My suggested substitution was x=a*sin(u). That means u=arcsin(x/a). How does that turn into arcsin(sqrt(a^2-x^2)/a)? I really think you should work this problem out for yourself. You KNOW the answer isn't zero, right?
     
  9. Mar 18, 2009 #8
    I don't see where I've gone wrong up to getting:

    [tex]\frac{-2b}{a}\int{\frac{u^2}{\sqrt{a^2 - u^2}}}[/tex]

    And unless my calculus book has an incorrect integral in it:

    [tex]\int{\frac{u^2}{\sqrt{a^2 - u^2}}} = \frac{1}{2}(-u\sqrt{a^2 - u^2} + a^2\arcsin{\frac{u}{a}})[/tex]

    Can you give me a little hint as to WHY one can use that substitution? I tried to set up a right triangle and can get that [tex]x=a\arcsin{\frac{u}{a}}[/tex] which yields [tex]u=a\arcsin{\frac{x}{a}}[/tex]. Is there an identity involved. You don't have to tell me which one.
     
    Last edited: Mar 18, 2009
  10. Mar 18, 2009 #9

    Dick

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    The problem is that your substitution u=sqrt(a^2-x^2) isn't a one-to-one correspondence between x and u. Since x=a and x=-a both correspond to u=0. That means you need to be a little careful. But they are one-to-one for x between 0 and a. Just evaluate the antiderivative between u=0 and u=a and double it to account for the x negative part, like Mark suggested. Or do the original integral using the substitution x=a*sin(u), dx=a*cos(u). You wind up having to figure out the integral of cos(u)^2*du. But that's just a double angle formula.
     
  11. Mar 18, 2009 #10

    Dick

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    Doesn't your integral table have an entry for sqrt(a^2-x^2)*dx?? That's the integral we are really trying to do, right? If I put x=a*sin(u), dx=a*cos(u)*du, I get a*sqrt(1-sin(u)^2)*a*cos(u)*du. 1-sin(u)^2=cos(u)^2. So now we've got integral a^2*cos(u)^2*du.
    Now use the double angle formula cos(u)^2=(1+cos(2u))/2.
     
  12. Mar 18, 2009 #11
    Thank you Dick! I just realized that I didn't understand your first suggestion because we have yet to cover trig substitution. One more question: Is that integration rule easily provable? I don't want it to look like I was lazy but I could think of no way to do it with the knowledge I have now. Thanks again!
     
    Last edited: Mar 18, 2009
  13. Mar 18, 2009 #12

    Dick

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    Which integration rule?
     
  14. Mar 18, 2009 #13
    [tex]\int{\frac{u^2}{\sqrt{a^2 - u^2}}}[/tex]
     
  15. Mar 18, 2009 #14

    Dick

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    It's another trig substitution. Which you haven't covered yet. Same idea as sqrt(a^2-u^2). Substitute u=a*sin(t).
     
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