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Area of the Region Inside the Curve.

  1. Mar 5, 2009 #1
    Would it be possible if I could get help getting started? I don't have any work yet, but I am really trying to figure it out..

    Problem:
    Determine the area of the region inside the curve x^2+y^2-2y=0 and outside y(x^2+4)=5.

    It has a graph and I'll upload a picture if needed. Basically? It's the area under the top of the circle but above the curve.


    We have been discussing parametric, but I can't wrap my head around it because he didn't explain it all too well. Obviously it is symmetric about the y-axis. Other than that, I just need help getting started...
     
  2. jcsd
  3. Mar 5, 2009 #2
    Since the circle is the upper bound the the other curve is the lower bound, you will be finding the difference of the integral of the circle and that of the curve bounded by their intersections on the left and right, namely [tex]\int^{b}_{a}[(x^{2}+y^{2}-2y)-(y(x^{2}+4))]dx[/tex], where a is their intersection on the left and b is their intersection on the right. I haven't down any calculation for them yet so you will need to find out the a and b. :)
     
  4. Mar 5, 2009 #3
    So, I will set, for example, [tex]\sqrt{2y-y^2}[/tex]=[tex]\sqrt{\frac{5}{y}-4}[/tex]?
    Then solve it from there, to find a and b?
     
  5. Mar 5, 2009 #4
    I suggest you solving for x instead, since the x values are what you need for a and b. Try to rewrite the equations in terms of x, and then set then equal and solve for x. You should be getting two distinct values, which the lower one will be a and the other one be b.
     
  6. Mar 5, 2009 #5
    How would you solve for x when in the equation for the circle we're given "y^2-2y"? I know I should know this, but I don't... :(
     
  7. Mar 5, 2009 #6
    Point a and b are the points of intersections of the two equations you are given.
     
  8. Mar 5, 2009 #7
    I know that, I just don't know how to solve for x in the situation where there is a "y^2-2y) situation...
     
  9. Mar 5, 2009 #8
    Then you may wish to solve for y first, and then plug the values of y into either equations to get two separate x values. :)
     
  10. Mar 5, 2009 #9
    **I totally meant in terms of x... Wow. That was silly.
     
  11. Mar 5, 2009 #10
    ([tex]\frac{5}{x^2+4}[/tex])2-2([tex]\frac{5}{x^2+4}[/tex]=-x2
    which then gives me

    I have ([tex]\frac{5}{x^2+4}[/tex])2-([tex]\frac{10}{x^2+4}[/tex])=-x2
    But when I graphed it, I don't really think that's right....
     
  12. Mar 5, 2009 #11
    For this question, solving for x directly is quite complicated; on the other hand, solve for y is relatively easier. Try solving for y then plug the values of y into either equation to get x, that will make things a whole lot easier.. :D
     
  13. Mar 5, 2009 #12
    I already tried that, and then it wasn't easily factored... So that's when I got stuck.
     
  14. Mar 5, 2009 #13
    First you write it as 0=y^3-2y^2-4y+5, then you do a bit of trial and error and find that (x-1) is a factor by doing long division, which leaves you (y-1)(y^2-y-5), then I'm sure you know what to do to find the root in (y^2-y-5). ;)
     
  15. Mar 5, 2009 #14
    Oh my gosh... I'm dumb as a box of rocks. That totally makes sense.
    So, I get y=-2, 1, and 3... But I'll be using -2 and 3, correct? And it is y, right? Hah.

    I really appreciate your help.. I'm still going to need some more, I'm sure... But I seriously do appreciate it.
     
  16. Mar 5, 2009 #15
    Ok, so, now I'm trying to figure out the integral.
    Going from:
    with a=-2 and b=3
    [tex]\int[/tex]((x^2+y^2-2y)-(y(x^2+4))dx
    obviously I have to solve for y still, correct?

    That's why I'm confused, because when solving for y... The circle has a y^2-2y and how would I solve for y in that case?
     
  17. Mar 5, 2009 #16
    The circle's equations in terms of x are y=1+sqrt(1-x^2) and y=1-sqrt(1-x^2). Good luck! :)
     
  18. Mar 5, 2009 #17
    Thank you VERY much! I think that will definitely help... I'll post my answer (if I get it done) on here. Thanks!!!
     
  19. Mar 5, 2009 #18
    One more question, I think...
    Because a half circle is involved, will I use [tex]\frac{\pi r^2}{2}[/tex] for the equation of the circle?
     
  20. Mar 5, 2009 #19
    If that half involved is centered in the origin vertically, then yes for this question once the r is known (in which case it is). :)
     
  21. Mar 5, 2009 #20
    When I use:

    [tex]\int[/tex][(1+[tex]\sqrt{1-x^2}[/tex])-([tex]\frac{5}{x^2+4}[/tex]]^2 dx
    a=-2 and b=3
    (my professor is letting us use our calculator)
    When I put it in, it tells me that it's an unreal answer? (And I know we aren't supposed to get them).
     
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