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Homework Help: Definate integral for area

  1. Aug 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region bounded by the positive x-axis,
    the positive y-axis and the curve:
    (x^2) +[(y^2)/4] = 1
    using vertical and horizontal strips.

    2. Relevant equations
    Basically I just tried to use integration to find the area, but I suspect I have made a mistake about what the interval for the integration should be.

    3. The attempt at a solution
    Vertical Strips -
    Transposing the equation gives me: y=Sqrt[1-(x^2)]
    So basically I know that the area would be the integral of this but not 100% sure between what intervals. I suspect, from a graph obtained from a graphics calculator, that is between 1 and -1.
    Hence area= Integral of Sqrt[1-(x^2)] between 1 and -1.
    The integral I got to be Sqrt[1-(x^2)]x + arcsin(x), from a online integration calculator. Between 1 and -1 this would give a area of Pi.

    Horizontal Strips -
    Transposing the equation gives me: x=Sqrt[1-[(y^2)/4])
    Now the integral of this I got to be [(1/4)Sqrt(4-(y^2))].y + arcsin(y/2)
    From the same graph I assume the interval for this one should be from 2 to 0. But substituting these values gives me a area of (Pi/2).

    Obviously the areas need to be the same and they aren't! I'm positive that it is the intervals for integration that I am having trouble with, but where the mistake is and why I am not sure. My only guess is that becaise it says positive x and y axis that the interval for the vertical strips should be 1 to 0, but not sure and would this approach change anything I did for the horizontal? Any help or advice would be great. Thanks guys!
  2. jcsd
  3. Aug 11, 2007 #2
    here is the mistake
    you forgot 4
    and interval is from 0 to 1
    can you see why?
  4. Aug 11, 2007 #3
    and just a recommendation: It is better to get use to find integrals by hand rather than calculator;)
  5. Aug 11, 2007 #4
    Yeah sorry. Between 0 to 1 would make sense, as both horizontal and vertical would give me an area of (Pi/2) now, but using the same sort of approach is my horizontal method still right?
    Thanks for that matness, and yeah I really should stop relying so heavily on my calculator.
  6. Aug 11, 2007 #5
    Sorry i did a mistake also :
  7. Aug 11, 2007 #6
    you can trust calculators but i want to say it is better doing by hand in order to learn and useful for other calculations
    p.s.: Stop relying heavily what other people say:)
  8. Aug 11, 2007 #7
    Yeah I get what you mean matness, thanks for everything mate. I just not really confident with what I do and why, like in this question. Still not sure what I did was right for the horizontal strips.
  9. Aug 12, 2007 #8
    The equation says the curve is an ellipse.(see the image below)
    if x = 0 then y = 2 or-2 so graph intersect the y-ax's at 2 and =2
    if y=0 then x=1 or -1 similarly this means the graph intersects x axis at this points
    Anyway if you understood the graph no problem ? it is 1&4 of an ellipse

    first of all integral means infinitesimall sum. Roughly you can either sum up the areas of horizontal strips or vertical strips .
    lets look at the horizontal sum case
    one side of the rectangle is dy and the other side is x which is equal to Sqrt[1-[(y^2)/4]
    and you have a sum from 0 to 2 because your region is given in this way
    so you have
    area=integral from 0 to 2 of Sqrt[1-[(y^2)/4]dy
    vertical strips case is similliar

    you can check your result by using area formula for ellipses.you have to find 1/4 of the area of the ellipse

    p.s.what i wrote is not so formal things i just try to explain intuitively
    i hope this does not make you confuse

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