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Sketch Region enclosed

  1. Sep 26, 2004 #1
    Sketch the region enclosed by [tex]x+y^2=12[/tex] and [tex]x+y=0[/tex]. Decide whether to integrate with respect to [tex]x[/tex] or [tex]y[/tex]. Then find the area of the region.



    first thing i did was solve for x for each equation then set them to each other and got:
    [tex]12-y^2=-y[/tex]
    [tex]y^2-y-12=0[/tex]

    found the points of intersection of y at 4,-3
    plugged in for y and found the x intersections and got:

    (-4,4) & (3,-3) for the points of intersections


    ok time to set up the integral with respect to y

    [tex]\int (-y) -(12-y^2)[/tex] (a=-3, b=4, dont know how to set those up using latex)

    add the terms together to make it more neat...

    [tex]\int y^2-y-12[/tex] (a=-3, b=4)

    integrate...

    [tex] \frac{y^3}{3} - \frac{y^2}{2} - 12y[/tex] (a=-3,b=4)


    [tex] \frac{4^3}{3} - \frac{4^2}{2} - 12(4)[/tex]
    subtract the above part by...
    [tex] \frac{-3^3}{3} - \frac{-3^2}{2} - 12(-3)[/tex]


    and got the answer -57.16667, which i know is way off. what am i doing wrong??
     
    Last edited: Sep 27, 2004
  2. jcsd
  3. Sep 26, 2004 #2

    Pyrrhus

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    Homework Helper

    At first glance, i'll say check your signs for the f(b) - f(a).
     
  4. Sep 27, 2004 #3
    k checked that part and got the same signs, so what is wrong with my answer? did i set up the equation correctly? (is there a Ti-83 plus program i can use to check my answer? or is there any math program for the computer that would do this for me?
     
    Last edited: Sep 27, 2004
  5. Sep 28, 2004 #4
    well you can integrate in terms of x or try a change of variable, but its easy to get the area directly.

    first you have the area enclosed between the straight line y=-x and the horizontal parabola y^2=-x+12. to find the points of intersection substitute y=-x in the parabola ecuation and you will get a second grade ecuation for x that haves the solutions -4 and 3, then in x=-4 and x=3 are the intersections.

    second, make a sketch of the situation and think the area of the enclosed region between -4 and 0 is the integral of the parabola minus the integral of the line. and between 0 and 12 are 2 times the integral of the parabola minus a part of this (the two times is simply because with the integral of the parabola you only get the area above x axis and below the parabola and by symmetry the area between 0 and 12 is the same above x and below x).

    third, the part that you must substract is the area enclosed between the y=x x=0 and the parabola that it's trivial to get.

    then: where int(([a,b],f(x)) is integral of f(x) between a and b and sqrt square root:

    A=int([-4,0],sqrt(-x+12)+x)+2*int([0,12],sqrt(-x+12))-int([0,3],sqrt(-x+12)-x)

    A=(2/3)*(16^(3/2)+9^(3/2))-(5/2)=58,1666=58,2
     
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