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Evaluate the integral over the bounded closed region

  1. Apr 8, 2005 #1
    Evaluate the integral over the bounded closed region B

    [tex] \int \int_{B} x^2 y^3 dx dy [/tex] B is the region bounded by y=x^2 and y =x

    certainly the x=0 to x=1 will be the limits for the x part
    but what about the y does it go from root y to y?

    please help! i neeed to understand how to setup these little buggers
     
  2. jcsd
  3. Apr 8, 2005 #2

    StatusX

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    No, for the y part you are integrating over some range of y at a given x value. You are more or less getting the lengths of all the line segments parellel to the y-axis, which you then integrate over x to get the area. So your y bounds have to be in terms of x. Another way to think of it is that you need a constant at the end, so each integration has to eliminate the variable you are integrating over. If you had a y in the bounds, you would be left with an answer in terms of y, which doesn't make any sense.
     
  4. Apr 8, 2005 #3

    so i am supposed to find the lengths of all the segments parallel to the y axis?? Wouldnt this simply be given by x(1-x) with a lower bound of zero?
     
  5. Apr 8, 2005 #4

    StatusX

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    I'm sorry, let me correct myself. You need to find all the line integrals and then integrate over them. I was thinking of a double integral to find the area.

    I was trying to explain what a double integral means so you'd understand why the bounds have to be in terms of x. A double integral is just two single integrals. The inner one (over y) gives you a function of x, which represents the integral over each of the vertical line segments. You then have to integrate this over x to get the total integral over the entire area.

    So, for example, let's say you were trying to find the area. First you need the "area" of each line segment. For a give value of x, the length of the line segment between the functions is given by:

    [tex]\int_{x^2}^{x} dy [/tex]

    If you wanted to integrate a function over the region, you would instead put that function in the integral here. Now, to get the total area, you need to integrate over all the lines:

    [tex]\int_0^1 \left(\int_{x^2}^{x} dy \right) dx[/tex]
     
    Last edited: Apr 8, 2005
  6. Apr 8, 2005 #5
    You want your outer limits to be constant, and your inner limits (usually) to be a function of the outer integral.

    In this case the outer integral is dy, so the limits of y will be constant. The limits of the inner integral, dx, will be a function of y.

    Your bounds are x^2 to x, because thats where you are integrating from.
    Graph the domain, this is always a good step.

    To find your x limits, find the x value for the points of intersection.
     
  7. Apr 8, 2005 #6
    First integrate to y where the lower bound of y is x² and the upper bound is x

    Then integrate to x where x goes from 0 to 1

    marlon
     
  8. Apr 8, 2005 #7
    Graph the domain, get in a habit of doing this.
     
  9. Apr 8, 2005 #8
    Just draw a picture. y=x² is a parabole, symmetric to the y-axis. y=x is the bissectrice of the first quadrant. Both intersect at the origin and at x=1.

    y = x lies ABOVE the parabola in this region (where x goes from 0 to 1)

    If you are not certain, just make a sign chart of both x-x² and x²-x for x between 0 and 1, you will see that one of the two will be positive and the other negative. For example : x²-x will be <0 thus x²<x
    x - x² will be >0 thus x > x²

    marlon
     
  10. Apr 8, 2005 #9
    that woulkd hte domain

    so according to what statusX says i need to find the area of the yellow region by using [tex]\int_{x^2}^{x} dy [/tex] and thereafter integrate the x by using the points of intersection
    so this is the basic setup for something like this ??
    first i have to find the area of the intersecting region and then integrate over the given x values which are the points of intersection of the two surfaces?
     

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  11. Apr 8, 2005 #10

    StatusX

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    No no no, I'm sorry for confusing you. Finding the area is a different problem, it is equivalent to integrating the function "1" over the region. To integrate a different function over the region, it needs to be inside the inner integral:

    [tex]\int_0^1 \left( \int_{x^2}^x f(x,y) dy \right) dx [/tex]

    When I showed how to get the area, that was just an example to show you how the double integral method works.
     
    Last edited: Apr 8, 2005
  12. Apr 8, 2005 #11
    another one

    ok thanks for the help on that one really appreciated it

    now for this one
    [tex] \int \int_{B} \frac{1}{\sqrt{1-y^2}} dx dy [/tex] where B is bounded by y=1, y=sin x for [itex] 0 \leq x \leq \frac{\pi}{2} [/itex] and y=-sin x for [itex] - \frac{\pi}{2} \leq x \leq 0 [/itex] (This is the description, below is my thought on its solution)

    thats the region in the attachment (i hope i got this right!)
    so there will be two parts to this integral
    forh te left hand part of thi region
    x goes from -pi/2 to 0 and y goes from 0 to sin x?
    for the right hand part
    x goes from 0 to pi/2 and y goes from 1 to 1 which makes an integral of zero which is valid??
     

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    Last edited: Apr 8, 2005
  13. Apr 8, 2005 #12
    Hold on, just to be sure (but then your figure is wrong) y goes from sinx to 1 for x going from 0 to 1 right ? In that case, you are integrating over the 'yellow triangle' ABOVE the sine !!!

    And yes, you must split up the integral...because of the x-range

    marlon
     
  14. Apr 8, 2005 #13

    StatusX

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    Since the function only depends on y, the two regions will have the same integral (assuming your figure is wrong and the description is right), so you could just do one and then double it to get the total.
     
  15. Apr 8, 2005 #14
    Edited

    i think i see waht you guys mena
    both are identical so
    x from 0 to pi/2 and y goes from sin x to 1 is that good. and the whole integral times two?
     

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    Last edited: Apr 8, 2005
  16. Apr 8, 2005 #15
    is htat last one i posted correct or am i still wrong in which case i might as well just jump from a bridge
     
  17. Apr 8, 2005 #16

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    Yea, that looks right. if by "the whole integral times two" you mean double the result of a single region to get the total integral.
     
  18. Apr 8, 2005 #17
    that is what i meant by the whole integral times two
    thank you very much i think i understand them now!
     
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