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Double integral over circle region

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    So I have to use the type I type II region formula to find the volume under the equation (2x-y) and over the circular domain with center (0,0) and radius 2. Do I have to split this circle into semi-circles and treat it as 2 type I domains? I got the following limits for the top half, but I get stuck when integrating:

    3. The attempt at a solution

    y limits:
    Upper: Sqrt(2 - x^2) from the equation 2 = y^2 + x^2
    Lower: 0

    X limits:
    Upper: 2
    Lower: -2

    So I have to find the integral with respect to y of 2x-y with limits 0 to Sqrt[2-x^2]

    After integrating with respect to Y I got:

    2x(Sqrt[2-x^2]) - 1 + (x^2)/2

    Is this correct to start with? Then integrate with respect to x from -2 to 2?
     
  2. jcsd
  3. Jun 10, 2010 #2

    lanedance

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    the geomtric picture isn't very clear... can you explain type I & II regions, & how is this a volume if you're only integrating over 2 variables, is it actually an area?

    the line y=2x passes through the origin, so will split the circle you decribe in half... is there any ther bound? or is it the region above y = 2x, and bounde dabove by the circle?

    if you are integrating over the circle, polar coordinates always make life easy...
     
  4. Jun 11, 2010 #3
    Type 1 regions are two variable domains where the top and bottom of the boundaries of the domain (in the xy plane) are functions h1(x) and h2(x) and the left and right are constants e.g x = 1, x=2... Type II regions are two variable domains where the left and right boundaries are functions h1(y) h2(y) and the top and bottom are constants e.g y = 1 y = 2.

    In this question you have to use the double integral formula for type 1 or type 2 regions because it is in part of the textbook where polar coordinates have not yet been discussed. The answer is zero.

    The domain is a circular region so I was trying to cut it into two semi-circles of type 1 where the top semi-circle has functions y = 0 and y = Sqrt[2 -x^2] and then the sides would just be -2 and 2. Then the bottom would be the same but the y limits switched and -Sqrt[2 - x^2] instead of +ve. Is it possible to just say they will cancel out because of symmetry involved? That is,

    y limits of top semicircle are [0, Sqrt[2-x^2]] and bottom semicircle [-Sqrt[2-x^2],0] So could you just flip ones limits over by making the entire integral negative and they both cancel to zero??

    The double integral formula for volumes above Type 1 regions and below surfaces in 3D space are:

    [tex]\int\int[/tex]D f(x,y) dA = [tex]\int[a,b][/tex][tex]\int[h1(x),h2(x)][/tex] f(x,y) dy dx

    Type 2 regions are the same except dx dy and obviously g1(y) and g2(y)....
     
  5. Jun 11, 2010 #4
    Also it is under the plane 2x -y... hope that is clear :/
     
  6. Jun 11, 2010 #5

    vela

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    There's no reason to split the region into two. In fact, it's easier to show the integral is 0 if you don't.

    Hint: Consider the even or oddness of the integrand.
     
  7. Jun 11, 2010 #6
    I see my mistake :P I dont know why I didnt just treat it as a type I with y limits -Sqrt[4-x^2] and +Sqrt[4-x^2]...... I was making it too difficult for myself. My lecturer helped me out with this and obviously it cancels down to 0 when you integrate with respect to x.

    Thanks for your help anyway!! :)
     
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