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Double integral of y^3

  1. Jul 31, 2005 #1
    Double integral of y^3, where D is the triangular region with vertices (0,0), (1,2), and (0,3). I can't figure out what the limits are.

    D={(x,y)|0<=x<=3.....is this even half way right?
  2. jcsd
  3. Jul 31, 2005 #2
    e. g. {0<x<1, 2x<y<-x+3}
  4. Jul 31, 2005 #3


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    Just add [tex]y^3[/tex] into the double integral and evaluate it. Keep in mind for simple structures like a triangle you can verify that your answer is correct simply but using trigonometric formulas for area.

    In this case area of a triangle is= [tex]\frac{(b*h)}{2}=\frac{3*1}{2}[/tex]
    Last edited: Oct 8, 2005
  5. Jul 31, 2005 #4


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    No, because the "3" in (0, 3) is a y-value not x. The first thing you should do is draw a picture- like cronxeh did. One side of the triangle is the line x= 0 (from (0,0) to (0,3)). It should be easy to see that the line from (0,0) to (1,2) is y= 2x (A line through (0,0) is always of the form y= mx and m= 2 makes y= 2(1)= 2. If you don't see that you can still calculate that the slope is [tex]\frac{2-0}{1-0}= 2[/tex]. )
    For the line from (0,3) to (1,2), you can argue that the equation of a line is always of the form y- y0= m(x- x0). [tex]m= \frac{2-3}{1-0}= -1[/tex]. Taking (x0,y0) to be (0, 3), that gives
    y- 3= 1(x-0) or y= x+ 3. If you take (x0,y0) to be (1, 2) instead, you get y- 2= 1(x- 1) which reduces to y= x+ 3 also.

    Now, since x can range from 0 to 2, the "outer integral" will be [tex]\int_0^3 dx[/tex]. For every x, y ranges from the lower of the two lines (the one from (0,0) to (1,2)), which is y= 2x, to the higher (the one from (0,3) to (1,2)), which is y= -x+ 3 or, if you prefer, y= 3- x. The "inner integral" will be [tex]\int_{y= 2x}^{3-x} dy[/tex].

    Since the integrand is y3, the integral is
    [tex]\int_{x=0}^2\int_{y= 2x}^{3-x} y^3 dy dx[/tex].
    Last edited by a moderator: Jul 31, 2005
  6. Jul 31, 2005 #5


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    You could, by the way, do this "the other way around" (I wouldn't recommend it here- it's much more complicated. My point is to show how you can get the limits of integration.)

    Overall, y ranges from 0 to 3. If we choose to use the y integral as the "outer integral", the limits would be from 0 to 3. For each y, the left sides is always x= 0 and so the lower limit for the "inner integral" is x= 0. However, there are two different lines that make up the right side: y= 2x and y= 3-x or, since we are using x as the variable for the inner integral and the limits of the inner integral cannot depend on x, x= (1/2)y and x= 3- y. Which of those we use as the upper depends on what y is. If y< 2, the right side is x= (1/2)y and if y> 2, the right side is x= 3- y. The simplest thing to do is to do that as two separate integrals and add them:
    [tex]\int_{y=0}^2\int_{x=0}^{\frac{y}{2}}y^3 dxdy+ \int_{y=2}^3\int_{x=0}^{3-y}y^3dxdy[/tex].
    Last edited by a moderator: Jul 31, 2005
  7. Jul 31, 2005 #6
    Thanks. I really didn't think about it enough. :grumpy:
  8. Dec 9, 2009 #7
    is there a #1 answer rating?
  9. Mar 29, 2011 #8
    Why does x range from 0 to 2 and why does this give the outer integral from 0 to 3?
  10. Mar 29, 2011 #9


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    No doubt, that was a typo (In post #4).

    The limits of integration for x should be from 0 to 1.
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