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Math Help - Double Integral and trying to find the region 'D'

  1. #1
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    Double Integral and trying to find the region 'D'

    I have the double integral of...
    ∫∫y^3 dA

    and the region 'D' are the points:
    (0,2),(1,1),(3,2)

    I graphed it, it makes a triangle but how in the world can I set up my integral like this? I really don't see an upper and lower equation, but I do see the ends of the region I can take.

    If I fix 'x' I can go from 0 to 3
    If I fix 'y' I can go from 1 to 2

    but now what do I find? or how do I find it for that matter...
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by RET80 View Post
    I have the double integral of...
    ∫∫y^3 dA

    and the region 'D' are the points:
    (0,2),(1,1),(3,2)

    I graphed it, it makes a triangle but how in the world can I set up my integral like this? I really don't see an upper and lower equation, but I do see the ends of the region I can take.

    If I fix 'x' I can go from 0 to 3
    If I fix 'y' I can go from 1 to 2

    but now what do I find? or how do I find it for that matter...
    Always go back to the general form of the double integral

    \int_a^b dx \int_{g_1(x)}^{g_2(x)} dy

    Now draw the triangle. We clearly see that the top function here is the line  y = 2 . What about a bottem function? Well we can see there are
    2 parts here, one from x=0-->x=1 and one from x=1-->x=3

    Let us note that we can model sides of a triangle via the line y=mx+b

    by doing this we get

    y_1 = -1x + 2 for the region  0 \le x \le 1

    y_2 = \frac{1x}{2} + \frac{1}{2} for the region  1 \le x \le 3

    Now we can integrate our original function from these bounds.

    So we get

    \iint y^3 dA = \int_0^1 dx \int_{-x + 2}^2 y^3 dy + \int_1^3 dx \int_{ \frac{1x}{2} + \frac{1}{2} }^2 y^3 dy

    This should render the desired result
    Last edited by AllanCuz; April 23rd 2010 at 04:38 PM.
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  3. #3
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    Quote Originally Posted by AllanCuz View Post
    Always go back to the general form of the double integral

    \int_a^b dx \int_{g_1(x)}^{g_2(x)} dy

    Now draw the triangle. We clearly see that the top function here is the line  y = 2 . What about a bottem function? Well we can see there are
    2 parts here, one from x=0-->x=1 and one from x=1-->x=3

    Let us note that we can model sides of a triangle via the line y=mx+b

    by doing this we get

    y_1 = -1x + 2 for the region  0 \le x \le 1

    y_2 = \frac{1x}{2} + \frac{1}{2} for the region  1 \le x \le 3

    Now we can integrate our original function from these bounds.

    So we get

    \iint y^3 dA = \int_0^1 dx \int_{-x + 2}^2 y^3 dy + \int_1^3 dx \int_{ \frac{1x}{2} + \frac{1}{2} }^2 y^3 dy

    This should render the desired result
    thanks, but why are the integrals from the y=mx+b equations to '2' ?
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by RET80 View Post
    thanks, but why are the integrals from the y=mx+b equations to '2' ?
    We need both a floor and top function. By drawing the triangle we can see that the top function is y=2. Naturally when we follow the general double integral we are looking for a function in terms of x. But in this case y=2+0x which still has an "x" function if you want to think about it like that, but it has a co-efficient of zero. It's also of importance to note we are bounded by this region, by the lines (as given by y=mx+b) and y=2. This is what we want! Thus, our top is y=2.
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  5. #5
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    Quote Originally Posted by RET80 View Post
    thanks, but why are the integrals from the y=mx+b equations to '2' ?
    Two of the points, (0, 2) and (3, 2) have y-coordinate 2. One of the things you should have learned long ago is that two points determine a line and a line can be written as y= mx+ b. Taking x= 0 and y= 2 gives 2= b. Taking x= 3 and y= 3 gives 2= 3m+ b. Subtracting the first equation from the second gives 3m= 0 so m= 0 and b= 2. The equation of the top line is "y= 2".
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