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Math Help - Double integral question.

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    Double integral question.

    When you want to take a double integral of some function involving x and y.

    Say for example 0<x<1 and 0<y<1
    That means that you are integrating over a triangle really. And the slope of the line is
    y = 1-x

    My question is, what would you use as the bounds of the integral? My prof said that the bounds of x are from 0 to 1, but the bounds of y are from 0 to 1-x? Or something like that. Can someone please explain this, I don't understand why the bounds of y just cant be 0 to 1.

    Thanks.
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    First, draw the region of integration. You are correct that it is a triangle, and the equation of the line is \displaystyle y = 1-x or \displaystyle x = 1 - y

    Second, decide if you are going to take horizontal or vertical strips.

    If you take horizontal strips, then your \displaystyle x values range from \displaystyle 0 to the graph, in other words \displaystyle 1 - y, and your \displaystyle y values range from \displaystyle 0 to \displaystyle 1.

    So by taking horizontal strips, your double integral would be

    \displaystyle \int_0^1{\int_0^{1-y}{f(x,y)\,dx}\,dy}.


    Can you work out what it would be if you took vertical strips instead?
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    okay kind of makes sense. In the horizontal strips, the y values go from 0 to 1 but its kind of weird visualizing it. I can see that the x values go till the line and the y values just drop down from 1 to 0. But in the vertical strips, the y values go from 0 till the line and the x values range from 0 to 1.
    Thanks.
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    Quote Originally Posted by Kuma View Post
    okay kind of makes sense. In the horizontal strips, the y values go from 0 to 1 but its kind of weird visualizing it. I can see that the x values go till the line and the y values just drop down from 1 to 0. But in the vertical strips, the y values go from 0 till the line and the x values range from 0 to 1.
    Thanks.
    Correct, and your line has what equation when you have vertical strips?
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    the y values go to the line y = 1-x. Another question though, why can't you use the same equation for the line in both cases? When you took horizontal strips, can't you say that the x values went to the line y = 1-x? It is the same thing as x = 1-y.
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    It's because when you do the integration, you require your final integral to only be in terms of one variable.

    So if you integrate with respect to \displaystyle x first (horizontal strips), your second integral needs to be only in terms of \displaystyle y.

    Similarly if you integrate with respect to \displaystyle y first (vertical strips), your second integral needs to be only in terms of \displaystyle x.
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    Quote Originally Posted by Kuma View Post
    When you want to take a double integral of some function involving x and y.

    Say for example 0<x<1 and 0<y<1
    That means that you are integrating over a triangle really. And the slope of the line is
    y = 1-x

    My question is, what would you use as the bounds of the integral? My prof said that the bounds of x are from 0 to 1, but the bounds of y are from 0 to 1-x? Or something like that. Can someone please explain this, I don't understand why the bounds of y just cant be 0 to 1.

    Thanks.
    Dear Kuma,

    Refer the attached figure. We have to integrate over the area bounded by the traingle, y=1-x. If we take the function as f(x,y); to get the integration over the element dx, you have to consider, \int_{0}^{y=1-x}f(x,y)dy (keeping x fixed). Now since the whole area from x=0 to x=1 must be considered \int_{0}^{1}{\int_{0}^{y=1-x}f(x,y)dy}dx

    Hope this will help you to understand.
    Attached Thumbnails Attached Thumbnails Double integral question.-sp.pdf  
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