1. ## 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.

2. 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?

3. 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.

4. Originally Posted by Kuma
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?

5. 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.

6. 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 $\displaystyle x$.

7. Originally Posted by Kuma
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 $\int_{0}^{1}{\int_{0}^{y=1-x}f(x,y)dy}dx$