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I am trying to find double integral of function $F(x,y)=3y$ over the triangle with vertices $(-1,1), (0,0)$, and $(1,1)$. Here is picture of this triangle:
enter image description here

I guess that we should integrate it by $x$ and then by $y$,but i am confused on region bound ,I think for $y$ lowest bound it $0$ and highest $1$,and for $x$ lowest is $-1$ and $1$ (or $-y$ and $y$)?am I correct or wrong? please help.

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    $\begingroup$ Does it help already if I tell you that you can evaluate it using either one of the double integrals $$\int_{0}^{1} \int_{-y}^{y} F(x,y)\,dx\,dy = \int_{-1}^{1} \int_{|x|}^1 F(x,y)\,dy\,dx\quad?$$ Does that make sense? You're talking about the first variant, in fact. $\endgroup$ Commented Jul 30, 2011 at 12:41
  • $\begingroup$ i have one question we know that interval of x is [-1 1] and for y is [0 1] why we can't use directly these number and dont use -y or y? or 1 and x? $\endgroup$ Commented Jul 30, 2011 at 12:48
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    $\begingroup$ The trouble is that by evaluating $\displaystyle \int_{0}^{1}\int_{-1}^{1} F(x,y)\,dx\,dy$ you integrate over the entire square, not just over the triangle. You can look at it this way: You're "slicing the triangle horizontally". If you're at height $y$, then $x$ may only vary between $-y$ and $y$, right? $\endgroup$ Commented Jul 30, 2011 at 12:51
  • $\begingroup$ yes it seems so ok thanks very much @Theo Buehler $\endgroup$ Commented Jul 30, 2011 at 12:55
  • $\begingroup$ @Theo: I didn't look at your comment :) $\endgroup$ Commented Jul 30, 2011 at 12:56

2 Answers 2

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Because the function $F(x,y)$ is in this case symmetric across the $y$ axis, I think I would almost automatically integrate over the right half of the triangle in the picture, then double the answer.

In this case there are various advantages, not the least of which is that we do not meet any negative numbers, which are a notorious source of error.

For any fixed $y$ with $0 \le y \le 1$, we make $x$ travel horizontally from $0$ to $y$. Then $y$ is made to travel from $0$ to $1$. Thus our integral is equal to $$2\int_{y=0}^{y=1}\left(\int_{x=0}^{x=y} 3y\:dx\right)dy.$$

The calculation is now easy. The inner integral is $3y^2$, so we end up calculating $2\int_{y=0}^{y=1} 3y^2\;dy$. This is $2$.

We could also integrate first with respect to $y$. Here again life is made easier if we integrate over the right half of the region, and then double. Now $y$ first travels from $x$ to $1$, and then $x$ travels from $0$ to $1$. So our integral is equal to $$2\int_{x=0}^{x=1}\left(\int_{y=x}^{y=1} 3y\:dy\right)dx.$$ The details of evaluation are a little more messy than with the first approach. In addition, if we do not use the symmetry, the integral will have to be broken up as two integrals, as follows: $$\int_{x=-1}^{x=0}\left(\int_{y=-x}^{y=1} 3y\:dy\right)dx +\int_{x=0}^{x=1}\left(\int_{y=x}^{y=1} 3y\:dy\right)dx.$$ Uglier, more work, much greater chance of minor errors.

Notational Comment: When we are expressing a double integral as an iterated integral, it is easy to lose track of who is going from where to where. That's why, for example, the integral in the second solution was not written as $$2\int_{0}^{1}\left(\int_{x}^{1} 3y\:dy\right)dx.$$

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Hint: $$\int\limits_{-1}^{1} \int\limits_{|x|}^{1} 3y \ dy \cdot dx = \int\limits_{-1}^{0}\int\limits_{-x}^{1} 3y \ dy \cdot dx + \int\limits_{0}^{1}\int\limits_{x}^{1} 3y \ dy \cdot dx$$

Or even you can do this: $$I = \int\limits_{0}^{1} \int\limits_{-y}^{y} 3y \ dx \cdot dy$$

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  • $\begingroup$ minor point of language: "even you can do this" can be taken to mean "this is so simple that anybody, even you, can do it." I suspect that what you meant is "you can even do this", which just says it is another option. From your previous posts I am sure you did not mean this disrespect. $\endgroup$ Commented Jul 30, 2011 at 23:28

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