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I’m really having problems getting suitable limits when using double integration. For example:

Let $E$ be a region defined by

$$E=\left\{(x,y): y-x \leq 2,\ x + y \geq 4,\ 2x + y \leq 8\right\}$$

Sketch the region $E$.

The sketching is never a problem. So for this one we have:

$y = 8-2x$, $y=x+2$ and $y=4-x$ - the triangle between all the lines is the region $E$.

Find the area of $E$ and calculate

$$\int\!\!\!\int_{E}\frac{1}{x}\,\mathrm{d} x\, \mathrm{d} y$$

The integration isn’t the problem. I’m really confused as to how to get the correct limits. Our notes say to fix a variable and then look at the boundary of the other. So if we fix $x\in (1,4)$ then taking $y=x+2$ and $y=8-2x$ will give me a greater area then needed....

I hope someone can explain this to me as I’m rather confused!

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@user4645: You've been ignoring my comments here:… and here:…. Please show more respect for the site. If you go on like this I may start flagging your questions. – joriki May 27 '11 at 16:27
@joriki: I posted an answer before I saw your comment. I think your comment is a good one, so deleted the answer. Let me think if you think I should repost. – Ross Millikan May 27 '11 at 16:43
@Ross: Thanks. My tendency would be to wait with reposting until user4645 has made some effort to cooperate. – joriki May 27 '11 at 16:48
@user4645: Find the corners of the triangle. I get $(1,3)$, $(3,5)$, and $(4,0)$. We have to decide whether to integrate first w.r.t. $x$ or w.r.t. $y$. Doing $x$ first sounds hard, you will get logs, which you need to later integrate. So try to integrate first w.r.t. $y$. The "top" boundary changes. So break up the region using a vertical line through $(3,5)$. In the left region, $y$ goes from $4-x$ to $x+2$. Integrate (trivial). Then integrate w.r.t. $x$, from $1$ to $3$. Please leave a message if the process is not clear. – André Nicolas May 28 '11 at 4:10
hi... ive corrected my mistakes. sorry about that! – user4645 May 30 '11 at 17:05

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