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I am bit unsure about the following problem:

Evaluate the double integral:

$-\iint_{A}(y+x)\,dA$

over the triangle with vertices $(0,0), (1,1), (2,0)$

OK, so I figured here that I would do this by first evaluating the integral over the region bounded by the vertices $(0,0), (1,1), (1,0)$ and then evaluate the integral over the region bounded by the vertices $(1,0), (1,1), (2,0)$ before adding the two answers together, and then reversing the sign of this answer (since there is a minus sign in front of the original double integral). Thus, I begin by finding:

$\int_{0}^{1}dx \int_{0}^{x}(y+x)\,dy$

When solved this gives me the answer $\frac{1}{2}$.

Next I solve:

$\int_{1}^{2}dx \int_{1}^{2-x}(y+x)\,dy$

When solved this gives me the answer $-\frac{7}{6}$.

I have verified both the integrals in Wolframalpha, and they give me the same answer. I would therefore believe that the final answer should be:

$-(\frac{1}{2} - \frac{7}{6}) = \frac{2}{3}$

However, the final answer should, according to the book, be $-\frac{4}{3}$.

Thus, obviously I do something wrong here. If anyone can help me out, I would greatly appreciate it. Is it perhaps that it is not allowed to "split up" this into two separate integrals? I couldn't find a way to solve this without doing this.

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    Ah, now I get it! Duh. Of course - the$y$lower bound is zero throughout. Thanks a lot all of you :). I just needed to get my head straight :)2012-07-15

1 Answers 1

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Your second integral should be $\int_{1}^{2}dx \int_{0}^{2-x}(y+x)dy.$ Your lower $y$ limit was 1 instead of 0.

Draw the triangle to see the area you are integrating over.

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    Thanks. Yes, I just figured it out. Stupid mistake on my part :).2012-07-15