I am bit unsure about the following problem:

Evaluate the double integral:


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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    $\begingroup$ I see one problem at least: The inner ($y$) integral in the second part should start at $0$. Anyway, you can avoid the splitting by doing the integrals with the $x$ integral on the inside. Try it! $\endgroup$ – Harald Hanche-Olsen Jul 15 '12 at 20:32
  • $\begingroup$ Thank you for your answer. But why should the second integral start at 0? The y-values being at 1 and end up at 0 (hence I chose y = 2-x for the upper value of the integral). $\endgroup$ – Kristian Jul 15 '12 at 20:39
  • $\begingroup$ Draw a picture. The whole triangle has its base at the $x$-axis, and so does each of the two pieces resulting from the split. $\endgroup$ – Harald Hanche-Olsen Jul 15 '12 at 20:41
  • $\begingroup$ You can also solve this with the Gauss integral theorem on $F(x,y) = xy$ and evaluate the integral on the boundary instead. Maybe this is easier, since the boundary is made of line segments. $\endgroup$ – Cocopuffs Jul 15 '12 at 20:42
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    $\begingroup$ It seems you may be confusing the two bounds. The bound that begins at $1$ and ends up at $0$ is the upper bound; the lower one is $0$ throughout. Also, you can tell from the sign that $-\frac76$ must be wrong, since the integrand is non-negative throughout the triangle. $\endgroup$ – joriki Jul 15 '12 at 20:45

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.

  • $\begingroup$ Thanks. Yes, I just figured it out. Stupid mistake on my part :). $\endgroup$ – Kristian Jul 15 '12 at 20:54

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