Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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! –  Harald Hanche-Olsen Jul 15 '12 at 20:32
    
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). –  Kristian Jul 15 '12 at 20:39
    
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. –  Harald Hanche-Olsen Jul 15 '12 at 20:41
    
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. –  Cocopuffs Jul 15 '12 at 20:42
1  
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. –  joriki Jul 15 '12 at 20:45
show 1 more comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
Thanks. Yes, I just figured it out. Stupid mistake on my part :). –  Kristian Jul 15 '12 at 20:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.