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.

How could you integrate the function $f(x,y) = x^2 + y^2$ over the triangle with vertices $(0,0)$, $(0,1)$ and $(1,0)$?


I define the set $D = \{(x,y)\; |\; 0\leq x\leq 1 \text{ and } 0\leq y\leq x\}$ and then calculate

$$\int_0^1 \int_0^x x^2 + y^2 \; \mathrm{d}y \; \mathrm{d}x = \frac{1}{3},$$

but apparantly the answer is $\frac{1}{6}$.

share|improve this question
    
The answer is definitely 1/6. The first integration will give you $\frac{2x^3}{3}$.... –  Simon Hayward Dec 11 '12 at 20:29
add comment

3 Answers 3

up vote 0 down vote accepted

I think you have the limits changed and wrong. I think it must be

$$\int_0^1dx\int_0^{1-x}(x^2+y^2)dy=\int_0^1\left.\left(x^2y+\frac{1}{3}y^3\right|_0^{1-x}\right)dx=\int_0^1\left(x^2(1-x)+\frac{1}{3}(1-x)^3\right)dx=$$

$$=\frac{1}{3}-\frac{1}{4}-\frac{1}{12}(0-1)=\frac{2}{12}=\frac{1}{6}$$

share|improve this answer
    
Thanks a lot. What about if we integrate over a different domain. For example the triangle with vertices $(0,0)$, $(1,0)$, $(1,2)$. How do we define $D$? $D=\{ (x,y)\;|\; 0\leq x \leq 1 \;\;\; 0\leq y \leq 2x\}$ Is this right? –  inttri Dec 11 '12 at 20:34
    
Yes, it is right in that case. –  DonAntonio Dec 11 '12 at 20:36
add comment

You are mistaken in defining the integration domain. Correct domain is $$D = \{(x,y) |\;\; 0\leqslant x\leqslant 1, \;\; \; 0\leqslant y\leqslant 1-x\}$$

share|improve this answer
add comment

Look at the graph of the region. You have described this region

However, the region in question is really this one.

Thus, your domain is $$D=\{(x,y)|0\le x \le 1 \text{ and } 0 \le y \le 1-x\}$$

I believe you can take it from here.

share|improve this answer
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.