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I have to calculate $$\int_D x^2(y+1)\,\mathrm{d}x\,\mathrm{d}y\,,$$

where $$D=\left\{(x,y)\in \mathbb R^2 : |x|+|y|\le 1\right\}$$

Now I went on to try to write this $D$ as a set of the form $$ E = \left\{(x,y) \in \mathbb{R^2}: a < x < b \,,\, \alpha(x)<y<\beta(x)\right\} \,,$$ since in these cases we know how to calculate the integral. But there are many possible solutions to the inequation in D, namely:

from Wolfram Alpha

and I wouldn't know which to take as borders of the integral because I don't think they all give the same result.

Moreover I tried to use symmetry. The set in my case looks like this

from Wolfram Alpha

so maybe we could assume it is symmetric and the Integral is equal to zero. But I am very unsure. This topic was explained very briefly in the lecture therefore I am having a lot of problems.

Thank you in advance

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You need to integrate over the square region as you have sketched. Equation of the lines of the square are $x+y = -1, y-x = 1, x-y=1, x+y=1$.

So,

$-1-x \leq y \leq 1+x, -1 \leq x \leq 0$

$x-1 \leq y \leq 1-x, 0 \leq x \leq 1$

Here is one way to set it up -

$\displaystyle \int_{-1}^{0} \int_{-1-x}^{1+x} x^2(y+1) \ dy \ dx + \int_{0}^{1} \int_{x-1}^{1-x} x^2(y+1) \ dy \ dx$

But please also note that $x^2y$ is an odd function wrt. $y$ and so due to symmetry above and below $x-$axis, the integral will cancel out and be zero. So you can just choose to integrate $x^2 \ dy \ dx$ which comes to $\frac{1}{3}$.

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  • $\begingroup$ Thank you for your answer. So the cancling out due to symmetry has also to do with the function which we are integrating, is not only dependent on the set? $\endgroup$ – Annalisa Jan 17 at 11:49
  • $\begingroup$ If $f(x,y) = x^2y, $ you can see that $f(x,-y) = -x^2y = - f(x,y)$. So it is odd wrt $y$. So it will cancel out if there is symmetry along x-axis. In this case there is. So in other words, the function you are integrating has to be odd and it also depends on the region you are integrating over. $\endgroup$ – Math Lover Jan 17 at 11:51
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Values at a finite number of points have no effect on the integral. It is not even necessary to cosnider $x>0, x<0$ etc.

What we have is $\int_{-1}^{1}\int_{-|1-|x|}^{1-|x|} x^{2}(y+1)dy dx$. Note that $x^{2}$ can be pulled out of the inside integral and the integral of $y$ is $0$ becasue it is an odd function. Hence, we are left with $\int_{-1}^{1} 2x^{2}(1-|x|)dx$. Once again, note that this is twice the integral from $0$ to $1$. I will let you finish.

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  • $\begingroup$ Thank you that was very helpful but I have two little questions left. First could you please give me a bit information on how you found the boundaries? That's my main problem in these excercises. Ans secondly, what about the symmetry thing? In the lecture we had a case where the Inequation in D was $x^2+y^2 \leq 9 $ and this was symmetric so we said directly that Integral is zero, isn't this also the case? I mean, The set D seems pretty symmetric. $\endgroup$ – Annalisa Jan 17 at 11:46
  • $\begingroup$ For fixed $x$, the inequality $|x|+|y| \leq 1$ translates to $|y|\leq 1_|x|$ or $-(1-|x|) \leq y \leq (1-|x|)$. The integral of an odd function over a symmetric set is $0$. But the function here is not odd. @Annalisa $\endgroup$ – Kavi Rama Murthy Jan 17 at 11:49

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