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

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) since in these cases we know how to calculate the integral. But there are many possible solutions to the inequation in D, namely:

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

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.

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 $$\frac{1}{3}$$.

• 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? – Annalisa Jan 17 at 11:49
• 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. – Math Lover Jan 17 at 11:51

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.

• 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. – Annalisa Jan 17 at 11:46
• 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 – Kavi Rama Murthy Jan 17 at 11:49