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

Question

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:

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.

Thank you in advance

Answer 1

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 $\frac{1}{3}$.

Answer 2

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.