I'm struggling figuring out how to tackle down this exercise. I don't know how to approach this question. I guess the answer needs to be $I > 0$, because D is alwasy greater or equal then 1. And because inside the integral all the variables are not related to A, so you can see them as constants. So a solution would be $A^2/2 + C$. But I have the feeling that this approach might be wrong. The exercise goes like this.
Let $D=\{(x,y) \in \mathbb{R}^2 |1 \leq x^2 +y^2 \leq3 \}$ if $$I = \iint_D(|x+y|-x-y)dA$$ which of the following is true: $$I = 0$$ $$I > 0$$ $$I < 0$$
If you know the answer and would like to share it, it would be very much appreciated!
Thanks in advance
$ \displaystyle I = \iint_D(|x+y|- (x+y)) \ dA$
$D=\{(x,y) \in \mathbb{R}^2 |1 \leq x^2 +y^2 \leq3 \}$
$D$ is region between two circles centered at the origin with radius $1$ and $\sqrt3$.
Now in the given region, let's check the sign of the integrand $|x+y| - (x+y)$. Given the absolute term, it is simpler to check the sign in each quadrant separately.
In first quadrant, $x, y \geq 0$ and so integrand is zero and so is the integral.
In second quadrant, $x$ is negative and $y$ is positive. So we divide the region in two -
i) when $y \geq |x|$, the integrand is zero.
ii) when $y \leq |x|$, the integrand is $- 2(x+y)$, which is positive.
So in second quadrant, integral is positive.
Similarly, show that the integral is positive in quadrant $3$ and $4$ as well.
So, $I \gt 0$.