Symmetry in a Double Integral I was working on the following two part question:


*

*Evaluate $$ \frac{5}{2} \iint_D x\left[1+y(x^2+y^2)\right] dA$$


where D is the region of integration bounded by the lines $y=1, x=1, \text{and the curve } y=x^3$.
I used the bounds $D = \left\{(x,y): -1\le x \le 1, x^3 \le y \le 1\right\}$
This was fairly simple to do as the double integral itself evaluates to $\frac{-2}{5}$, which makes sense since when combined with the factor of to $\frac{5}{2}$ we get $-1$ which feels like a crisp clean answer.
The second part of the problem is where I got stuck:


*

*If $f$ is a continuous function of a single variable, evaluate $$\frac{5}{2} \iint_D x\left[1+yf(x^2+y^2)\right] dA$$
Hint: Use the symmetry properties of the integral
I have come to the conclusion of two possible answers:


*

*It is $-1$, because if we let $f(t)=t$ and then let $t=x^2+y^2$, we arrive back to the first part of the question no matter what $f$ is? However, I do feel there is some heavily flawed logic in this argument, but it's what first catches my attention when comparing the two situations.

*It is $0$, because if we were to expand the integrand as so: $$\frac{5}{2} \iint_D x+xyf(x^2+y^2)\ dA$$ it seems like the appearance of $xy$ and $x^2+y^2$ is where the symmetry comes along because switching $x\text{ and } y$ which might mean it evalutes to $0$ for some reason? Or that the whole second term cancels out somehow and we are left with only $x$ in the integrand. 
Somewhere there must arise a situation of invariant transformations and it looks to be something simple, but its at the "tip of my tongue".
Any thoughts or help on this is greatly appreciated.
P.S I thought about changing to polar coordinates because $x^2+y^2=r^2$ seems like it might do something, but it led me nowhere.
 A: So the current region is a little annoying to integrate over. However, note that $-D=\{(-x,-y):(x,y)\in D\}$ is disjoint from $D$ (except at the boundary, which won't affect the integral) and $D\cup (-D)=[-1,1]\times [-1,1]$. Let $R=[-1,1]\times [-1,1]$.
Then,
$$\frac{5}{2} \iint_{-D} x + xy f(x^2+y^2) \,dA
=\frac{5}{2} \iint_{D} -x + xy f(x^2+y^2) \,dA.
$$
Let this value be denoted by $I_{-D}$. Similarly let $I_D=\frac{5}{2} \iint_{D} x + xy f(x^2+y^2) \,dA$ and let $I_R=\frac{5}{2} \iint_{R} x + xy f(x^2+y^2) \,dA$. 
Then $I_D-I_{-D} = 5 \iint_D x \,dA.$
On the other hand, $I_D+I_{-D}=I_R$. 
As long as we can work out $I_R$, we can solve for $I_D$.
Now, because the region, $R$, is symmetric about the $x$-axis, replacing $y$ by $-y$ and integrating over the region will yield the same result. However when we substitute in $-y$, we get 
$$\frac{5}{2} \iint_D x -xy f(x^2+y^2) \,dA.$$
Then we have 
$$2I_R = \frac{5}{2} \iint_R x -xy f(x^2+y^2) \,dA + \frac{5}{2} \iint_R x +xy f(x^2+y^2) \,dA = 5 \iint_R x \,dA, $$
or
$$I_R = \frac{5}{2} \iint_R x \,dA = 0, $$
since $R$ is symmetric around the $y$-axis, so positive $x$ values cancel
with negative $x$ values.
Therefore $I_D+I_{-D}=0$, so $-I_D=I_{-D}$.
Together with $I_D-I_{-D} = 5\iint_D x\,dA$, we have $I_D=\frac{5}{2}\iint_D x\, dA$. 
