I am interested in calculating the Second Moment of Area integrals $$\iint_A x^2 \,\mathrm{d}A, \quad \iint_A y^2\,\mathrm{d}y, \quad \iint_A xy \,\mathrm{d}A$$ for the area $A$ which is the area enclosed by the graphs and $y=x$ and $y= x^2 - 20$.
$\qquad \qquad \qquad \qquad \qquad $ 
First of all, I found the points where the two graphs intersect given by $x^2 - x - 20 = 0$, which are $A(-4,-4)$ and $B(5,5)$.
For the latter two integrals, I think I am correct in the calculation, as : $$\iint_A y^2 \,\mathrm{d}A = \int_{-4}^5 \int_{x^2-20}^x y^2 \,\mathrm{d}y \, \mathrm{d}x \quad \text{and} \quad \iint_A xy \, \mathrm{d}A = \int_{-4}^5 x \left(\int_{x^2-20}^x y \,\mathrm{d}y \right)\,\mathrm{d}x$$
Correct me if I am mistaken for the integral limits I have set above, but my thought proccess was that $y$ moves alongsides $x^2-20$ until it reaches $y=x$ given that $x\in[-4,5]$.
I am not sure though, how to approach the first integral. Solving for $x$, one yields that of course $x=y$ and $x = \pm \sqrt{y + 20}$. Since we have negative values for $x$ as well, I am not sure how to set the integral limits : $$\iint_A x^2 \,\mathrm{d}A = \int_{-4}^5 \int_y^{\sqrt{y+20}} x^2 \,\mathrm{d}x \,\mathrm{d}y \quad \Large?$$ I would greatly appreciate any clarification in the given calculations/approaches and any trick that my help me generally in making it out clean of such integrations.
Why don't you approach the $x^2$ integral the same as for the other? You integrate $dy$ then integrate $x^2dx$: $$\iint_Ax^2dA=\int_{-4}^5x^2dx\int_{x^2-20}^xdy$$