Is $\int_E \frac{1}{(x^2+y^2)^2}dxdy$ convergent? I have to tell whether this integral is convergent:
$$\int_E \frac{1}{(x^2+y^2)^2}dxdy$$
where $E=\{0\leq y \leq x^a\} \cap \{x^2+y^2\leq 1\} $.
I'm asked for which $a \geq 0$ the integral converges. How am I supposed to act when I find this kind of integrals? I mean, these domains determined by intersections of a curve with $[0,1]^2$ or with $B_n(0,0)$.
Thanks in advance.
EDIT: I suppose I should do it for $x \geq 0$ even if not specified.

The curves $y=x^a, y=\sqrt {1-x^2}$ intersect at some $(b,b^a),$ where $0<b<1.$ Thus $E= \{(x,y): 0\le x \le b,0\le y\le x^a\}.$ We don't need the exact value of $b$ to do the problem.
The integral is thus
$$\int_0^b\int_0^{x^a}\frac{1}{(x^2+y^2)^2}\,dy\,dx.$$
In the inner integral, let $y=xt.$ The integral becomes
$$\tag 1 \int_0^b\frac{1}{x^3}\int_0^{x^{a-1}}\frac{1}{(1+t^2)^2}\,dt\,dx.$$
If $a\le 1,$ then $x^{a-1} > 1.$ So $(1)$ is at least
$$\int_0^b\frac{1}{x^3}\int_0^{1}\frac{1}{(1+t^2)^2}\,dt\,dx.$$
The inner integral is a constant $C,$ so $(1)$ is at least $C\int_0^b\dfrac{1}{x^3}\,dx = \infty.$
If $a>1,$ then $x^{a-1} <1. $ $(1)$ is then bounded above by
$$\int_0^b\frac{1}{x^3}\int_0^{x^{a-1}}1\,dt\,dx = \int_0^b \frac{1}{x^3}\cdot x^{a-1}\,dx = \int_0^b x^{a-4}\,dx. $$
This converges iff $a-4>-1,$ or $a>3.$
It follows that the given integral in the problem converges iff $a>3.$
