How to perform a double integration Suppose you are trying to find the integral of $x^2 + y^2$ such that $(x^2 + y^2) \leq 1$. How would you do this?
Attempt: I know the radius is 1 but I am stuck when trying to determine the limits of integration. 
 A: ncmathsadist's answer is perfect. I want to expand on it a bit, hinting at the parts where I think you are stuck).
We do not need to do polar coordinates (although you'll likely find while doing the integral that you'll use a trig substitution to compute it, ironically doing the polar coordinate bit surreptitiously). Usually, the way to think of these problems is to find the boundary curves. 
We recognize $x^2 + y^2 = 1$ as the unit circle. So we are integrating over the unit circle. Now let's set up our boundary curves. We're going to write our integral as $\iint (\text{stuff}) \mathrm{d} 1 \mathrm{d} 2$, where the $1$ and the $2$ are $x$ and $y$ in some order. What order today? Usually, this is done by considering which direction is easier.
Suppose we wanted to integrate with respect to $x$ first (so that the $1$ in the above integral was $x$). This seems good, because we note that for every $x$, the boundary curves are always the same (so we don't need to split up the integral or do anything fancy). What are the boundaries? Well, it goes as high as the top of the circle and as low as the bottom of the circle. The top has formula $x = \sqrt{1 - y^2}$, and the bottom $-\sqrt{1 - y^2}$. So the inner integral in this case reads 
$$\int_{-\sqrt{1 - y^2}}^{\sqrt{1 -  y^2}} (\text{stuff}) \;\mathrm{d}x$$
Now we have collapsed the $x$ direction. How far does $y$ extend over this boundary? It goes from $-1$ to $1$. That's how we get the second set of limits.
Does that make sense?
(I deliberately chose the opposite order as ncmathsadist, because they can both be done).
A: Let's take advantage of symmetry and stay in the first quadrant. Integrate as follows
$$4\int_0^1 \int_0^{\sqrt{1 - x^2}}(x^2 + y^2)\,dy\,dx$$
We obtain the limits of integration because the first quadrant of the unit disk is described by the inequalities $0\le x \le 1$ and $0\le y \le \sqrt{1 - x^2}$.
