Meaning of bounds in $ \iint_{0\leq x^2+y^2\leq 4} \, dx\, dy $ If the bounds are written as
$$
\iint_{0\leq x^2+y^2\leq 4} \, dx\, dy
$$
What does it mean? I want the bounds written as
$$
\int_{x_\text{lower}}^{x_\text{upper}}\int_{y_\text{lower}}^{y_\text{upper}} \, dx\,dy
$$
Same for a triple integral, if
$$
\iiint_{2\leq x^2+y^2+z^2 \leq 6} \, dx\,dy\,dz
$$
What is $
\int_{x_\text{lower}}^{x_\text{upper}}\int_{y_\text{lower}}^{y_\text{upper}} \int_{z_\text{lower}}^{z_\text{upper}} \, dx\,dy \,dz
$?
Thanks!
 A: When you write $\displaystyle \int_\bullet^\bullet \int_\bullet^\bullet \cdots\, dx\,dy, $ then one integral is INSIDE the other: $\displaystyle \int_\bullet^\bullet \left( \int_\bullet^\bullet \cdots \, dx \right) \, dy.$
So the variable $y$ goes from $-2$ up to $2$, but in the INSIDE integral, for every fixed value of $y$, the other variable $x$ goes from something to something.  You have $x^2\le 4-y^2$, so $-\sqrt{4-y^2} \le x \le \sqrt{4-y^2}$.  Thus you have
$$
\int_{-2}^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} \cdots\,dx\,dy.
$$
A: It mean Area of the disc $\mathcal D=\{(x,y)\mid x^2+y^2\leq 4\}$. You can change it as $$\int_{0}^{2\pi}\int_0^2 r\,\mathrm d r \, \mathrm d \theta.$$
A: $0 \leq x^2 + y^2 \leq 4$ defines a region in $\mathbb{R}^2$, specifically the closed disk centered at the origin with radius $2$.  
At a given $x$, notice that $y$ ranges from $-\sqrt{4 - x^2}$ to $\sqrt{4 - x^2}$, and further $x$ can take values between $-2$ and $2$.  Therefore:
$$\iint_{0 \leq x^2 + y^2 \leq 4} \ dx\,dy = \int_{-2}^2 \int_{-\sqrt{4 - x^2}}^{\sqrt{4-x^2}} \ dy \,dx$$
Note that the disk is symmetric in $x$ and $y$, so if you prefer, the order of integration can be swapped by everywhere replacing $x$ with $y$ and vice-versa.  I'd recommend not doing it this way though; the integral looks nicer with polar coordinates, where $0 \leq r \leq 2$ and $0 \leq \theta \leq 2 \pi$.  Or we can apply Green's theorem to arrive at:
$$\iint_{0 \leq x^2 + y^2 \leq 4} \ dx\, dy = \frac{1}{2} \oint_C \ x\,dy - y\,dx$$
where $C$ is the circle of radius $2$ centered at the origin.
The easiest method for this problem, however, is recognizing that, for a given region $R \subset \mathbb{R}^n$, the integral $\displaystyle \int_R \ dx_1\,dx_2 \cdots dx_n$ gives the volume of $R$.

For the second integral, evaluating $\displaystyle \iiint_{2 \leq x^2 + y^2 + z^2 \leq 6} \ dx\, dy\, dz$ is much easier using spherical coordinates, where we'd have $0 \leq \phi \leq \pi$ and $0 \leq \theta \leq 2 \pi$ and $2 \leq \rho \leq 6$.  Thus:
$$\iiint_{2 \leq x^2 + y^2 + z^2 \leq 6} \ dx\, dy\, dz = \int_0^{2 \pi} \int_0^\pi \int_2^6 \rho^2 \sin(\phi) \ d \rho\, d \phi\, d \theta $$
(As before, the easiest method would've been to calculate the volume of this region without calculus.)
A: $x^2 + y^2 = 4$ is the equation for a circle with radius 2, centered at (0,0). So the bounds $x^2+y^2\le 4$ refers to the set of all points within that circle.
That means:
$$
\iint_{0 \leq x^2 + y^2 \leq 4} dx\, dy = \int_{-2}^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}dx\, dy
$$
Similarly, 2 ≤ x2 + y2 + z2 ≤ 8 refers to the the set of all points inside a sphere of radius 2, but outside a sphere of radius 3√2. (Both spheres are centered at 0,0)
