How to evaluate the integrals in the cylindrical coordinates 
Evaluate the following integral in cylindrical coordinates
  $$\int^{1}_{-1}\int^{\sqrt{1-x^2}}_{0}\int^{2}_{0}\dfrac{1}{1+x^2+y^2}dzdydx$$

My try:
I first took the boundaries as $$-1\le x\le1\\0\le y\le\sqrt{1-x^2}\\0\le z\le2$$
and I know the formula that $$D=\{(r,\theta,z):g(\theta)\le r\le h(\theta),\alpha\le\theta\le\beta,G(x,y)\le z\le H(x,y)\}$$$$\int^{}_{}\int^{}_{D}\int^{}_{}f(r,\theta,z)dV=\int^{\beta}_{\alpha}\int^{h(\theta)}_{g(\theta)}\int^{H(r\cos\alpha,r\sin\theta)}_{G(r\cos\theta,r\sin\theta)}f(r,\theta,z)dzdrd\theta$$
But how to apply this formula and change the boundaries of the integrals?
Can anyone please explain this.
 A: First of all, your $z$ goes from $0$ to $2$ so we just have to look at the $xy$ plane. In the $xy$ plane, you have the upper half circle with radius $1$. So $0\le \theta\le \pi$ and $0\le r\le 1$. 
So:
$$\int_0^{\pi}\int_0^1\int_0^2\frac{1}{1+r^2}rdzdrd\theta$$
This integral should be easy enough, use a $u$ sub for the $r$ part.
A: I think your primary confusion here arises from the fact that you think you must "apply" that formula. What I believe that formula is telling you is the general form for a cylindrical variable transformation.
Side note: I believe the argument of the integral requires an additional $r$ in it; i.e. it should say $$\iiint_D f(r,\theta,z)r\ dzdrd\theta$$
The main takeaway from your formula should be that after your transformation, the $z$ bounds will depend on $r$ and $\theta$, the $r$ bounds will depend on $\theta$, and the $\theta$ bounds will be constants.
In order to determine the specific bounds for your problem, you should try to picture what the domain $D$ of this integral looks like. Here, it is just a half-cylinder of height 2, so all of your bounds will be constants, as Zachary has already pointed out.
A: Since we see that the limits of $z$ are independent of $x$ and $y$, we can rewrite the given integral as:$$\int^{2}_{0} \left( \int^{1}_{-1}\int^{\sqrt{1-x^2}}_{0}\dfrac{1}{1+x^2+y^2}dydx\right)dz$$
Let $$I=\int^{1}_{-1}\int^{\sqrt{1-x^2}}_{0}\dfrac{1}{1+x^2+y^2}dydx$$
Now, observing the limits of $x$ and $y$ it can be seen that they span over the area of a semicircle (as shown in the image below):

Therefore, to evaluate $I$ we'll convert the integral into polar cooordinates (which are cylindrical coordinates in this case). 
We substitute $x=r\cos\theta$ and $y=r\sin\theta$ so that $x^2+y^2=r^2$
Now we have to find $dA=dxdy$ in these coordinates. In polar coordinates let us consider a strip of width $dr$ in radial direction $d\theta$ in angular direction. Thus the length of small differential element is $rd\theta$ and therefore differential area element is $rdrd\theta$.That is, $$dA = dxdy = rdrd\theta$$

We can see that the limits for $r$ and $\theta$ for the semicircle will be,
$$0\le r\le1\\0\le \theta\le\pi$$
Therefore,
$$I=\int^{1}_{-1}\int^{\sqrt{1-x^2}}_{0}\dfrac{1}{1+x^2+y^2}dydx$$
$$I=\int^{1}_{0}\int^{\pi}_{0}\dfrac{1}{1+r^2}rdrd\theta$$
A: Like you wrote, and correctly, we have that
$$0 \leq x \leq 1$$
$$0 \leq y \leq \sqrt{1 - x^2}$$
$$0 \leq z \leq 2$$
and we also have that for cylindrical coordinates, these relations hold
$$\begin{cases}
x = \rho \cos\theta \\
y = \rho \sin\theta \\
z = z
\end{cases}$$
so we immediately get that in cylindrical coordinates, $0 \leq z \leq 2$. For $\rho$ and $\theta$ consider this sketch:

from $0 \leq y$ I painted the red region; from $-1 \leq x \leq 1$ I painted the blue region and from $y \leq \sqrt{1 - x^2} \iff y^2 + x^2 \leq 1$ I painted the green region. From the sketch you get $0 \leq \rho \leq 1$ and $0 \leq \theta \leq \pi$.
Also, note that from here for example, you get

where the $E$ from the image is your $D$. The point is, when you change coordinates, an extra $\rho$ should appear in the integral. Hence, your integral will become
$$\int_0^\pi \int_0^1 \int_0^2 \frac{1}{1 + \rho^2}\rho\ dzd\rho d\theta $$
Can you proceed with the integration?
