How to evaluate two integrals (double and triple)? One with a conic boundary and the other with a square root boundary. 
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*Calculate the integral: $$\iiint_V z^2 dx dy dz$$
Where $V \subset R^3$, bounded by: $y = \sqrt z$ rotated around $Oz$, and the plane $z = h$ $(h>0)$.


My try:
Introduce the polar coordinates ($\rho \ge$0, $\phi \in[0, 2\pi]$, $\theta \in [0, \pi]$)  $x = \rho \cos\phi \sin \theta$
 $y = \rho \sin\phi \sin \theta$
 $z = \rho \cos \theta$
Because of the square root, I guess both y and z need to be non-negative, so the bounds of integration for $d \theta$ integral are from 0 to $\pi/2$ (so the cos is positive), the bounds for $\phi$ are 0 to $\pi$ (so y is positive).Is that right? What are the bounds for $\rho$? I guess some of these bounds needs to include the z = h plane, but I don't see how. 


*Estimate
$$\iint_D (x^2 - y^2)dxdy$$
where D is an area in plane $Oxy$ bounded by $x^2 + y^2 = 2x$.
My try:
I introduce the polar coordinates again, $x =  \rho \cos \phi$, $y = \rho \sin \phi$ ($\rho \ge 0$, $\phi \in [0, 2\pi]$).
Because the left of $x^2 + y^2 = 2x$ is non-negative, x also needs to be $\ge 0$, so the bounds of integration for $\phi$ are $-\pi/2$ to $\pi/2$.Now, since that equation is the equation of a circle with radius 1, I put $\rho$'s bounds to be from 0 to 1, but the solution says it should be from 0 to 2$cos\phi$. Where did that come from? Also, what do I use to "estimate" the value of the integral? The offered solutions are:a) $\pi/2 < I < 2\pi$;   b) $\pi/2 < I < 4\pi$;  c) $-\pi/2 < I < 4\pi$;   d) $-\pi/2 < I < 2\pi$;

 A: For the first one in cylindrical coordinates $x^2+y^2 =r^2$ with $dx dy = r dr$:
$$\int_0^h dz\int_0^{\surd z} rdr \int_0^{2\pi} d\phi z^2=2\pi \int_0^h z^2dz \int_0^{\surd z} r dr = 2\pi\int_0^h z^2 z/2 dz = 2\pi \left(z^4/4\right)_{\mid 0} = 2\pi h^4/8.$$
A: PROBLEM 1:
$$\begin{align}
I&\equiv \int_V z^2\,dV\\\\
&=\int_{0}^{2\pi}\int_0^{h^{1/2}}\int_{\rho^2}^{h}z^2dz\,\rho d\rho d\phi\\\\
&=2\pi\,\int_0^{h^{1/2}}\int_{\rho^2}^{h}z^2dz\,\rho d\rho \\\\\
&=2\pi\,\int_0^{h}\frac13\left(h^3-\rho^6\right)\rho d\rho\\\\
&=2\pi\,\left(\frac{h^4}{6}-\frac{h^4}{24}\right)\\\\
&=\frac{\pi}{4}h^4
\end{align}$$

PROBLEM 2:
First, we note that $x^2+y^2=2x\implies (x-1)^2+y^2=1$.  We can immediately translate coordinates through the translation $x\to x'+1$.  The region $S$ defined by $(x-1)^2+y^2\le 1$ becomes the new region $S'$ defined by $x'^2+y^2=1$.  Then we can write
$$\begin{align}
I&\equiv \int_S (x^2-y^2)\,dS\\\\
&=\int_{S'}((x'+1)^2)-y^2)dS'\\\\
&=\int_0^{2\pi}\int_0^{1}((\rho \cos \phi-1)^2-\rho^2\sin^2\phi)\,\rho d\rho d\phi\\\\
&=\int_0^{2\pi}\int_0^{1}(\rho^2 \cos 2\phi-2\rho\cos \phi +1)\,\rho d\rho d\phi\\\\
&=\pi
\end{align}$$
since the integrals from $0$ to $2\pi$ of $\cos 2\phi$ and $\cos \phi$ are equal to zero.
