Wanting to know if my set up for these integrals using spherical and cylindrical coords is correct. So as usual I've been doing problems out my book for practice, only to find one of the answers pages with this section torn out... I don't expect to get answers for all of them here, or even full answers for a few, but I was hoping maybe some could tell me if the set up on the few I've done is correct, after all, I can check the evaluation on symbolab. At least then when trying other problems I'll have something more to go off when it comes to the set up. If this isn't an ok question or should be broken up into separate ones, just let me know, just hoping to get all the practice I can.
Using Cylindrical:

$(A)$
$\int\int\int (z) dv$ where E is enclosed by the paraboloid $z=x^2+y^2$ and the plane $z=4$
From the fact that it's a paraboloid I have theta between $0$ and $\pi$, r between $0$ and 2 by plugging the given z value into the paraboloid equation, and z between 4 and $r^2$ which I got from the x,y equation swapping to polar, which is the part I'm not really sure about. 
$\int^\pi_0\int_0^2\int^4_{r^2}(zr)dzdrd\theta$

$(B)$
$\int\int\int_E (x) dv$ where E is the solid in the first octant that lies under the paraboloid $z=4-x^2-y^2$."
Since it's the first octant I have $0<=\theta<=\pi/2$, r between 0 and 4 from the z equation, and zero through the the z equation for z:
$\int^{\pi/2}_0\int^4_0\int^{4-r^2}_0[(4-r^2)r]dzdrd\theta$

Using Spherical:

$(C)$
$\int\int\int(xe^{x^2+y^2+z^2})dv$ where E is the portion of the unit ball $x^2+y^2+z^2<=1$ that lies in the first octant.
From the first octant part I have theta between 0 and $\pi/2$, and from $\rho^2=x^2+y^2+z^2$ I have $\rho$ between zero and 1. I'm not sure how to determine $\phi$ in this one though going off a similar problem I seen I have it between zero and $\pi/4$. Replacing all the x and y's with the matching cartesian to spherical identities:
$\int^{\pi/2}_0\int_0^{\pi/4}\int_0^1(\rho sin(\phi)e^{\rho^2}\rho^2 sin(\phi))d\rho d\phi d\theta$

...And I think I'll leave it there. Have a couple more I tried in spherical but this seems like enough for one post. 
 A: *

*For (a), the intersection with the paraboloid and the plane is $x^2+y^2=4$; in particular, the projection onto the $xy$-plane is indeed the full circle $x^2+y^2=4$. For this reason, we should have $0 \le \theta \le 2\pi$ and not just the semicircle $0 \le \theta \le \pi$.
$$
\int^{2\pi}_0\int_0^2\int^4_{r^2} rz  \; dz \,  dr \, d\theta
$$

*For (b), again the projection onto the $xy$-plane looks like $x^2+y^2 =4$ (only in the first quadrant). Thus, need to change upper bound for $r$ to 2. Also, in polar coordinates $x= r \cos \theta$, so your integral looks like
$$
\int^{\pi/2}_0\int^2_0\int^{4-r^2}_0 [r \cos \theta] r \; dz \, dr \, d\theta
$$

*For (c), again the first octant restricts $0 \le \theta \le \pi/2$ since you only have $x \ge 0, \ y  \ge 0$. Correct that $0 \le \rho \le 1$ since you're integrating over the unit ball. The first octant restricts $\phi$ to $0 \le \phi \le \pi/2$, since $\phi$ will range from 0 (straight up) to the $xy$-plane (which is $\phi = \pi/2)$. See http://mathworld.wolfram.com/SphericalCoordinates.html for an illustration.
Alternatively, note that $z = \rho \cos \phi = \cos \phi$, so this equals one when $\phi=0$ and equals zero when $\phi = \pi/2$.
Finally, $x = \rho \sin \phi \cos \theta$ in spherical, which goes in the integrand along with $e^{\rho^2}$.
$$
\int^{\pi/2}_0\int_0^{\pi/2}\int_0^1(\rho 
\sin \phi \cos \theta) e^{\rho^2}\rho^2 \sin \phi \; d\rho \,  d\phi \, d\theta
$$


Let me know if you have questions.
