Polar coordinates for $xz$-plane: $z = r\sin\theta$ ? [Stewart P1091 16.7.25] 


$1.$ The unit disk is projected onto the xz-plane, so shouldn’t $x = 1\cos \theta$ and $\color{red}{z = 1 \sin \theta} $?

User Semsem below kindly identified the problem:  The normal to the disk is on the direction $-j$ so we have to reverse the orientation as follows. 

$2.$ Would someone please explain why the orientation must be reversed? By "reverse", does Semsem mean the following, that thee $xz$-plane should be viewed in the direction of the green arrow (instead in that of the red arrow, which was my problem)?

 A: I will show you how the double integral works out no matter what you select for $z$.
Consider $$\begin{align}I &= \int_0^{2\pi}\!\int_0^1 \!\! r^3 + 2r^3f^2(\theta)\, \mathrm{dr} \, \mathrm{d}\theta \\ &=  \int_0^{2\pi}\!\dfrac{1}{4} + \dfrac{1}{2}f^2(\theta)\;\mathrm{d}\theta\end{align}$$
$f(\theta)$ is either $\cos(\theta)$ or $\sin(\theta)$ so $f^2(\theta) = \dfrac{1\pm\cos(2\theta)}{2}$.
$$\begin{align} I &= \int_0^{2\pi}\!\dfrac{1}{4} + \dfrac{1}{2}f^2(\theta)\;\mathrm{d}\theta \\&= \dfrac{\pi}{2} + \dfrac{1}{2}\int_0^{2\pi}\!\dfrac{1\pm\cos(2\theta)}{2}\mathrm{d}\theta \\ &= \dfrac{\pi}{2} + \dfrac{1}{4}\int_0^{2\pi}\!1\pm\cos(2\theta)\;\mathrm{d}\theta \\&= \dfrac{\pi}{2} + \dfrac{1}{4}\int_0^{2\pi}\!1\;\mathrm{d}\theta\pm\int_0^{2\pi}\!\cos(2\theta)\;\mathrm{d}\theta \\ &=\pi\pm \int_0^{2\pi}\!\cos(2\theta) \; \mathrm{d}\theta \\&= \pi\end{align}$$
So you can see that it does work out whether or not you set $z = \sin(\theta)$ or $z = \cos(\theta)$.
A: In this question you have to integrate over the whole disk. The best way there is to use the polar coordinates. The normal to the disk is on the direction $-j$ so we have to reverse the orientation as follows  
 
so we take $x=r\sin \theta, z=r\cos \theta, x^2+z^2=r^2$ and integrate from $r=0$ to $r=1$. So the integrand is
$$-(x^2+z^2)-2z^2=-[x^2+z^2+2z^2]=-[r^2+2z^2]=-[r^2+2r^2\cos^2\theta]$$
If we take $r=1$ we only consider the boundary of the disk while we have to consider the interior also.
A: If we parametrize with $x=r\cos\theta$ and $z=r\sin\theta$, then when $\theta$ increases from $0$ to $2\pi$, the graph created is a circle, but it is oriented clockwise whereas a standard (coherent) orientation would be anticlockwise, this is why we get the normal $-j$ rather than $j$.
But if we re parametrize with $x=r\sin\theta$ and $z=r\cos\theta$. When $\theta$ increases from $0$ to $2\pi$  the graph created is a circle, but this time it is  oriented anticlockwise, and thus has the normal $j$.
A coherent orientation is one where, when we move around the boundary, if we were "standing" on the boundary, the enclosure must always be on our left hand side, so for a circle the coherent orientation is anticlockwise.
Imagine the orientation was clockwise, then when we "stand" on the bounday, and move around the circle clockwise, the interior of the circle is no longer on our left.
With regards to "anticlockwise", I was looking from the green direction, because this corresponds to the normal $n=-j$.
