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)?


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 $z = \cos(\theta)$.
