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)?


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.
