I have been working on the same problem for hours and originally I thought I finally achieved the answer and then I realized I converted the function to cylindrical coordinates improperly.
Now, on my next try I ended up with a negative value and I'm not sure if that is a reasonable answer (I feel like it isn't) or if I integrated to the wrong bounds.
$I = \iiint z(x+y) dV$, E is the region bounded by:
$x^2 + y^2 = 1$
$x^2 + y^2 = 4$
$z = x^2 + y^2$
the z-y plane and to the left of y-z plane (integrating the portion that lies in the second and third quadrant)
MUST BE CONVERTED FROM RECTANGULAR TO CYLINDRICAL THEN EVALUATED.
Converting to cylindrical:
$x = rcos\theta$ , $y = rsin\theta$
$z(x+y) = z(rcos\theta + rsin\theta)$
Integral (NOT DEFINITE IF THESE ARE CORRECT):
$I = \int_{\pi/2}^{3\pi/2} \int_1^2 \int_0^{r^2} z(rcos\theta + rsin\theta) rdzdrd\theta$
$I = \int_{\pi/2}^{3\pi/2} \int_1^2 \int_0^{r^2} zr^2(cos\theta + sin\theta)dzdrd\theta$
$I = \frac{1}{2}\int_{\pi/2}^{3\pi/2} \int_1^2 z^2 |^{z=r^2}_{z=0} r^2(cos\theta + sin\theta)drd\theta$
$I = \frac{1}{2}\int_{\pi/2}^{3\pi/2} \int_1^2 r^6(cos\theta + sin\theta)drd\theta$
$I = \frac{1}{14}\int_{\pi/2}^{3\pi/2} r^7|^{r=2}_{r=1} (cos\theta + sin\theta)d\theta$
$I = \frac{127}{14}(sin\theta – cos\theta)|_{\theta=\pi/2}^{\theta=3\pi/2}$
$I = \frac{127}{14}(sin(\frac{3\pi}{2}) – cos(\frac{3\pi}{2}))-(sin(\frac{\pi}{2}) – cos(\frac{\pi}{2}))$
$I = \frac{127}{14}(-1 – 0)-(1 – 0)$
$I = -\frac{127}{7}$ <— negative?!?
Best Answer
It seems that the set up should be
$$\int_0^1 \int_{\pi/2}^{3\pi/2} \int_1^{2} z(r\cos\theta + r\sin\theta) r\,dr\,d\theta\,dz+ \int_1^4 \int_{\pi/2}^{3\pi/2} \int_{\sqrt z}^{2} z(r\cos\theta + r\sin\theta) r\,dr\,d\theta\,dz$$
or as an alternative by your set up
$$\int_{\pi/2}^{3\pi/2}\int_1^{2} \int_0^{r^2} z(r\cos\theta + r\sin\theta) r\,dz\,dr\, d\theta=-\frac{127}7$$
and the result seems correct indeed we are assuming $x\le 0$ in that region.