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)


Converting to cylindrical:

$x = rcos\theta$ , $y = rsin\theta$

$z(x+y) = z(rcos\theta + rsin\theta)$


$I = \int_{\pi/2}^{3\pi/2} \int_1^2 \int_0^{r^2} z(rcos\theta + rsin\theta) rdzdrd\theta$

$I = \int_{\pi/2}^{3\pi/2} \int_1^2 \int_0^{r^2} zr^2(cos\theta + sin\theta)dzdrd\theta$

$I = \frac{1}{2}\int_{\pi/2}^{3\pi/2} \int_1^2 z^2 |^{z=r^2}_{z=0} r^2(cos\theta + sin\theta)drd\theta$

$I = \frac{1}{2}\int_{\pi/2}^{3\pi/2} \int_1^2 r^6(cos\theta + sin\theta)drd\theta$

$I = \frac{1}{14}\int_{\pi/2}^{3\pi/2} r^7|^{r=2}_{r=1} (cos\theta + sin\theta)d\theta$

$I = \frac{127}{14}(sin\theta - cos\theta)|_{\theta=\pi/2}^{\theta=3\pi/2}$

$I = \frac{127}{14}(sin(\frac{3\pi}{2}) - cos(\frac{3\pi}{2}))-(sin(\frac{\pi}{2}) - cos(\frac{\pi}{2}))$

$I = \frac{127}{14}(-1 - 0)-(1 - 0)$

$I = -\frac{127}{7}$ <--- negative?!?

  • $\begingroup$ What does it mean " and to the left of y-z plane"? $\endgroup$
    – user
    Nov 25, 2019 at 16:49
  • $\begingroup$ @user Means that the portion of the solid to be integrated is the portion to the left of the y-z plane. So I mentioned in parentheses that it would be the portion in the second and third quadrant. $\endgroup$
    – Amai
    Nov 25, 2019 at 16:52
  • $\begingroup$ Yes finally I realized what you mean, maybe it should be stated the region with $z\ge 0$ and $x\le 0$. If I'm not wrong with the intepretation, I've proposed a set up for that. Take a look and let me know. $\endgroup$
    – user
    Nov 25, 2019 at 16:53
  • $\begingroup$ Negative value is fine since we are assuming $x\le 0$ in the region. $\endgroup$
    – user
    Nov 25, 2019 at 17:15

1 Answer 1


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.

  • $\begingroup$ From the one you are showing, isn't the second integral for theta rather than r? And I'm not understanding overall why that is the setup. $\endgroup$
    – Amai
    Nov 25, 2019 at 16:56
  • $\begingroup$ @Amai Yes sorry I need to reorder the variables. $\endgroup$
    – user
    Nov 25, 2019 at 17:02
  • $\begingroup$ Is it not possible to do it in the order $dzdrd\theta$? That is how all the images I drew and what I was taught to integrate by. $\endgroup$
    – Amai
    Nov 25, 2019 at 17:08
  • $\begingroup$ @Amai Yes of course we can also do in that other way! I add that one $\endgroup$
    – user
    Nov 25, 2019 at 17:10
  • $\begingroup$ Thank you! I thought it couldn't be a negative value. Do you happen to know what exactly I'm integrating for? I believe double integrals would be volume, but I'm not very certain about triple. Just out of curiousity since it is possible to get a negative value. $\endgroup$
    – Amai
    Nov 25, 2019 at 17:22

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