# Is it possible to evaluate $\iiint \frac{2x^2+z^2}{x^2+z^2} dxdydz$ using cylindrical coordinates instead of spherical?

I know that this integral is way easier with spherical coordinates, but I would like to understand my mistakes; evaluate $$\iiint_D \frac{2x^2+z^2}{x^2+z^2} dxdydz$$ Where $$D=\{(x,y,z)\in\mathbb{R}^3 \ \text{s.t.} \ 1 \leq x^2+y^2+z^2 \leq 4, \ x^2-y^2+z^2 \leq 0\}$$.

Letting $$x=\rho \cos \theta$$, $$y=y$$ and $$z=\rho \sin \theta$$ it follows that $$\iiint_E (2\cos^2 \theta+\sin^2 \theta)\rho d\rho dyd\theta=\iiint_E (1+\cos^2 \theta)\rho d\rho dyd\theta$$ Where $$E=\{(\rho,y,\theta)\in\mathbb{R}^3 \ \text{s.t.} \ 1 \leq \rho^2+y^2 \leq 4, \rho^2 \leq y^2\, \rho \geq 0, 0 \leq \theta < 2\pi\}$$.

The point is that now I have a lot of conditions on $$y$$, because $$\sqrt{1-y^2} \leq \rho \leq \sqrt{4-y^2}$$, $$-y\leq\rho\leq y$$ and $$\rho \geq 0$$.

From the existence conditions of the roots we get $$-1 \leq y \leq 1$$ and $$-2 \leq y \leq 2$$, so it follows that $$-1 \leq y \leq 1$$.

So it remains to discuss the cases of $$\max\left\{\sqrt{1-y^2},-y\right\} \leq \rho$$ and $$\rho \leq \min\left\{y,\sqrt{4-y^2}\right\}$$; it is $$y \leq \sqrt{4-y^2}$$ for $$-1 \leq y \leq 1$$ and it is always $$\sqrt{1-y^2} \leq \sqrt{4-y^2}$$, we have that $$\max\left\{\sqrt{1-y^2},-y\right\}=\begin{cases} -y, \ \text{if} -1 \leq y \leq -\frac{1}{\sqrt2} \\ \sqrt{4-y^2}, \ \text{if} \ -\frac{1}{\sqrt2} \leq y \leq 1 \end{cases}$$ So I end up with $$\iiint_E (1+\cos^2 \theta)\rho d\rho dyd\theta=\int_0^{2\pi} \left(\int_{-1}^{-\frac{1}{\sqrt2}} \left(\int_{-y}^{\sqrt{4-y^2}} (1+\cos^2 \theta)\rho d\rho\right)dy \right)d\theta+$$ $$+\int_0^{2\pi} \left(\int_{-\frac{1}{\sqrt2}}^{1} \left(\int_{\sqrt{1-y^2}}^{\sqrt{4-y^2}} (1+\cos^2 \theta)\rho d\rho\right)dy \right)d\theta$$

But I get the wrong answer, am I missing some more conditions (maybe the discussion of $$\rho \geq 0$$ too) or am I making other mistakes? Thanks.

• I think the integration region is not right. $D$ is disconnected if it is $x^2-y^2+z^2 \leq 0$. May 20, 2020 at 19:11
• Thanks for your answer, can you prove it? It is an exercise from an exam, so I hope it is not written badly (I've checked and I have copied it right on here). May 20, 2020 at 19:19
• (I wrote this as an answer, but as it was even less than a hint I think a comment is more appropriate.) It is better to graw a picture. You will need four integrations instead of two: i.stack.imgur.com/dCZnc.png May 21, 2020 at 13:00

In the $$y-\rho$$ plane the condition $$1\le \rho^2+y^2\le 4$$ defines a ring centered at $$(0,0)$$ with inner radius 1 and outer radius 2. The conditions $$\rho^2 and $$\rho\ge 0$$ subselect two 45$$^\circ$$ sections delimited by $$0\le \rho \le y$$. $$I =\int\int\int\frac{2x^2+z^2}{x^2+z^2} dxdydz$$ $$= \int_0^{2\pi} d\theta \int_0^{\sqrt 2} \rho d\rho \int_{1\le \rho^2+y^2\le 4} dy \frac{2\rho^2\cos^2\theta+\rho^2\sin^2\theta}{\rho^2}.$$ Because there is no dependence on $$y$$ in the integrand and the limits are symmetric functions of $$y$$, we may look only at $$y\ge 0$$ and introduce a factor 2 to collect the part of $$y<0$$: $$I = 2\int_0^{2\pi} d\theta \int_0^{\sqrt 2} \rho d\rho \int_{1\le \rho^2+y^2\le 4,y\ge 0} dy (2\cos^2\theta+\sin^2\theta)$$ $$= 2\int_0^{2\pi} d\theta \int_0^{\sqrt 2} \rho d\rho \int_{1\le \rho^2+y^2\le 4,y\ge 0} dy (1+\cos^2\theta)$$ $$= 6\pi \int_0^{\sqrt 2} \rho d\rho \int_{1\le \rho^2+y^2\le 4,y\ge 0} dy$$ The ring geometry suggests to chop this into 2 sections of $$\rho$$ where the $$y-$$limits differ as they are delimited by the circles and the diagonal: $$= 6\pi [ \int_0^{1/\sqrt 2} \rho d\rho \int_{\sqrt{1-\rho^2}}^{\sqrt{4-\rho^2}} dy +\int_{1/\sqrt 2}^{\sqrt 2} \rho d\rho \int_\rho^{\sqrt{4-\rho^2}} dy ]$$ $$= 6\pi [ \int_0^{1/\sqrt 2} \rho d\rho (\sqrt{4-\rho^2} -\sqrt{1-\rho^2}) +\int_{1/\sqrt 2}^{\sqrt 2} \rho d\rho (\sqrt{4-\rho^2}-\rho) ]$$ $$= 6\pi [ \int_0^{\sqrt 2} \rho d\rho \sqrt{4-\rho^2} -\int_0^{1/\sqrt 2} \rho d\rho \sqrt{1-\rho^2} -\int_{1/\sqrt 2}^{\sqrt 2} \rho^2 d\rho ]$$ $$= 6\pi [ \frac12 \int_0^ 2 du \sqrt{4-u} -\frac12 \int_0^{1/2} du \sqrt{1-u} -\frac{7\surd 2}{12} ]$$ $$= 6\pi [ \frac12 (-\frac{4\surd 2}{3}+\frac{16}{3}) -\frac12 (-\frac{\surd 2}{6}+\frac23) -\frac{7\surd 2}{12} ] = 6\pi [ \frac{7}{3}-\frac{7\surd 2}{6} ] =7\pi(2-\sqrt 2)$$
• Not sure, but do you confuse the correct sector of the annulus? $\rho<y$ and $1<\rho^2+y^2<4$ should give rise to the following integrals $$\int_0^{1/\sqrt{2}} {\rm d}\rho \int_{\sqrt{1-\rho^2}}^{\sqrt{4-\rho^2}} {\rm d}y + \int_{1/\sqrt{2}}^{\sqrt{2}} {\rm d}\rho \int_\rho^{\sqrt{4-\rho^2}} {\rm d}y \, .$$ Feb 16, 2022 at 16:45
• There are 3 regions for $\rho$. The diagonal hits the inner circle at $\rho=1/\sqrt 2$. The inner circle hits the horizontal line at $\rho=1$. The diagonal hits the outer circle at $\rho=\sqrt 2$. The outer circle hits the horizontal at $\rho=2$. If you integrate $\rho$ only to $\surd 2$ you miss a spherical cap of the outer circle. Feb 16, 2022 at 16:51
• Yeah, I know what area you are computing. Maybe I'm wrong, but as far as I see it atm, you should use the complement of the quarter annulus as you did so far. Your area atm corresponds to $\rho > y$. Feb 16, 2022 at 17:36
• I see; you're right. Indeed I need to revise my answer, because so far it corresponds to the condition $y< \rho$ but should use $y>\rho$. Feb 16, 2022 at 17:54