2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I think the integration region is not right. $D$ is disconnected if it is $x^2-y^2+z^2 \leq 0$. $\endgroup$
    – Hw Chu
    May 20, 2020 at 19:11
  • $\begingroup$ 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). $\endgroup$
    – ZaWarudo
    May 20, 2020 at 19:19
  • $\begingroup$ (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 $\endgroup$
    – Hw Chu
    May 21, 2020 at 13:00

1 Answer 1

1
$\begingroup$

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 $\rho^2<y^2$ 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: $$ = 6\pi [ \int_0^{1/\sqrt 2} \rho d\rho \int_{\sqrt{1-\rho^2}}^{\sqrt{4-\rho^2}} dy +\int_{1/\sqrt 2}^{\sqrt 2} \rho d\rho \int_\rho^{\sqrt{4-\rho^2}} dy ] $$ $$ = 6\pi [ \int_0^{1/\sqrt 2} \rho d\rho (\sqrt{4-\rho^2} -\sqrt{1-\rho^2}) +\int_{1/\sqrt 2}^{\sqrt 2} \rho d\rho (\sqrt{4-\rho^2}-\rho) ] $$ $$ = 6\pi [ \int_0^{\sqrt 2} \rho d\rho \sqrt{4-\rho^2} -\int_0^{1/\sqrt 2} \rho d\rho \sqrt{1-\rho^2} -\int_{1/\sqrt 2}^{\sqrt 2} \rho^2 d\rho ] $$ $$ = 6\pi [ \frac12 \int_0^ 2 du \sqrt{4-u} -\frac12 \int_0^{1/2} du \sqrt{1-u} -\frac{7\surd 2}{12} ] $$ $$ = 6\pi [ \frac12 (-\frac{4\surd 2}{3}+\frac{16}{3}) -\frac12 (-\frac{\surd 2}{6}+\frac23) -\frac{7\surd 2}{12} ] = 6\pi [ \frac{7}{3}-\frac{7\surd 2}{6} ] =7\pi(2-\sqrt 2) $$

$\endgroup$
4
  • $\begingroup$ 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 $$\int_0^{1/\sqrt{2}} {\rm d}\rho \int_{\sqrt{1-\rho^2}}^{\sqrt{4-\rho^2}} {\rm d}y + \int_{1/\sqrt{2}}^{\sqrt{2}} {\rm d}\rho \int_\rho^{\sqrt{4-\rho^2}} {\rm d}y \, .$$ $\endgroup$
    – Diger
    Feb 16, 2022 at 16:45
  • $\begingroup$ 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. $\endgroup$ Feb 16, 2022 at 16:51
  • $\begingroup$ 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 $\rho > y$. $\endgroup$
    – Diger
    Feb 16, 2022 at 17:36
  • $\begingroup$ 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 $y>\rho$. $\endgroup$ Feb 16, 2022 at 17:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .