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Let $D=[(x,y,z)\in R^3: x^2+y^2+z^2\le 1, z\ge 0, z^2-x^2-y^2\le0]$ and let $f(x,y,z)=\frac {z} {\sqrt{x^2+y^2}}$. The exercise is about calculating the integral: $\iiint_D f(x,y,z) dxdydz$. The text tells to be careful because it is an improper integral. Now, I know how to use integration "by wires", that is used when we can write the domain $D$ as $D=[(x,y,z)\in R^3: g_1(x,y)\le z\le g_2(x,y); (x,y)\in E]$, and "by layers", that is used when we can write the domain $D$ as $D=[(x,y,z)\in R^3:h_1\le z \le h_2; (x,y)\in E(z)]$. I also know the Jacobians of spherical coordinates [$J=r^2cos(\phi)$] and of cylindrical coordinates [$J=r$] and how to use the change of variables in a triple integral. I know, at last, that $z^2-x^2-y^2\le 0$ is an infinite cone with vertex in the origin. Still, I don't know where to start the exercise.

  • Hi matteo! You mentioned you know integration by wires so maybe you can start from there since you already have a nice low boundary for $z$. Remember that in this exercises, and in many other cases, it's very useful to visualize the domain of integration $D$ and since you already know that the equation $z^2-x^2-y^2=0$ represents a cone: what part of the $3D$ space do you keep to visualize $D$, the inside of the cone or the outside? Also you know $z \geq 0$, and you have $z^2+x^2+y^2 \leq 0$ which is the inside and the boundary of the sphere of radius $1$. – Don Abbondio Apr 13 '23 at 16:13
  • Since it is $z^2-x^2-y^2\le 0$ i think it is the inside of the cone. I will try to play with integration by wires and see what I can get. –  Apr 13 '23 at 17:25

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By applying the cylindrical coordinates $$x=r\cos t,\ y=r\sin t$$ $D$ can be described by $$0\le t< 2\pi,\ 0\le r\le 1,\ 0\le z\le \min(r,\sqrt{1-r^2})$$ Since the integral is improper due to the factor $r=\sqrt{x^2+y^2}$ in the denominator. Therefore we will integrate over the region described in cylindrical coordinates by $$0\le t< 2\pi,\ \delta\le r\le 1,\ 0\le z\le \min(r,\sqrt{1-r^2})$$ where $\delta>0.$ Then we will calculate the limit when $\delta\to 0^+.$

The integral over the restricted region is equal $$\int\limits_0^{2\pi} \int\limits_\delta^1\int\limits_0^{\min(r,\sqrt{1-r^2})} {z\over r}\,r\,dz\, dr\, dt\\ =\pi \int\limits_\delta^1\min(r^2,{1-r^2})\,dr\underset{\delta\to 0^+}{\longrightarrow}\int\limits_0^1\min(r^2,{1-r^2})\,dr$$

Ryszard Szwarc
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  • Thanks, this answer is very elegant. But considering my level in mathematics it is probably an overkill. –  Apr 13 '23 at 17:27
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    I do not think it is an overkill. You have declared that you are familiar with cylindrical coordinates. The region of integration is bounded. After changing to new coordinates the integral becomes proper. Formally we should integrate with respect to $r$ from $\delta$ to $1,$ and then take a limit when $\delta\to 0^+.$ – Ryszard Szwarc Apr 13 '23 at 18:06
  • I asked to a professor about your answer and he said to think carefully about it. I think I will, thanks. –  Apr 13 '23 at 19:30