Lebesgue measure of $\{(x, y, z) \in \mathbb{R^3}: z\sqrt{x^2 + y^2} < 2, \sqrt{x^2 + y^2} < z + 1, z > 0 \}$

Question

Find Lebesgue measure of $M := \{(x, y, z) \in \mathbb{R^3}: z\sqrt{x^2 + y^2} < 2, \sqrt{x^2 + y^2} < z + 1, z > 0 \}$.

My approach: Use substitution: $\rho(r,t,s) = (r\cos(t), r\sin(t), s)$ on $(0, \infty)\times(0, 2\pi)\times(0, \infty)$. Jacobian of $\rho$ is $r$.

Then by applying substitution we get:

$$ \begin{align*} r &\lt \,\frac{2}{s} \\ r &\lt s + 1 \\ s &\gt \,0 \end{align*} $$

$r > 0$, that gives: $0 < r < \frac{1}{s} + \frac{s}{2} + \frac{1}{2}$, $0 < t < 2\pi$, but I have issues with specifying upper bound for variable $s$ to form an integral. Lower bound is apparently 0 (based on third inequality and the fact that $r$ is non negative from first inequality).

Bounds for $r$ were calculating adding first 2 equations and dividing by 2.

Also another approach is merging first two inequalities before substitution, so we get:

$$\sqrt{x^2 + y^2} < \frac{z + 3}{z + 1}$$ but here $z$ can go to $\infty$, which is I suppose not the correct upper bound.

Restriction: I can only use substitution of either spherical or cylinder polar coordinates. So $\rho$ is either in the form: $$(r\cos(t), r\sin(t), s)$$ or: $$(r\cos(t)\cos(s), r\sin(t)\cos(s), r\sin(s))$$ with possible modifications such as $(\sqrt{2}r\cos(t), r\sin(t) + 1, s)$ etc.

Hints are welcome.

Thanks.

Answer 1

Using your substitution:

\begin{align} V &= \int_{0}^{\infty} \int_{0}^{2\pi} \int_{0}^{\infty} r \mathbf 1_{r\le \frac2s}\mathbf 1_{r \le s+1}\mathrm dr\mathrm dt \mathrm ds\\ &= 2\pi \left(\int_{0}^{\infty} \int_{0}^{\infty} r \mathbf 1_{r\le \frac2s}\mathbf 1_{\frac2s \le s+1}\mathrm dr \mathrm ds + \int_{0}^{\infty} \int_{0}^{\infty} r \mathbf 1_{r\le s+1}\mathbf 1_{\frac2s > s+1}\mathrm dr \mathrm ds\right)\\ &= 2\pi \left(\int_{0}^{\infty} \int_{0}^{\infty} r \mathbf 1_{r\le \frac2s}\mathbf 1_{s^2 + s - 2\ge 0}\mathrm dr \mathrm ds + \int_{0}^{\infty} \int_{0}^{\infty} r \mathbf 1_{r\le s+1}\mathbf 1_{s^2 + s - 2\le 0}\mathrm dr \mathrm ds\right)\\ &= 2\pi \left(\int_{0}^{\infty} \frac12 \times \frac{4}{s^2}\mathbf 1_{s^2 + s - 2\ge 0}\mathrm ds + \int_{0}^{\infty} \int_{0}^{\infty} \frac12(s+1)^2\mathbf 1_{s^2 + s - 2\le 0}\mathrm ds\right)\\ &= \pi \left(4\int_{a}^{\infty} \frac1{s^2}\mathrm ds + \int_{0}^{a} (s+1)^2\mathrm ds\right)\\ &= \pi \left(\frac4a + \frac13(a+1)^3 - \frac13\right) \end{align}

Where is the positive solution of $s^2 + s - 2=0$, $$a = \frac{-1 + \sqrt{1+8}}{2} = 1$$ Then, $V=\frac{19}3\pi$.