Thread: The measure of a subset of the real plane given by inequalities

1. The measure of a subset of the real plane given by inequalities

I have to calculate the Lebesgue measure of

\left\{(x,y,z)\,:\,x^2+y^2\leq 1,\,(1-x^2-y^2)^{k}z^2\leq(x^2-y^2)^2\right\}

for every $\displaystyle k\in\mathbb{Z}$
After the substitution I get $\displaystyle \int_U r drd\alpha dz,$ where

$\displaystyle U=\left\{0<r<1,\,\alpha\in(0,2\pi),\,(1-r^2)^kz^2<r^4\cos2\alpha\right\}.$

I can't calculate it. Everything I try leads to difficult integrals. Could you please give me an idea of what the right way of doing this is?

PS: Latex wouldn't work, so I left the code. If any moderator could fix this, I would be very grateful.

2. Originally Posted by ymar
I have to calculate the Lebesgue measure of
$\displaystyle \left\{(x,y,z)\,:\,x^2+y^2\leq 1,\,(1-x^2-y^2)^{k}z^2\leq(x^2-y^2)^2\right\}$
Now we can see the formula.

3. Thank you, Fernando!

PS I have just noticed that the title of my thread is misleading. Of course the set is in $\displaystyle \mathbb{R}^3$. I made a mistake.

4. Originally Posted by ymar
I have to calculate the Lebesgue measure of

$\displaystyle \left\{(x,y,z)\,:\,x^2+y^2\leq 1,\,(1-x^2-y^2)^{k}z^2\leq(x^2-y^2)^2\right\}$

for every $\displaystyle k\in\mathbb{Z}$
After the substitution I get $\displaystyle \int_U r drd\alpha dz,$ where

$\displaystyle U=\left\{0<r<1,\,\alpha\in(0,2\pi),\,(1-r^2)^kz^2<r^4\cos^{\color{red}2}2\alpha\right\}.$ Note the square there!

I can't calculate it. Everything I try leads to difficult integrals. Could you please give me an idea of what the right way of doing this is?
The condition on z for a point to lie in U is $\displaystyle -z_0<z<z_0$, where $\displaystyle z_0 = \frac{r^2|\cos2\alpha|}{(1-r^2)^{k/2}}.$ The integral for negative z is the same as for positive z, and so

$\displaystyle \int_U r\, dzd\alpha dr = 2\int_0^1\int_0^{2\pi}\int_0^{z_0} r\, dzd\alpha dr = 2\int_0^1\int_0^{2\pi}\frac{r^3|\cos2\alpha|}{(1-r^2)^{k/2}}d\alpha dr$ (doing the z integral first).

The variables conveniently separate and give you $\displaystyle 2\int_0^{2\pi}|\cos2\alpha|\,d\alpha\int_0^1\frac{ r^3}{(1-r^2)^{k/2}}\,dr.$ Check that $\displaystyle \int_0^{2\pi}|\cos2\alpha|\,d\alpha = 8\int_0^{\pi/4}\cos2\alpha\,d\alpha = 4.$ For the other integral, substitute $\displaystyle s=r^2$ and get the integral in the form

$\displaystyle \tfrac12\int_0^1\frac s{(1-s)^{k/2}}\,ds = \tfrac12\int_0^1\frac {1-(1-s)}{(1-s)^{k/2}}\,ds = \tfrac12\int_0^1\bigl((1-s)^{-k/2} -(1-s)^{1-(k/2)}\bigr)\,ds,$

which you can easily integrate. I get the answer for the measure of U to be $\displaystyle \frac{16}{(2-k)(4-k)}$ (but don't trust my accuracy there). Note that this becomes infinite when k=2. I guess that U must have infinite measure whenever $\displaystyle k\geqslant2$.

5. Opalg, thank you. I would never notice the lacking square. I think the problem wouldn't be solvable without it, am I right?