$A= \{(x,y,z) \in \mathbb{R}^3:x^2+2y^2+z^2<4z\}$. Calculate the limit:
$$\lim_{n \to \infty}\frac{1}{n} \int_A \frac{y^2z}{ln(x^2+2y^2+n) - ln(n)} \ d\lambda_3$$
Solution:
First, I can parametrize the set A with:
$$x = r cos \alpha$$$$\sqrt{2} y = \ r sin \alpha \implies y = \frac{\sqrt{2}}{2} \ r sin \alpha$$
The jacobian of that parametrization is equal to $\frac{\sqrt{2}}{2} r$
From that I get:$$ x^2+2y^2+z^2<4z \iff r^2 cos^2 \alpha + r^2 sin^2 \alpha + z^2 < 4z \iff z^2 - 4z + r^2 < 0$$
So we get that: $x \in (2 - \sqrt{4 - r^2}, 2 + \sqrt{4 - r^2})$
Then:$$\lim_{n \to \infty}\frac{1}{n} \int_A \frac{y^2z}{ln(x^2+2y^2+n) - ln(n)} \ d\lambda_3 = \lim_{n \to \infty} \int_A \ \frac{y^2z}{ ln \left( 1+ \frac{x^2+2y^2}{n} \right)^n } \ d\lambda_3$$
Here I would like to use Dominated convergence theorem. Therefore I need...
A function that would bound the function under my limit. That can be: $f(x,y,z) = y^2 z$ since the logarithm in the denominator is smaller than 1 since it goes to $x^2 + 2y^2$ from the bottom up and on the given set A, the value of $x^2 + 2y^2$ will be smaller than $1$. Function $f(x,y,z)$ needs to be integrable over A and I need to check that.
A second function that would be the limit of functions functions under my limit. Therefore I can take: $g(x,y,z) = \frac{y^2 z}{x^2 + 2y^2}$. Function $f(x,y,z)$ needs to be integrable over A and I need to check that:$$\frac{\sqrt{2}}{2} \int_0^{2 \pi} \int_0^{\infty} \int_{2 - \sqrt{4 - r^2}}^{2 - \sqrt{4 + r^2}} r cos^2 \alpha \cdot z \ dz \ dr \ d \alpha$$
... and that looks like a total mess so I'm not sure if it's any good.
