Calculate $\int_{\mathbb R^3} x_3^2 e^{-\lVert x \rVert _2} \lambda_3(dx)$

Question

I want to calculate $$\int_{\mathbb R^3} x_3^2 e^{-\lVert x \rVert _2} \lambda_3(dx)$$ where $x = (x_1, x_2, x_3)$.

Attempt:
Parametrize $\mathbb R^3$ a.e. via the diffeomorphism $$\Psi:\,]0, \infty[ \,\times \, ]0, \pi[ \,\times\, ]-\pi, \pi[ \to \mathbb R^3 \setminus(\mathbb R_0^- \times \{0\} \times \mathbb R) \\ (r, \theta , \phi) \mapsto (r \cos \theta \sin \phi, r \sin \theta \sin \phi, r \cos \theta)$$ (change to spherical coordinates). Since $\lVert x \rVert_2 = r$ we get $$\int_{\mathbb R^3} x_3^2 e^{-\lVert x \rVert _2} \lambda_3(dx) = \int_0^\infty \int_0^\pi \int_{-\pi}^\pi r² \cos^2 {\theta} \,e^{-r}r^2 \sin \theta \, d\phi d\theta dr = \\ =2\pi \int_0 ^\infty r^4e^{-r}\int_{-\pi}^\pi \cos^2 \theta \sin \theta \, d\theta dr$$ Now substituting $u = \cos \theta$ the last integral evaluates to $$[-\frac{1}{3}\cos(\theta)]_{\theta = 0}^\pi = \frac{2}{3}.$$ With the other integral I am not sure. Of course, I can integrate by parts four times, but certainly there is a better way, maybe by expressing it through a Gauss function.
Anyhway, using Wolfram Alpha it evaluates to $24$ and we get
$$\int_{\mathbb R^3} x_3^2 e^{-\lVert x \rVert _2} \lambda_3(dx) = 32\pi.$$ Is this correct and is there a smart way to evaluate the $r^4 e^{-r} $ integral?

Answer 1

if you're familiar with Laplace transforms you can notice that : $$F(s) = \int_{0}^{+\infty}x^4e^{-sx}dx = \frac{4!}{s^5}$$

and you just want $F(1) = 4! = 24$

Answer 2

By symmetry, your integral is just

$$ \frac{1}{3}\iiint_{\mathbb{R}^3}\|x\|_2^2 e^{-\|x\|_2} \, d\mu_3 = \frac{1}{3}\int_{0}^{+\infty} 4\pi \rho^2\cdot \rho^2\cdot e^{-\rho}\,d\rho=\frac{4\pi}{3}\cdot 4!=\color{red}{32\pi}.$$