Expected value of the area of an annulus Consider the following problem:

Let $X,Y,Z,T$ be independent random variables with standard normal
distribution ($N(0,1)$). Consider the circles with center $(0,0)$ and radius
$(X,Y)$ and $(Z,T)$, respectively. Let $A$ be equal to the area of the
annulus defined by these circles (i.e. the region covered between the two concentric circles).
Calculate $\mathbb{E}(A)$ (the
expected value of the random variable $A$).

My attempt:
Since these random variables are independent, their joint probability density function can be written as:
$$ f(x,y,z,t) = \prod_{cycl} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} = \frac{1}{4\pi^2} e^{-\frac{1}{2}(x^2 + y^2 + z^2 +t^2)} $$
On the other hand, the area of the annulus can be calculated the following way:
$$ A= \pi |X^2+Y^2-Z^2-T^2| =: g(X,Y,Z,T) $$
So this means that
$$ \mathbb{E}(A) = \mathbb{E}(g(X,Y,Z,T)) = \iiiint\limits_{\mathbb{R}^4} g(\mathbf{x})f(\mathbf{x}) d\mathbf{x} =  $$
$$ = \frac{1}{4\pi}\iiiint\limits_{\mathbb{R}^4} |x^2+y^2-z^2-t^2|e^{-\frac{1}{2}(x^2 + y^2 + z^2 +t^2)} \,dx\,dy\,dz\,dt = $$ $$= \frac{1}{2\pi} \iiiint\limits_{x^2+y^2>z^2+t^2} (x^2+y^2-z^2-t^2)e^{-\frac{1}{2}(x^2 + y^2 + z^2 +t^2)} \,dx\,dy\,dz\,dt =$$
$$ =\frac{1}{2\pi}\int\limits_{-\infty}^{\infty} e^{-\frac{x^2}{2}} \int\limits_{-\infty}^{\infty}e^{-\frac{y^2}{2}} \int\limits_{-\infty}^{\infty} e^{-\frac{z^2}{2}} \int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} (x^2+y^2-z^2-t^2)e^{-\frac{t^2}{2}} \,dt\,dz\,dy\,dx$$
The inner integral can be written as
$$\int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} (x^2+y^2-z^2-t^2)e^{-\frac{t^2}{2}} \,dt =$$
$$= (x^2+y^2-z^2)\int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} e^{-\frac{t^2}{2}} \,dt - \int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} t^2e^{-\frac{t^2}{2}} \,dt$$
which is fine, but I'm not able to evaluate it because of the limits...
Is my approach wrong? Thanks for any help in advance.

The length of the radius is $U := \sqrt{X^2+Y^2}$ and $U^2=X^2+Y^2\sim\chi^2(2)$ therefore the random area is $A_{X,Y}=\pi U^2\sim \Gamma(1,2\pi)$. Same for $A_{Z,T}=\pi H^2$. The characteristic function of $\pi(U^2-H^2)$ is
$$E[e^{i\pi\theta(U^2-H^2)}]=E[e^{i\pi\theta U^2}]E[e^{-i\pi\theta H^2}]=(1-2i\pi\theta)^{-1}(1+2i\pi\theta)^{-1}=\frac{1}{1+4\pi^2\theta^2}$$
We manipulate this a bit:
$$\frac{1}{1+4\pi^2\theta^2}=\frac{1}{4\pi}\frac{\frac{1}{\pi}}{\frac{1}{4\pi^2}+\theta^2}$$
We get the antitransform (i.e. the density)
$$f_{\pi(U^2-H^2)}(y)=\frac{1}{4\pi}e^{-\frac{|y|}{2\pi}}$$
The expected area of the annulus is
$$E[|\pi(U^2-H^2)|]=\int_\mathbb{R}|y|f_{\pi(U^2-H^2)}(y)dy=2\int_{[0,\infty)}y\frac{1}{4\pi}e^{-\frac{|y|}{2\pi}}dy=2\pi$$
