variance in the time domain versus variance in frequency domain
Variance is never defined as power. For a wide-sense stationary random process $X(t)$ with zero mean

$$\mu_X=E\{X(t)\}=0\tag{1}$$

the variance of $X(t)$ equals its power.

The autocorrelation of $X(t)$ is defined by

$$R_X(\tau)=E\{X^*(t)X(t+\tau)\}\tag{2}$$

The power of $X(t)$ is

$$P_X=E\{|X(t)|^2\}=R_X(0)\tag{3}$$

The variance of $X(t)$ is

$$\sigma^2_X=E\{|X(t)-\mu_X|^2\}=E\{|X(t)|^2\}-|\mu_X|^2=R_X(0)-|\mu_X|^2\tag{4}$$

Clearly, if $\mu_X=0$, we have $\sigma^2_X=R_X(0)=P_X$.

Note that the autocovariance of $X(t)$ is defined by

$$C_X(\tau)=E\{[X(t)-\mu_X]^*[X(t+\tau)-\mu_X]\}=R_X(\tau)-|\mu_X|^2\tag{5}$$

Consequently,

$$\sigma^2_X=C_X(0)\tag{6}$$

But since the power spectral density $S_X(\omega)$ is the Fourier transform of $R_X(\tau)$, the integral over $S_X(\omega)$ equals the power of $X(t)$ and not its variance, unless $\mu_X=0$:

$$P_X=R_X(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_X(\omega)d\omega\tag{7}$$

Also take a look at this related answer.

For the difference between the definition of autocorrelation in statistics and in signal processing look here.