Question: Assume I have two random independent standard normal variables $X,Y\sim N(0,1)$, How can I find the distribution of $Z=X^2+Y^2$? 

I thought integrating the convolution, i.e $F\_Z(z)=\displaystyle\int\_{-\infty}^{\infty}\int\_{-\infty}^{\infty}f\_{X^2}(t)f\_{Y^2}(x-t)dt$. If X is a normal variable, does also $X^2$ is a normal variable (then I could find the density by simply substituting)?

Answer: 
Your distribution is the $\chi^2$ distribution with $2$ degrees of freedom, aka the exponential distribution with mean $2$. You should be able to calculate the cdf of $X^2+Y^2$ directly by switching to polar coordinates.

Or else you can use your convolution idea. The random variable $X^2$ is not normally distributed, but you can find its distribution in various ways, such as the method of transformations. Or else you can express the probability that $X^2\le w$ as an integral of the normal density function, and then differentiate under the integral sign (fundamental theorem of calculus).
First, both $X^2$ and $Y^2$ are known distributions, namely a chi-squared distribution. See <https://en.wikipedia.org/wiki/Chi-squared_distribution>.

If they are independent, to add them just convolve the pdf's. See: <https://en.wikipedia.org/wiki/Convolution_of_probability_distributions>