Assume $X_1$ and $X_2$ are independent exponential random variables with respective parameters $\lambda_1=1$ and $\lambda_2=2$ and let $Z=\frac{X_1}{X_2}$. I am trying to show that $F_Z(z)=\frac{z}{z+2}$. The first method I have seen is $$P(\frac{X_1}{X_2}\leq z)=P(X_1\leq zX_2)=\int_0^{\infty}\int_0^{zx_2}f_{X_1,X_2}(x_1,x_2)dx_1dx_2=\cdots=\frac{z}{z+2}$$ and thus gives the desired result. But when I try $$P(\frac{X_1}{X_2}\leq z)=P(X_2\geq \frac{X_1}{z})=\int_0^{\infty}\int_{\frac{x_1}{z}}^{\infty}f_{X_1,X_2}(x_1,x_2)dx_2dx_1=\cdots=-\frac{z}{2}$$ I get a definitely wrong answer. Now, my first question is what can be wrong about the 2nd method.
But, I would like to get the general idea i.e. why do we have to integrate two times for $P(g(X,Y)\leq z)$ if $g(\cdots)$ some function and why the order of integration matters? Is there any thumb rule on what to do in these cases and what to expect about the result to judge if it's correct or wrong?
In general, if $X$ and $Y$ are continuous random variables, then the density of $W:=\frac XY$ is given by $$ f_W(w) = \int_\mathbb R |y|f_{X,Y}(wy,y)\ \mathsf dy. $$ Here $X_1$ and $X_2$ are independent so the joint density is the product of the marginal densities, and hence \begin{align} f_Z(z) &= \int_0^\infty x_2\lambda_1e^{-\lambda_1zx_2}\lambda_2 e^{-\lambda_2 x_2}\ \mathsf dx_2 \\ &= \int_0^\infty \lambda_1\lambda_2 x_2 e^{-(z\lambda_1 +\lambda_2)x_2}\ \mathsf dx_2\\ &= \frac{\lambda _1 \lambda _2}{\left(\lambda _2+\lambda _1 z\right){}^2}. \end{align} The distribution function of $Z$ would then be $$ F_Z(z) = \int_0^z\frac{\lambda _1 \lambda _2}{\left(\lambda _2+\lambda _1 t\right){}^2}\ \mathsf dt = \frac{\lambda_1z}{\lambda_1z+\lambda_2}. $$ Substituting $\lambda_1=1$ and $\lambda_2=2$, we have $$ F_Z(z) = \frac z{z+2}. $$