# find pdf (using bivariate transformation) of X/Y where X,Y ~ uniform(0,1) independent

I know how to find the distribution of X/Y when they are independent uniform(0,1) by drawing the integration area.

$$P(X/Y \leqslant t) = \\ \frac{1}{2} t, t\leqslant1\\1-\frac{1}{2t},t>1$$
But when I use bivariate transformation $$U=\frac{X}{Y}$$ and $$V=Y$$ I find the following problem.
$$f_{U,V}(u,v) = v \text{ where }0\leqslant v \leqslant 1, 0\leqslant u \leqslant \frac{1}{v}$$
$$f_U(u) = \int_{0}^{1} f_{U,V}(u,v)dv=\int_{0}^{1} vdv=\frac{1}{2}$$
which is the pdf when $$t\leqslant1$$. How to get the other part?