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Since the oil is $3$ feet deep, we want to find the ratio of the area of the part of the circle covered with oil (part under the horizontal line of the figure below) to the entire area of the circle. The two radii drawn makes a $120$ degree angle, so the area of the fraction of the circle covered by the oil is $\frac{2}{3}$ of the circle in addition to the isosceles triangle. We can find the length of half the base of the isosceles triangle by using the Pythagorean theorem on the smaller right triangle. Setting half the length of the base to $x$, we have $x^2+1=4$, so $x=\sqrt{3}$ and the length of the base is $2\sqrt{3}$. Therefore, we have that the area of the triangle is $\frac{1}{2}\cdot 1\cdot 2\sqrt{3}=\sqrt{3}$. So, the area of the part of the circle that's covered in oil is $\frac{2}{3}\cdot 4\pi +\sqrt{3}=\frac{8}{3}\pi +\sqrt{3}$.

Thus, we have that the oil takes up $\frac{\frac{8}{3}\pi +\sqrt{3}}{4\pi }\approx \frac{10.11}{12.57}\approx 0.805$ of the cylinder.

With the cylinder upright, the fraction of the cylinder the oil covers is the same as the fraction of the height the oil covers. Therefore, the oil would be $15\text{ feet}\cdot 0.805\approx 12.08\approx \boxed{12.1}$.