No, I'm crap at calculus.

I'm still a bit confused by responses like this. Even if you do it the hard way, it only needs simple trigonometry, not calculus (in UK terms, it's a GCSE question not an A-level question).

Consider a circular cross-section of the tank. The fuel comes halfway from the bottom to the centre. So by looking at the triangle formed by the centre of the tank, the point on the fuel's surface directly below the centre, and the "shore" of the fuel where it touches the edge, you can work out the angle at the centre, i.e. the angle that one side of the fuel subtends at the centre. It's the arc-cosine of 6/12, or 60 degrees. So the whole surface of the fuel subtends twice that, or 120 degrees, at the centre of the tank.

So what area of the cross-section is taken up by fuel? Well, it's the area of the 120-degree pie-slice shape, minus the two triangles. In other words, it's 1/3 of pi*12*12, minus twice 6*(12 sin 60). If you divide that by the total area of the cross-section, pi*12*12, you get the fraction of the total volume occupied by fuel. Multiply that by the capacity and you get the volume of remaining fuel.

Now all this needs a calculator, of course. But if the trucker's smart, he'll do this only once and calibrate his dipstick in actual fuel units.

Peter