I picture this as a cylinder radius 18" and 60" long lying on its curved side, so that the water level is 6" deep maximum.
If we take a disc of radius 18" representing a cross-section of the tank or its base or top (vertically orientated), we need the area of a sector bound by a chord and the circumference.
The end points of the chord are A and B, and the centre of the circle is O, so OA=OB=18" because they're radii. If N is where the tank touches the ground, then ON=18" is also a radius. ON bisects AB at X, so OX=18-6=12". The area of the bound sector= area of sector AOB - area of isosceles triangle AOB. The height of this triangle is OX=12".
Triangle AOB is made of two back to back right triangles with 18" hypotenuses. By Pythagoras:
OX2+AX2=OA2, 144+AX2=324, AX2=180, AX=√180=√(62×5)=6√5 inches. Right triangles OXA and OXB each have an area of ½(6√5)(12)=36√5 square inches. Combined area is 72√5=161.00in2.
cosAÔX=cosBÔX=OX/OA=OX/OB=12/18=⅔. Measure of AÔX=BÔX=arccos(⅔)=0.8411 radians approx. Therefore measure of AÔB=AÔX+BÔX=1.6821 radians. Area of sector AOB=½(OA)2(1.6821)=272.51in2.
(Another way of doing this working with degrees instead of radians: area of circle is πr2=182π=324π in2. The ratio of the sector area A to the area of the circle is the same as the ratio of angle AOB to 360°. AÔX=BÔX=arccos(⅔)=48.19°, AÔB=96.38°, so A/(324π)=96.38/360, A=96.38(324π)/360=272.51in2.)
The part of the sector bound by the chord AB therefore has an area of 272.51-161.00=111.51in2. The volume of water is this area times the length: 60×111.51=6690.6 cubic inches approx.