You want to see the computations, or just the answer?

If you are into Calculus, integration, already, the volume in terms of the depth, h, of the fuel is

The origin (0,0) is at the bottom of the circle, so that the depth of the fuel is h always.

The center of the circle is at (0,1.45).

The radius of the cirle is 2.9/2 = 1.45 m

The equation of the circle is x^2 +(y-1.45)^2 = (1.45)^2

So, x = sqrt[(1.45)^2 -(y-1.45)^2]

dV = (2x)(dy)(7.5)

dV = 15(x)dy

dV = 15sqrt[(1.45)^2 -(y-1.45)^2]dy

Integration is from y=0 to y=h

V = (15)INT.(0 --> h)[sqrt[(1.45)^2 -(y-1.45)^2]dy

V = 15{(1/2)(y-1.45)sqrt[(1.45)^2 -(y-1.45)^2] +(1/2)(1.45)^2 *arcsin[(y-1.45)/1.45]} |(0-->h)

V = (7.5){(h-1.45)sqrt[(1.45)^2 -(h-1.45)^2] +[(1.45)^2]*arcsin[(h-1.45)/1.45]}

That is it.