A hemispherical tank of radius 6 feet is positioned so that its base is circular. a) How much work is required to fill the tank with water through a hole in the base if the water source is at the base?

b) How much work is required to half fill the tank with water through a hole in the base if the water source is at the base?

i got part a). i let x be the distance from the base to the top of the water layer and r be the radius of the water layer since each layer is a disc. so by pythagorean theorem, r = sqrt(36-x^2). so since the weight of water is 62.5 lbs/ft^3 the work done is integral (from 0 to 6) (62.5)(pi)(x)(36-x^2)dx and i got 20250pi ft lbs which was correct.

however i can't seem to get my teachers answer for part b). i found the volume of the hemisphere which is 144pi and since we want to fill it to half, the volume of water will be 72pi. i then set up the equation 72pi = 144pi - (1/2)(4/3)(pi)(36-x^2)^(3/2). what i did was subtract the smaller hemisphere from the larger hemisphere since when you fill the half fill the tank the volume of the water won't be a hemisphere. i solved for x (the height of depth of water) and i got 3.649 ft. but my teacher's answer for the depth was 2.083 ft. how did she get that?