using shells of thickness dy, we have
v = ∫[0,8.3] 2πrh dy
where r = 15-y and h = 1.6√y - (-1.4y)
v = ∫[0,8.3] 2π(15-y)(1.6√y - (-1.4y)) dy = 4474.15
using discs of thickness dx, we have to switch boundary curves at x=0, so
v1 = ∫[-11.9,0] π(R^2-r^2) dx
where R = 15-y and r = 15-8.3
v1 = ∫[-11.9,0] π((15+x/1.4)^2-(15-8.3)^2) dx = 2867.18
v2 = ∫[0,1.6√8.5] π(R^2-r^2) dx
where R = 15-y and r = 15-8.3
v2 = ∫[0,1.6√8.5] π((15-(x/1.6)^2)^2-(15-8.3)^2) dx = 1606.97
v = v1+v2 = 4474.15
Determine the volume of a solid produced by rotating about the line y=15 the region bounded by the indicated equations.
x=-1.4y , y=8.3, x=1.6 y1/2
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