Find the volume of the solid obtained by rotating the region bounded by y=x^6, y=1 about the line y=4

Question

this is a volumes question for calculus 2

Answer 1

The graph y=x^6 can be inverted and presented as x=+/-y^1/6. Instead of showing y in terms of x, the inverse shows x in terms of y. The area to be rotated about the line y=4 (parallel to the x axis) has a crucible shaped cross-section. Consider a hollow cylinder thickness dy lying on its side over the x axis with the line y=4 passing through its centre. The circumference of this cylinder will be 2(pi)(4-y) where (4-y) is the radius. At the origin (0,0) the radius will be 4 and at the point where the graph is bounded by the line y=1 the radius will be 3. The height of the cylinder will be 2x=2y^1/6, because it will be the difference between the positive and negative values of y^1/6. The volume of the cylinder is therefore 4(pi)(4-y)y^1/6dy. Integrating this over the range 0<=y<=1 we get 4(pi)int(4y^1/6-y^7/6dy)=4(pi)[24y^7/6)/7-(6y^13/6)13] 0<=y<=1 which gives us 24(pi)(4/7-1/13)=24(pi)(45/91)=1080(pi)/91=37.285.