did you make a sketch?
did you notice that (6,8) is the farthest right-most point of your region?
So I will find the volume of the cylinder with radius 8 and height of 6, then subtract the region we don't want
cylinder = π(8^2)(6) = 384π
radius of region we don't want = 8 -(x^3/27)
volume of that rotated region
= π∫(8 - x^3/27)^2 dx from x = 0 to x = 6
= π∫(64 - (16/27)x^3 + x^6 /729) dx
= π [64x - (4/27)x^4 + (1/5103)x^7 ] from 0 to 6
= π(384 - 192 + 384/7 - 0 - 0 - 0)
= (1728/7)π
volume as asked for = 384π - 1728π/7 = 960π/7
better check my arithmetic, easy to make errors here without writing the solution down on paper.
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the?
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 8.
27y = x^3, y = 0, x = 6
V=
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