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(a) See the attached sketch. Each shell will have a radius <em>y</em> chosen from the interval [2, 4], a height of <em>x</em> = 2/<em>y</em>, and thickness ∆<em>y</em>. For infinitely many shells, we have ∆<em>y</em> converging to 0, and each super-thin shell contributes an infinitesimal volume of
2<em>π</em> (radius)² (height) = 4<em>πy</em>
Then the volume of the solid is obtained by integrating over [2, 4]:
(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - <em>x</em> (this is the distance between a given <em>x</em> value in the orange shaded region to the axis of revolution) and a height of 8 - <em>x</em> ³ (and this is the distance between the line <em>y</em> = 8 and the curve <em>y</em> = <em>x</em> ³). Then each shell has a volume of
2<em>π</em> (9 - <em>x</em>)² (8 - <em>x</em> ³) = 2<em>π</em> (648 - 144<em>x</em> + 8<em>x</em> ² - 81<em>x</em> ³ + 18<em>x</em> ⁴ - <em>x</em> ⁵)
so that the overall volume of the solid would be
I leave the details of integrating to you.
