Posted by Ted on Tuesday, July 24, 2012 at 10:35am.
I will assume that the x-axis was the other boundary of the region we are rotating
Using discs ...
radius = y = x^3
volume = π∫ y^2 dx from 0 to 2
= π∫ x^6 dx from 0 to 2
= π[ (1/7)x^7 ] from 0 to 2
= π((1/7)(2^7) - 0 ]
= 128π/7
It is not clear to me how you would do this without using calculus :)
Graph a sketch of it first. The corners are at (0,0) , (2,0) and (2,8)
Lets add up a bunch of thin cylinders with axes along x axis
volume of thin cylinder = pi y^2 dx
so
in from x = 0 to x = 2 of pi (x^3)^2 dx
or pi x^6 dx
or pi (1/7)x^7
or 1/7 (128)
about 18.3
I forgot to multiply by pi
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