The region under the curves #y=xe^(x^3), 1<=x<=2# is rotated about the x axis. How do you find the volumes of the two solids of revolution?

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Answer 1 · 2017-05-05T14:13:18

There is only on solid of revolution generated by this description.

The volume is #pi int_1^2 y^2 dx#

#pi int_1^2 (xe^(x^3))^2 dx = pi int_1^2 x^2e^(2x^3) dx#

Use the substitution #u = 2x^3# to get:

# = [pi/6 e^(2x^3)]_1^2#

# = pi/6(e^16-e^2)#

Use a calculator if you want a decimal approximation.