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?

Answer 1

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.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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