How do you find the volume of the solid generated by revolving the region bounded by the curves #y=2x^2 -x^3# and y = 0 rotated about the y-axis?

1 Answer
Truong-Son N.
Jul 27, 2015

This is what you will be revolving.

#y = 2x^2 - x^3#:
graph{2x^2 - x^3 [-0.1, 10, -0.3, 5]}

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What we could do here is use the shell method, which is far more convenient in this case than the traditional revolution method.

#V(x) = sum_(x = a)^b 2pixf(x)Deltax#

#= int_a^b 2pixf(x)dx#

where:
#x# is the radius of the shell
#f(x)# is the height of the shell
#dx# is the thickness of the shell
#2pi# indicates the radian measure of the "circumference" of the shell (not a true circular circumference, hence the quotes)

Now where is the x-intercept? We can see it's #2#, so let's show that.

#0 = 2x^2 - x^3#

#0 = x^2(2 - x)#

#x = 2, 0#

So we have:

#= 2piint_0^2 x(2x^2 - x^3)dx#

#= 2piint_0^2 2x^3 - x^4dx#

#= 2pi [x^4/2 - x^5/5]|_(0)^(2)#

#= 2pi {[(2)^4/2 - (2)^5/5] - 0}#

#= 2pi [16/2 - 32/5]#

#= 2pi [40/5 - 32/5]#

#= color(blue)((16pi)/5 "rad") ~~ 10.0531 "rad"#

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