How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region # y = x^3#, #y = 0#, #x = 2# rotated about the x axis?

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Answer 1 · 2015-05-23T10:18:25

The answer is #V = (128pi)/7#.

Firstly, you should graph your three functions to see clearly what is the delimited region :

enter image source here

We are rotating our plane region around the x-axis, which means we have to write our function #f(x) = x^3# in the form #f(y) =root(3)(y)#.

The radius of any cylinder in our final volume will be given by #y#.

And the width will be #2-root(3)(y)#.

(Look here for more details or diagrams about the cylinders)

Therefore, the cross-sectional area of any cylinder will be :

#A(y) = 2piy(2-root(3)(y)) = 4piy - 2piy^(4/3)#

The radius of the cylinders, according to the delimited region, goes from :

#f(x) = 0 = y# (with #x = 0#) to #f(x) = 8 = y# (with #x=2#).

So the volume of our solid, which is the sum of all the cross-sectional area of the cylinders, is :

#V = int_0^8A(y)dy#.

The antiderivative of #ay^n# is given by #a*1/(n+1)y^(n+1)#.

Thus, the antiderivative of #A(y)# is :

#1/2 4piy^2 - 3/7 2piy^(7/3) = 2piy^2(1-3/7root(3)y) #

Now we can calculate the integral :

#V = int_0^8A(y)dy = [2piy^2(1-3/7root(3)y)]_0^8#

#= 2pi [y^2(1-3/7root(3)y)]_0^8 = 2pi(8^2(1-(3*2)/7)-0)#

#= 2pi*64/7 = (128pi)/7#.