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=1#, #y=x^2#, and #x=0# rotated about the line #y=2#?

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Answer 1 · 2015-09-22T21:13:02

#(28pi)/15# cubic units

Since we are revolving around a horizontal line using the method of shells we will integrate with respect to #y#.

We are bounded by the #y# axis, the horizontal line #y=1#, and the function #y=x^2#

Solve #y=x^2# for #x#

#x=sqrt(y)#

We are in quadrant #I# so we do not have to worry about the negative square root.

Our representative cylinder height is our function #sqrt(y)#

Our representative radius is #2-y# over the interval #0<=y<=1#

The integral for the volume is

#2piint_0^1(2-y)(y^(1/2))dy#

#2piint_0^1 2y^(1/2)-y^(3/2)dy#

Integrating we get

#2pi[4/3y^(3/2)-2/5y^(5/2)]#

Evaluating we get

#2pi[4/3-2/5-0]#

#2pi[20/15-6/15]=2pi[14/15]=(28pi)/15# cubic units