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?

1 Answer
Sep 22, 2015

(28pi)/15 cubic units

Explanation:

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