Question
Use the cylindrical shell method to find the volume of the solid generated by revolving the area bounded by the given curves
(x-3)^2 + y^2 = 9, about y-axis.
(x-3)^2 + y^2 = 9, about y-axis.
Answers
Reiny
We are rotating a circle with centre at (3,0) and radius 3 around the y-axis
Because of symmetry we can use the part only above the x-axis, then double it.
see Khan for a nice explanation of the shell method
https://www.khanacademy.org/math/integral-calculus/solid-revolution-topic/shell-method/v/shell-method-for-rotating-around-vertical-line
by the shell method:
V = 2π[integral] x f(x) dx from 0 to 6
y^2 = 9 - (x-3)^2
= 9 - x^2+ 6x -9
= 6x - x^2
y = ± √(6x-x^2)
using only the upper part
y = f(x) = √(6x-x^2)
volume of upper part
= 2π [integral] x√(6x -x^2) dx from 0 to 6
nasty integral, Wolfram shows this:
http://www.wolframalpha.com/input/?i=integral+x%E2%88%9A(6x+-x%5E2)+from+0+to+6
Because of symmetry we can use the part only above the x-axis, then double it.
see Khan for a nice explanation of the shell method
https://www.khanacademy.org/math/integral-calculus/solid-revolution-topic/shell-method/v/shell-method-for-rotating-around-vertical-line
by the shell method:
V = 2π[integral] x f(x) dx from 0 to 6
y^2 = 9 - (x-3)^2
= 9 - x^2+ 6x -9
= 6x - x^2
y = ± √(6x-x^2)
using only the upper part
y = f(x) = √(6x-x^2)
volume of upper part
= 2π [integral] x√(6x -x^2) dx from 0 to 6
nasty integral, Wolfram shows this:
http://www.wolframalpha.com/input/?i=integral+x%E2%88%9A(6x+-x%5E2)+from+0+to+6
Steve
As a check, the theorem of Pappus says that the volume is the area times the distance traveled by the centroid, which is 9π*6π = 54π^2.