Calculus: Early Transcendentals, Chapter 6, 6.3, Section 6.3, Problem 18

With the method of cylindrical shells we sum up the volumes of thin cylinders.
The volume of a cylinder is
2pi*r*h*dr,
where h is the height, r is the radius of a cylinder (the distance from the axis of rotation to the argument) and dr is the thickness.
y=x^2 and y=2-x^2 intersect at the points x=-1, y=1 and x=-1, y=1. Between x=-1 and x=1 2-x^2gtx^2, so the height h is equal to 2-x^2-x^2=2(1-x^2).
Also r=1-x.
So the volume is (remove odd functions integrating from -1 to 1)
2pi int_(-1)^1 (1-x)*2(1-x^2) dx=4pi int_(-1)^1 (1-x-x^2+x^3) dx =
=4pi int_(-1)^1 (1-x^2) dx=8pi int_0^1 (1-x^2) dx=
=8pi (x-1/(3)x^3)_0^1=8pi(1-1/3)=16/3 pi.