Need help with computing the volume of a solid by revolving lines

[db2dz]

Homework Statement

the directions are:

compute the volume of the solid formed by revolving the given region about the line

region bounded by y=2x, y=2 and x=0 about (a) the y-axis; (b) x=1

Homework Equations

V=∫A(x)dx,a,b

The Attempt at a Solution

From what my teacher showed in class this is how i think it is worked out
(a)
V=∫A(x)dx,0,2

=∫(pi)[r(x)]^2,0,2

=∫(pi)[2x-2]^2,0,2

=∫(pi)[4x^2-a],0,2

=(8(pi))/3

so i am not sure if this is right or if i have to times(8(pi))/3 by 1/3 because the volume of a cone is (1/3)(pi)(r^2)(h)

and i don't know where to begin on (b), i think when it is revolved it will be Cylinder with a cut out. please help

 

[Dick]

You want to think about what a cross section of solid perpendicular to the rotation axis. In this case, it's easier to integrate dy than dx, since that's the rotation axis. The volume is then integral A(y)*dy where A(y) is the cross sectional area at y. In a) the cross section is a disk? So yes, the area is pi*R(y)^2. What's R(y) as a function of y? In b) the cross section is a disk with a hole in the middle (a 'washer'), but it's the same general idea.

 

[db2dz]

so i used dy to integrate and for a i got 2(pi)/3 which i think is right because i checked it with V = (1/3) π r² h. But i am not sure how to get b. i know its a cylinder with the cone removed and i think its volume is 4(pi)/3 but i still have no idea.

 

[Dick]

You are right. To get b) using integration the cross sectional area is area of an outer circle minus the area of the inner circle that was removed. So it's pi*(outer radius)^2-pi*(inner radius)^2. Integrate that dy.