Find Area Using Washer and Disk Method | Revolve Curves Around Axis

[Carmen12]

Homework Statement

choose between washer and disk method, and find the area of the region bounded by the following curves by revolving around a) the y-axis b) the x-axis c) y=8, and x=2

Homework Equations

y=2x²
y=0
x=2

The Attempt at a Solution

So I set up a) from 0 to 8 [ 2-sqrt(y/2)]^2 dy (washer method)

I don't know how to do the notation around here, so I hope that is clear? And I did remember the PI out front in these. I got 16pi/3

b) from 0 to 2 (2x^2)^2 dx (disk method). for 128pi/5

c) from 0 to 2 (x-2x^2)^2 dx (washer method) for 184pi/15

d) from 0 to 8 (sqrt(y/2))^^2 dy (disk method) for 16pi


I have no clue if I'm doing this right... all the values seem so different... :(

 

[HallsofIvy]

The values should be different- they are solutions to different problems.

Yes, for rotation around the y-axis, since the y-axis is not a boundary of the region being rotated, use the "washer method". However, you have the integrand wrong. \pi (r_1- r_2)^2 is the area of a full circle of radius r_1- r_2. A "washer" with outer radius r_1 and inner radius r_2 can be thought of as the are of the outer circle, \pi r_1^2 and then subtract of the area of the inner circle, \pi r_2^2: the area of the washer is \pi(r_1^2- r_2^2).

The radius will be along the x- direction and, since y= 2x^2 but we only need x positive, x= y^{1/2}/\sqrt{2} and the area of the "washer" from that x to x= 2 is \pi(r_1^2- r_2^2)= \pi(4- y/2). The volume is \pi \int_0^8 (4- y/2)dy.