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 $\pi \int_0^8 (4- y/2)dy$.