Volume by washer or shell help

[TheKracken]

Homework Statement

Find the area given by y=x^2, y=4, x=0 and revolved about y = -2

Use either the washer or shell method

Homework Equations

Shell; integral from a to b of 2∏x(f(x))dx

The Attempt at a Solution

Alright so I tried to do this is terms of dy so since we are revolving about y=-2 which would be the horizonal axis. I got x= sqrt(y) and then I set up the integral from 0 to 4 of 2pi(y-2)(sqrt(y))dy

but this seems to be giving me a volume of 0 which means I've got to be doing something wrong!

I also attempted it with a dx and I decided to use the integral from 0 to 2 of 2 pi(x-2)(4-x^2)dx but received a value of 0 as well! I feel like this means that I am doing something consistently wrong which is kind of good, but I would like to fix that.

 

[TheKracken]

Ok so now I am really confused. I mean integral from 0 to 4 of 2pi(y+2)(sqrt(y)) and when I enter that in wolfram I don't get the correct value, but when I move the 2pi out of the integral and then multiply I get the right answer? I thought it wasn't supposed to matter...

 

[haruspex]

Ok so now I am really confused. I mean integral from 0 to 4 of 2pi(y+2)(sqrt(y)) and when I enter that in wolfram I don't get the correct value, but when I move the 2pi out of the integral and then multiply I get the right answer? I thought it wasn't supposed to matter...

I'd guess you are entering it into Wolfram wrongly. Maybe a problem with the parentheses?

 

[HallsofIvy]

Using the washer method, a vertical line from y= -2 through the figure, at fixed x, crosses it at $(x, x^2)$ and $(x, 4)$ so the inner and outer boundaries are at distance $x^2+ 2$ and $4+ 2= 6$. Those are the inner and outer radii of the "washer". The area of such a "washer" will be $\pi(x^2+ 2)^2- \pi(6)^2= \pi(x^4+ 4x^2- 32)$. The thickness of such a washer will be "dx" so the volume $\pi(x^4+ 4x^2- 32)dx$. Integrate that from x= 0 to x= 1.

Using the "shell method", at a given y, the "shell" will have length $x- 0= \sqrt{y}$ and will rotate around a radius $y- (-2)= y+ 2$, describig a circle of circumference $2\pi(y+ 2)$. The area of such a shell will be $2\pi(y+ 2)\sqrt{y}$. The thickness will be "dy" so the volume of $2\pi(y+ 2)\sqrt{y}dy$. Integrate that from y= 0 to y= 4.