Difficulty with washer method when revolving around axis other than y or x axis.

[bah]

Homework Statement

The region in the second quadrant bounded above by the curve y=-x^3, below by the x-axis, and on the left by the line x=-1, about the line x=-2

Homework Equations

It's basically the big radius squared minus the small radius squared, integrated in terms of y, and multiply that by pi, I think. But I have a hard time coming up the with expression for that when it's revolved around something other than the x or y axis... Please help?

The Attempt at a Solution

Well, the curve in terms of y is x=(-y)^(1/3). I have a hard time adjusting that so it is the right equation for the big radius, and adjusting the x=-1 so it is the small radius...

 

[bob1182006]

Wow that is an interesting washer o.o.

Well if it's about the line x=-2 and you want to change it to the y -axis. What is the y-axis? x=? how can you change x=-2 to x=(y-axis)?

also look at a graph of this function. What are the limits? x=-1, and x=? You also need these for when you change x=(y-axis).

 

[rocomath]

wow I'm really confused by this problem. it's been a while since I've done volumes but i figure i could at least get somewhere.

thinking thinking :)