How do I find the volume of integration when it is rotated around a line that is not an axis?

Question

The problem is to find the volume of the solid enclosed by y=x^4 and y=1, rotated about the line y=2. (Disk method)

Answer 1

OK, I had to pull out my calc book for this.

I think the "disk method" would be the "volume by cross-sections method" in my book. Let me know if not.

So, do you have a mental picture of what's going to happen?

x^4 meets y = 1 at x = 1 and x = -1, so the thing that's going to be rotated looks a bit like a squashed, colored-in letter U.

The method of cross sections (or disks) involves trying to find the area of an infinitely thin cross section, rotated around some other line.

So, if you can imagine rotating around y=2, going out from that line, you first run into y=1 and then into y=x^4. That is, y=1 is the "inside" function and y=x^4 is the "outside" function.

So imagine for a moment that y=1 isn't there and you're just rotating y=x^4 about y=2. Then the radius of the disk would be "2-x^4", and the area would be pi times (2-x^4)^2, right?

Now subtract out the "inner ring" formed by rotating y=1 around y=2. The radius would be 1, so the area would be just pi.

So... the function for the area of the cross section, A(x) would be:

A(x) = pi[(2-x^4)^2 - 1] = pi[x^8 - 4x^4 + 4 - 1]
A(x) = pi[x^8 - 4x^4 + 3]

And then you would just integrate this from x = -1 to x = 1.

If you need more help, I could take it further.

Disclaimer: I haven't had calc since 1991.