Find the volume of the solid obtained by rotating the region bounded by the curves...

The region bounded by: y=1 and y=x^8

Rotated around: y=4

I used to know how to do these, but I can't remember it appears and after about an hour of trying to figure it out by trying new things I've given up. I really just need the correct integral that comes out of it I suppose, not an actual number answer.

Thank you for your help.

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

I would use the washer method here, although the shell method is also fairly straightforward here. A good place to begin is with the formula for the volume of a washer, in terms of the outer radius, the inner radius and the thickness. What is this formula?

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

Integral from a to b of pi*(((outer)^2)-((inner)^2))

From what I can tell, the inner radius would simply be 6, no? And the outer would be 1-(x^8)?

I would believe it's from -1 to 1, but that doesn't seem to be working. I've tweaked here and there...but I can't find my mistake.

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

Inner is 3...silly me...hang on.

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

No beyond that I'm still missing something haha.

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

You are on the right track...the volume of a washer is:

$\displaystyle V_w=\pi\left(r_o^2-r_i^2 \right)h$

where $\displaystyle r_o$ is the outer radius, $\displaystyle r_i$ is the inner radius and $\displaystyle h$ is the thickness.

If we slice our solid of revolution vertically, i.e., perpendicular to the *x*-axis, the thickness of each slice we may call $\displaystyle dx$.

You are right that the inner radius is 3 units, since this is the distance between the upper bound of the region being rotated and the axis of rotation.

So what is the distance between the lower bound of the region being rotated and the axis of rotation?

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

Isn't it just the distance between y=1 and y= x^8? And seeing as y=1 is on top, it would be that minus y=x^8?

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

No, you want the distance between the axis of rotation $\displaystyle y=4$ and the lower bound of the rotated region, which is $\displaystyle y=x^8$.

What is this distance?

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

By the way, an invaluable aid to doing these problems is to sketch the graphs of the bounding functions and the axis of rotation. Have you made such a sketch?

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

So then it would be (4-(x^8)). And yeah...I've drawn it about 100 times now haha.

So then.....Integral of pi[((4-(x^8))^2)-(9)]

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

Okay, I just got it to work.

I see my mistake. I wasn't taking the outer radius, just the distance between the two given bounds...which is not how you do this haha. Now I just feel silly.

Thank you very much for your help.

Re: Find the volume of the solid obtained by rotating the region bounded by the curve

I like to state the volume of an arbitrary slice as:

$\displaystyle dV=\pi\left(\left(4-x^8 \right)^2-3^2 \right)\,dx$

Since the bounded region is symmetric across the *y*-axis, we may then state (after determining the limits as -1 and 1):

$\displaystyle V=2\pi\int_0^1\left(\left(4-x^8 \right)^2-3^2 \right)\,dx$

Then, of course expand the binomial within the integrand and integrate term by term.