1. ## Weird Rotational Solid

So I'm asked to find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis (obv.). I'm given:

x + 3 = 4y - y^2 , x = 0; about the x-axis

The book says the answer is (16π)/3

This seems weird to me because I'm imagining the graph will yield a solid just begging me to apply the washer method and use dx. Yet, unless my algebra is terrible (which wouldn't surprise me), I don't see how we can set an equation equal to one y value. Therefore, I tried devising a few approaches using dy and the disk method (I haven't seen a prototype for this problem, but I figured it couldn't be too hard to come up with my own approach to a textbook problem).

To start, I set x = -y^2 + 4y - 3. Then, I proceeded with π∫ (-y^2 + y - 3)^2 dy from 1 to 3 (value obtained by factoring). I ended up with (16/15)π. Way wrong.

I saw a problem that sort-of, kind-of resembled this one earlier in the book, which advised I solve the problem with the formula 2π∫ y[ u(y) - v(y) ]dy.

There wasn't an explanation though, so I figured maybe that y outside the brackets represented an equation composed with y variables. So I basically repeated my first attempt with 2π out front. Not surprisingly, wrong again.

The closest I came was trying 2π∫ (-y^2 + 4y - 3) dy (incorrectly assuming that maybe the 2 coefficient would account for the volume and I could discard the square outside the parentheses around the quadratic). I got 8π/3.

Then I figured that first y in 2π∫ y[ u(y) - v(y) ]dy could be replaced by the value obtained after applying the quadratic formula to -y^2 + 4y - 3. This led to me repeating that last equation with a 4 in front of the parentheses. I got 32π/3.

I guess I'd get 16π/3 if I'd have put a 2 in front of the parentheses, yet I can't figure out any good reason why I should.

I thought I was pretty decent at rotational solids, but this shook my confidence. I spent all afternoon thinking about this and I'm stumped. I apologize for the length of the post, but I figured you might appreciate the effort. Hopefully I presented everything clear enough and I hope someone can help!

2. ## Re: Weird Rotational Solid

Wait, disk method? I'm forgetting about the shell method, aren't I?

3. ## Re: Weird Rotational Solid

Always do it both ways.

$\displaystyle 2\pi\cdot\int_{1}^{3}y\cdot (4y-y^{2}-3)\;dy$

$\displaystyle \pi\cdot\int_{0}^{1}(\sqrt{1-x}+2)^{2}-(2-\sqrt{1-x})^{2}\;dx$

Just stare at them until they soak in. Well, okay, you can blink when necesary.

Note: This is unlikely to be a natural skill. It will take effort to get good at it.