Can I get some assistance with this intregral / area problem? The problem states:
Set up the integral needed to find the volume of the solid formed by revolving the area between $y = cosx$ and $y = x, x = 0$ around the $y$ axis.
The first thing I did was find the area of limitations:
$$\cos x = x$$ 
$$x - \cos x = 0$$
$$ x(1-\cos(1)) = 0 $$
$$x = 0, 0.46 \ .$$
I'm not sure which formula I should be using to find the area. Originally I thought I should use the washer formula, but now it looks like won't have a washer look to it. 
So where do I go from here?
 A: If a volume $V$ is rotationally symmetric, and created by rotating the area under $f(x)$ between $a$ and $b$ around the x-axis, then it's volume is $$
  V = \pi\int_a^b\left(f(x)\right)^2 \,dx
$$
Basically this works because you can imagine the object to consist of $n$ discs with their centers on the $x$-Axis, all having width $\frac{b-a}{n}$, and the discs with center at $x$ having radius $f(x)$.
If the object is created by rotating the area between $f(x)$ and $g(x)$ ($f(x) \geq g(x)$, otherwise this doesn't make much sense), then you can simply compute the volume you get by rotating the whole area under $f(x)$, and then subtract the volume you get by rotating the area under $g(x)$. Combined into one formula, you get $$
  V = \pi\int_a^b\left(f(x)\right)^2 - \left(g(x)\right)^2 \,dx
$$
This is what you refer to as the Washer formula, I think, and it's perfectly applicable here. Since the volume doesn't change if you rotate objects, it doesn't matter that the rotation is around the $y$-axis instead of around the $x$-axis. You do, however, have to transform your function to also be in the form $f(y)=x$ to use the formula.
In your case, if you rotate around the $y$-Axis, you do, in fact, rotate the whole area between $\cos x$ and the $y$-axis around that axis, so there's no need to cut away anything, i.e. no need for a function $g(x)$. However, if you view your outline as a function of $y$, it has two parts. The first one is simply $y=x$, which starts at $1$ and ends at the $y$ for which $y = \cos x = x$. The next part is $y=\cos x$ part, and continue until $y=0$. So what you have to do is to bring $y = cos x$ into the form $x = f(y)$, then you can use the Washer formula. You'll have to do two integrals, the first extends from 0 to the $y$ where $y = \cos x = x$, the next one extends from that $y$ to 1. I hope this now answers the question you intended to ask.
Btw, when doing such problems, the first (and most important!) step is to do a sketch of how your outline and your object looks like. If you continue having problems, I suggest you make such a sketch and include it in your question
