How to solve this volume by revolution about the y-axis, $y=2 \cos(x)$ 
The first step I took was to solve for $x$.
$y=2\cos(x) => x=\arccos(\frac{y}{2})$
Then I found the y values of the function and intersections between the graph and the axis $(0,2)$ and $(\frac{π}{2},0)$
I then integrated, $\int_0^2 π \cdot [\arccos(\frac{y}{2})]^2 \;dy$ and I solved this using the formula $\int \arccos(x) = x \cdot \arccos(x)-\sqrt{1-x^2}$ and from that I got a final answer of 7.28. 
However, my textbook has the answer as 7.17.
Can someone explain to me why I am close but still wrong?
 A: Well, since you're trying to find a volume of revolution, you should be using one of the two methods that are used to do that: either the disk method (which is a special case of the washer method) or the shell method. The setup for the disk method should look like this:
$$
V=\pi\int_{0}^{2}\left[\arccos\left(\frac{y}{2}\right)\right]^2\,dy\approx7.17\ cubic\ units.
$$
Wolfram Alpha check.
Or you can do it using the shell method which in this case seems to be easier in terms of integration:
$$
V=2\pi\int_{0}^{\frac{\pi}{2}}x(2\cos{x})\,dx=
4\pi\int_{0}^{\frac{\pi}{2}}x\cos{x}\,dx=\\
4\pi\bigg[x\sin{x}+\cos{x}\bigg]_{0}^{\frac{\pi}{2}}=
4\pi\bigg[\frac{\pi}{2}\sin{\frac{\pi}{2}}+\cos{\frac{\pi}{2}}-(0\cdot\sin{0}+\cos{0})\bigg]=\\
4\pi\left(\frac{\pi}{2}-1\right)=
2\pi(\pi-2)\approx 7.17\ cubic\ units.
$$
A: (You have edited your question so your indefinite integral no longer matches the definite integral. )
Cylindrical polar coordinates are appropriate since when the curve is rotated you will have a volume with cylindrical symmetry.
The problem is then
$$
z=2  \cos{r} \tag{1}\label{eq1}
$$
and
$$
V=\int_0^2 \pi r^2 dz \tag{2}\label{eq2}
$$
From $\eqref{eq1}$
$$
\frac{dz}{dr}=-2 \sin{r}
$$
Substituting this in $\eqref{eq2}$ and changing limits appropriately yields (using integration by parts )
$$
V=4\pi(\pi/2-1)\approx7.17
$$
