How to solve this volume by revolution about the y-axis, $y=2 \cos(x)$

Question

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?

Answer 1

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:

$$ V=2\pi\int_{0}^{\frac{\pi}{2}}x(2\cos{x})\,dx= 4\pi\int_{0}^{\frac{\pi}{2}}x\cos{x}\,dx=\\ 4\pi\bigg[x\sin{x}+\cos{x}\bigg]_{0}^{\frac{\pi}{2}}= 4\pi\bigg[\frac{\pi}{2}\sin{\frac{\pi}{2}}+\cos{\frac{\pi}{2}}-(0\cdot\sin{0}+\cos{0})\bigg]=\\ 4\pi\left(\frac{\pi}{2}-1\right)= 2\pi(\pi-2)\approx 7.17\ cubic\ units. $$

Answer 2

(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 ) $$ V=4\pi(\pi/2-1)\approx7.17 $$