# Question 6a05f

Apr 3, 2017

The volume is ${\pi}^{2} - 2 \pi$ cubic units

The integral using the method of disks is

$\pi {\int}_{0}^{1} {\left(\arccos \left(x\right)\right)}^{2} \mathrm{dy}$

#### Explanation:

From the figure I am assuming that the problem is rotating $\cos \left(x\right)$ (in the first quadrant) from $x = 0$ to $x = \frac{\pi}{2}$ about the $y$ axis.

Using the shell method our radius will be some value of $x$ over the interval $\left[0 , \frac{\pi}{2}\right]$

The shell height is just the function $y = \cos \left(x\right)$

So the integral using the shell method is

$2 \pi {\int}_{0}^{\frac{\pi}{2}} x \cos \left(x\right) \mathrm{dx}$

We can evaluate this integral using integration by parts.

Let $u = x$ then $\mathrm{du} = \mathrm{dx}$

Let $\mathrm{dv} = \cos \left(x\right) \mathrm{dx}$ the $\int \mathrm{dv} = \int \cos \left(x\right) \mathrm{dx}$
so $v = \sin \left(x\right)$

The parts formula is $u v - \int v \mathrm{du}$

Proceeding we have

$2 \pi \left[x \sin \left(x\right) - \int \sin \left(x\right) \mathrm{dx}\right]$

$2 \pi \left[x \sin \left(x\right) - \left(- \cos \left(x\right)\right)\right]$

$2 \pi \left[x \sin \left(x\right) + \cos \left(x\right)\right]$

Evaluating

2pi[(pi/2)sin((pi/2))+cos((pi/2)))-(0+1)]#

$2 \pi \left[\left(\frac{\pi}{2}\right) - 1\right]$

${\pi}^{2} - 2 \pi$

Now onward with the disk method.
We have to get things in terms of $y$

So $x = \arccos \left(y\right)$

We will integrate over the interval $y = 0$ to $y = 1$

Our radius is $x = \arccos \left(x\right)$

Using the method of disks the integral for the volume is

$\pi {\int}_{0}^{1} {\left(\arccos \left(y\right)\right)}^{2} \mathrm{dy}$