Question

Question #6a05f

Answer 1

The volume is `pi^2-2pi` cubic units

The integral using the method of disks is

`piint_0^1(arccos(x))^2dy`

Explanation:

From the figure I am assuming that the problem is rotating `cos(x)` (in the first quadrant) from `x=0` to `x=pi/2` about the `y` axis.

Using the shell method our radius will be some value of `x` over the interval `[0,pi/2]`

The shell height is just the function `y=cos(x)`

So the integral using the shell method is

`2piint_0^(pi/2)xcos(x)dx`

We can evaluate this integral using integration by parts.

Let `u=x` then `du=dx`

Let `dv=cos(x)dx` the `intdv=intcos(x)dx`
so `v=sin(x)`

The parts formula is `uv-intvdu`

Proceeding we have

`2pi[xsin(x)-intsin(x)dx]`

`2pi[xsin(x)-(-cos(x))]`

`2pi[xsin(x)+cos(x)]`

Evaluating

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

`2pi[(pi/2)-1]`

`pi^2-2pi`

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

So `x=arccos(y)`

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

Our radius is `x=arccos(x)`

Using the method of disks the integral for the volume is

`piint_0^1(arccos(y))^2dy`