Question #6a05f

1 Answer
Apr 3, 2017

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#