Question #6a05f

1 Answer
Apr 3, 2017

The volume is pi^2-2piπ22π cubic units

The integral using the method of disks is

piint_0^1(arccos(x))^2dyπ10(arccos(x))2dy

Explanation:

From the figure I am assuming that the problem is rotating cos(x)cos(x) (in the first quadrant) from x=0x=0 to x=pi/2x=π2 about the yy axis.

Using the shell method our radius will be some value of xx over the interval [0,pi/2][0,π2]

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

So the integral using the shell method is

2piint_0^(pi/2)xcos(x)dx2ππ20xcos(x)dx

We can evaluate this integral using integration by parts.

Let u=xu=x then du=dxdu=dx

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

The parts formula is uv-intvduuvvdu

Proceeding we have

2pi[xsin(x)-intsin(x)dx]2π[xsin(x)sin(x)dx]

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

2pi[xsin(x)+cos(x)]2π[xsin(x)+cos(x)]

Evaluating

2pi[(pi/2)sin((pi/2))+cos((pi/2)))-(0+1)]2π[(π2)sin((π2))+cos((π2)))(0+1)]

2pi[(pi/2)-1]2π[(π2)1]

pi^2-2piπ22π

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

So x=arccos(y)x=arccos(y)

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

Our radius is x=arccos(x)x=arccos(x)

Using the method of disks the integral for the volume is

piint_0^1(arccos(y))^2dyπ10(arccos(y))2dy