Question

Find the volume of the solid that lies between planes perpendicular to the x-axis at x = pi/4 and x = 5pi/4. The cross-sections between the planes are circular disks whose diameters run from the curve y = 2 cos x to the curve y = 2 sin x.

Answer 1

The volume of the solid = π²

Explanation:

As per the given data of the questions,

The diameter of each disk is

D = 2 sin(x) - 2 cos(x)

So its radius is

R = sin(x) - cos(x).

The area of each disk is

`= \pi *(sinx-cosx)^(2)`

`= \pi * [sin^(2)(x) - 2 sin(x) cos(x) + cos^(2)(x)]`

`= \pi[1-2sin(x)cos(x)]`

`= \pi[1-sin(2x)]`

Now,

Integrate from
`x=(\pi)/(4)\ \ \ to\ \ \ x=(5\pi)/(4)`, we get volume:

`V=\int_{(\pi)/(4)}^{(5\pi)/(4)} \pi[1-sin(2x)]dx`

After integrate without limit we get

`V=\pi[x+(cos2x)/(2)]`

Now after putting the limit, we get

V = π²

Hence, the required volume of the solid = π²