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

Answer:

The volume of the solid = π²

Step-by-step 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

$= \pi \times(sinx-cosx)^{2}$

$= \pi \times [sin^{2}(x) - 2 sin(x) cos(x) + cos^{2}(x)]$

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

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

Now,

Integrate from $x=\frac{\pi}{4}\ \ \ to\ \ \ x=\frac{5\pi}{4}$, we get volume:

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

After integrate without limit we get

$V=\pi[x+\frac{cos2x}{2}]$

Now after putting the limit, we get

 V = π²

Hence, the required volume of the solid = π²