#### To determine

**To find:**

The volume of the solid obtained by rotating the region bounded by the given curves about the x axis and sketch the region, the solid, and a typical disk or washer.

#### Answer

V= 384π5

The sketch of the bounded region, the solid, and a typical washer:

#### Explanation

**1) Concept:**

i. If the cross section is a washer with the inner radius rin and the outer radius rout, then the area of the washer is obtained by subtracting the area of the inner disk from the area of the outer disk.

A=π outer radius2 -π inner radius2

ii. The volume of a solid of revolution about the x-axis is

V= ∫abA(x)dx

**2) Given:**

y=6-x2, y=2 ; about the x- axis

**3) Calculation:**

The region bounded by y=6-x2 and y=2 and solid obtained by rotation about the x- axis are shown below.

Here, the region is rotated about the x – axis, so the cross-section is perpendicular to the x-axis.

A cross section of the solid is the washer with inner radius 2 and outer radius 6-x2.

So, its cross sectional area becomes

Ax=πouter radius2-πinner radius2 =π6-x22-π22

Ax=π6-x22-4=π36-12x2+x4-4

Ax=πx4-12x2+32

The region of integration is bounded by y=6-x2 and y=2,

At point of intersection of the curve and x-axis,

6-x2=2

x2=4

x=±2

The solid lies between x=-2 and x= 2

Therefore, the volume of a solid of revolution about x-axis is

V= ∫-22Axdx=∫-22π(x4-12x2+32)dx

As fx= 4x4-12x2+32 is an even function hence,

=2π∫02(x4-12x2+32)dx

By using the fundamental theorem of calculus part 2 and the power rule of integration,

V=2πx55-12.x33+32x02=2πx55-4x3+32x02

By substituting the limits of integration,

=2π 255-423+32(2)-(0)55-403+32(0)

=2π325-32+64-0

=2π 325+32=2π325+1605

=2π1925

V=384π5

Therefore, the volume of the solid obtained by rotating the region bounded by the given curves about the x axis is

V=384π5

**Conclusion:**

The volume of the solid obtained by rotating the region bounded by the given curves about the x axis is

V=384π5

Sketch of bounded region, the solid and a typical washer: